Equilibrium is reached when the rates of the forward and backward reactions become balanced.. This means that the concentrations of the reactants and products remain constant over time.
There are two main types of equilibrium:
Physical Equilibrium:
Involves the equilibrium between different phases of matter, such as solid-liquid, liquid-gas, or solid-gas.
Chemical Equilibrium:
Involves the equilibrium between reactants and products in a chemical reaction.
Factors that affect equilibrium include temperature, pressure, concentration, and the presence of a catalyst. Le Châtelier’s principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust to counteract the change and re-establish equilibrium.
The equilibrium constant Kc serves as a measure of the balance between reactants and products in a reversible reaction. It is calculated as the ratio of the product concentrations to the reactant concentrations, each raised to their stoichiometric coefficients.
Equilibrium is important in many areas of chemistry, including chemical reactions, industrial processes, and environmental science.
1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Ans : Analyzing the Effect of Increased Volume on a Liquid-Vapor Equilibrium
a) Initial Effect on Vapor Pressure:
When the volume of the container is suddenly increased, the vapor pressure will initially decrease. This is because the same amount of vapor is now spread out over a larger volume, leading to a lower concentration of vapor molecules.
b) Initial Changes in Rates of Evaporation and Condensation:
Evaporation: The rate of evaporation will initially increase as there is more space for the liquid molecules to escape into the vapor phase.
Condensation: The rate of condensation will initially decrease as the concentration of vapor molecules has decreased.
c) Final Equilibrium and Vapor Pressure:
In time, the system will attain a new equilibrium condition. At this point, the rate of evaporation will once again be equal to the rate of condensation.
Final Vapor Pressure: The final vapor pressure will be lower than the initial vapor pressure. This is because the increased volume has created a larger space for the vapor, and the equilibrium vapor pressure is determined by the temperature and the concentration of vapor molecules in the container.
In summary, increasing the volume of a container in equilibrium with its vapor initially decreases the vapor pressure. However, the system will eventually readjust to a new equilibrium state with a lower vapor pressure.
2 .What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) 2SO3(g)
Ans : The equilibrium constant (Kc) for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) can be calculated using the following formula:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
where:
[SO3] = equilibrium concentration of SO3
[SO2] = equilibrium concentration of SO2
[O2] = equilibrium concentration of O2
Substituting the given equilibrium concentrations, we get:
Kc = (1.90 M)^2 /
————————————–
((0.60 M)^2 * (0.82 M))
Kc ≈ 12.23
Therefore, the equilibrium constant (Kc) for the given reaction is approximately 12.23.
3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms I2 (g) 2I (g) Calculate Kp for the equilibrium.
Ans : To calculate Kp for the equilibrium I₂(g) ⇌ 2I(g), we first need to determine the partial pressures of I₂ and I.
Given:
Total pressure (P_total) = 105 Pa
Volume fraction of I atoms = 40%
Partial Pressures:
Partial pressure of I₂ (P_I₂) = 60% of total pressure = 0.60 * 105 Pa = 63000 Pa
Partial pressure of I (P_I) = 40% of total pressure = 0.40 * 105 Pa = 42000 Pa
Kp expression:
Kp = (P_I)² / P_I₂
Calculation:
Kp = (42000 Pa)² / 63000 Pa ≈ 2.8 × 10⁴ Pa
Therefore, the equilibrium constant Kp for the reaction I₂(g) ⇌ 2I(g) is approximately 2.8 × 10⁴ Pa.
4. Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl (g) 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3) 2 (s) 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5(aq) + H2O(l) CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) Fe(OH)3 (s)
(v) I2 (s) + 5F2 2IF5
Ans :
5. Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) CaO(s) + CO2(g); Kp= 167 at 1073 K
Ans : The equilibrium constants Kp and Kc are related to each other by the equation Kc = Kp(RT)^Δng, where Δng is the change in the number of moles of gas during the reaction.
The value of Kc can be calculated as follows:
(a) For the reaction 2NOCl(g) ⇌ 2NO(g) + Cl₂, Δng = 3 – 2 = 1.
Substituting the given values of Kp, R, and T, we get:
Kc = (1.8 x 10^-2 atm) / (0.0821 L atm K^-1 mol^-1 x 500 K)^1
= 4.4 x 10^-4 mol L^-1
(b) For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), Δng = 1.
the given values of Kp, R, and T, we get:
Kc = (167 atm) / (0.0821 L atm K^-1 mol^-1 x 1073 K)^1
= 1.9 mol L^-1
6. For the following equilibrium, Kc = 6.3 × 1014 at 1000 K NO (g) + O3 (g) NO2 (g) + O2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc , for the reverse reaction?
Ans : The ratio of the equilibrium constant for the forward reaction to the equilibrium constant for the reverse reaction is always 1.
Given:
Forward reaction = NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Kc (forward) =
6.3 × 10¹⁴
Therefore, Kc (reverse) =
1
—————————-
Kc (forward)
Kc (reverse) =
1
—————————————-
(6.3 × 10¹⁴)
Kc (reverse) ≈ 1.59 × 10⁻¹⁵
So, the equilibrium constant for the reverse reaction is approximately 1.59 × 10⁻¹⁵.
7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Ans : Pure liquids and solids are not included in the equilibrium constant expression because their concentrations remain essentially constant throughout the reaction.
Here’s a breakdown of why:
Incompressibility: Liquids and solids are relatively incompressible, meaning their volume changes very little with changes in pressure. This means that their concentration (moles per unit volume) remains nearly constant, even if the total volume of the system changes.
Constant Composition: The composition of pure substances (liquids and solids) is fixed. This means that the concentration of a pure substance remains constant as long as the temperature remains constant.
Therefore, since the concentrations of pure liquids and solids remain constant, they do not affect the equilibrium position and can be omitted from the equilibrium constant expression.
This is why the equilibrium constant expression only includes the concentrations of gaseous and aqueous species, which can change significantly during a reaction.
8. Reaction between N2 and O2– takes place as follows: 2N2 (g) + O2 (g) 2N2O (g) If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
Ans : Suppose x moles of N₂(g) participate in the reaction. Based on the equation’s stoichiometry, x/2 moles of O₂(g) will react to produce x moles of N₂O(g). The molar concentrations per liter of the various species before the reaction and at equilibrium are:
The equilibrium constant (Kc) for the reaction is 2.0 x 10^-37, which is a very small value. This indicates that the reaction proceeds to a very limited extent, with only a small amount of reactants being converted to products at equilibrium.
Kc, the equilibrium constant, is given by the equation:
Kc = [N₂O(g)]² / ([N₂(g)]²[O₂(g)])
Replacing the variables in the expression with the given equilibrium concentrations gives:
Kc = (x/10)² / ((0.482/10)² x (0.933/10))
= 0.01x² / 2.1676 x 10^-4
Solving for x², we get:
x² = 43.352 x 10^-40
Squaring both sides of the equation yields:
x = 6.6 x 10^-20
Since x is extremely small, it can be neglected in the equilibrium mixture.
Therefore, the equilibrium concentrations of N₂, O₂, and N₂O are:
N₂: 0.0482 mol L^-1
O₂: 0.0933 mol L^-1
N₂O: 0.1 x 6.6 x 10^-20 mol L^-1
= 6.6 x 10^-21 mol L^-1
9 .Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br2 (g) 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2 .
Ans : Calculating Equilibrium Amounts
Given:
Reaction: 2NO (g) + Br₂ (g) ⇌ 2NOBr (g)
Initial moles of NO: 0.087 mol
Initial moles of Br₂: 0.0437 mol
Equilibrium moles of NOBr: 0.0518 mol
ICE Table:
Species Initial Moles Change Equilibrium Moles
NO 0.087 -2x 0.087 – 2x
Br₂ 0.0437 -x 0.0437 – x
NOBr 0 +2x 0.0518
Equilibrium Constant Expression:
Kc = [NOBr]² / ([NO]² * [Br₂])
Substituting Equilibrium Concentrations:
Kc = (0.0518)² / ((0.087 – 2x)² * (0.0437 – x))
We know that at equilibrium, the moles of NOBr formed is 0.0518 mol. Therefore, x = 0.0518 / 2 = 0.0259 mol.
Substituting x back into the equilibrium concentration expressions:
[NO] = 0.087 – 2(0.0259) = 0.0352 mol
[Br₂] = 0.0437 – 0.0259 = 0.0178 mol
Therefore, at equilibrium:
The amount of NO remaining is 0.0352 mol.
The amount of Br₂ remaining is 0.0178 mol.
10. At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) 2SO3 (g) What is Kc at this temperature ?
Ans : The relationship between Kp (equilibrium constant in terms of partial pressures) and Kc (equilibrium constant in terms of molar concentrations) is given by the equation:
Kp = Kc * (RT)^(Δn_g)
where:
R is the ideal gas constant (0.08314 L bar K⁻¹ mol⁻¹)
T is the temperature in Kelvin (450 K in this case)
Δn_g is the difference in the number of moles of gaseous products and reactants
For the given reaction, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Δn_g = (2 – 3) = -1.
To solve for Kc, rearrange the equation
Kc = Kp / (RT)^(-1)
Substituting the given values:
Kc = (2.0 × 10¹⁰ bar⁻¹) / (0.08314 L bar K⁻¹ mol⁻¹ * 450 K)⁻¹
Kc ≈ 7.48 × 10¹¹ L/mol
Therefore, the value of Kc for the given reaction at 450 K is approximately 7.48 × 10¹¹ L/mol.
11. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? 2HI (g) H2 (g) + I2 (g)
Ans : Calculating Kp for the Reaction
Given:
Initial pressure of HI: 0.2 atm
Equilibrium partial pressure of HI: 0.04 atm
Reaction:
2HI (g) ⇌ H₂ (g) + I₂ (g)
ICE Table:
Species Initial Pressure Change Equilibrium Pressure
HI 0.2 atm -2x 0.04 atm
H₂ 0 atm +x x atm
I₂ 0 atm +x x atm
Equilibrium Constant Expression:
Kp = (P_H₂ * P_I₂) / (P_HI)²
Substituting equilibrium pressures:
Kp = (x * x) / (0.04)²
We know that at equilibrium, the partial pressure of HI is 0.04 atm. Therefore, x = (0.2 – 0.04) / 2 = 0.08 atm.
Substituting x into the Kp expression:
Kp = (0.08 * 0.08) / (0.04)²
Kp = 4
Therefore, the equilibrium constant Kp for the given reaction is 4.
12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Ans :The reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
The concentration quotient (Qc) is calculated using the formula:
Qc = [NH₃]² / ([N₂][H₂]³)
Substituting the given concentrations, we get:
Qc = (8.13/20 mol L⁻¹)² / ((1.57/20 mol L⁻¹)(1.92/20 mol L⁻¹)³)
= 2.38 x 10³
The equilibrium constant (Kc) for the reaction is 1.7 x 10⁻².
Given that Qc ≠ Kc, the reaction is not at equilibrium. If Qc exceeds Kc, the reaction shifts to the left to attain equilibrium.
13. The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
Ans : The balanced chemical equation represents a reaction with a stoichiometric coefficient of 4.
14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) H2 (g) + CO2 (g) Calculate the equilibrium constant for the reaction.
Ans : The initial number of moles of water is 1 mole. 40% of the water reacts, so 0.4 moles of water react. This leaves 0.6 moles of water remaining.
Based on the balanced chemical equation, 0.4 moles of water react with 0.4 moles of carbon monoxide to produce 0.4 moles of hydrogen and 0.4 moles of carbon dioxide.
Therefore, the molar concentrations per liter of the reactants and products before the reaction and at equilibrium are as follows:
Substance Initial Concentration (mol/L) Equilibrium Concentration (mol/L)
CO 0.5 0.1
H₂O 1 0.6
H₂ 0 0.4
CO₂ 0 0.4
15. At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) 2HI (g) is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
Ans : Setting Up the ICE Table and Calculating Equilibrium Concentrations
Given:
Reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
Kc = 54.8
Initial [HI] = 0.5 mol/L
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
H₂ 0 +x x
I₂ 0 +x x
HI 0.5 -2x 0.5 – 2x
Equilibrium Constant Expression:
Kc = [HI]² / ([H₂] * [I₂])
Substituting Equilibrium Concentrations:
54.8 = (0.5 – 2x)² / (x * x)
Simplifying and Solving for x:
54.8 = (0.25 – 2x + 4x²) / x²
54.8x² = 0.25 – 2x + 4x²
50.8x² + 2x – 0.25 = 0
Using the quadratic formula to solve for x:
x =
[-b ± √(b² – 4ac)]
—————————-
2a
x ≈ 0.0466 M
Equilibrium Concentrations:
[H₂] = x = 0.0466 M
[I₂] = x = 0.0466 M
Therefore, the equilibrium concentrations of H₂(g) and I₂(g) are both approximately 0.0466 mol/L.
16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?
2ICl (g) I2 (g) + Cl2 (g); Kc = 0.14
Ans : Calculating Equilibrium Concentrations
Given:
Reaction: 2ICl (g) ⇌ I₂ (g) + Cl₂ (g)
Kc = 0.14
Initial concentration of ICl = 0.78 M
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
ICl 0.78 -2x 0.78 – 2x
I₂ 0 +x x
Cl₂ 0 +x x
Equilibrium Constant Expression:
Kc = [I₂] * [Cl₂] / [ICl]²
Substituting equilibrium concentrations:
0.14 = x² / (0.78 – 2x)²
x ≈ 0.167 M
Equilibrium Concentrations:
[ICl] = 0.78 – 2x ≈ 0.446 M
[I₂] = x ≈ 0.167 M
[Cl₂] = x ≈ 0.167 M
Therefore, at equilibrium, the concentrations are:
ICl: 0.446 M
I₂: 0.167 M
Cl₂: 0.167 M
17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C2H6 (g) C2H4 (g) + H2 (g)
Ans : Calculating Equilibrium Concentration
Given:
Reaction: C₂H₆(g) ⇌ C₂H₄(g) + H₂(g)
Kp = 0.04 atm
Initial pressure of C₂H₆ = 4.0 atm
ICE Table:
Species Initial Pressure Change Equilibrium Pressure
C₂H₆ 4.0 atm -x 4.0 – x
C₂H₄ 0 atm +x x
H₂ 0 atm +x x
Kp Expression:
Kp = (P_C₂H₄ * P_H₂) / P_C₂H₆
Substituting equilibrium pressures:
0.04 = (x * x) / (4.0 – x)
Solving for x:
x² = 0.04(4.0 – x)
x² = 0.16 – 0.04x
x² + 0.04x – 0.16 = 0
Using the quadratic formula:
x ≈ 0.38 atm
Equilibrium Pressure of C₂H₆:
P_C₂H₆ = 4.0 – x ≈ 4.0 – 0.38 ≈ 3.62 atm
Therefore, the equilibrium concentration of C₂H₆ is approximately 3.62 atm.
18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc , for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Ans :
19.A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium? PCl5 (g) PCl3 (g) + Cl2(g)
Ans : The initial molar concentration of PCl5 is x mol/L.
At equilibrium, the concentration of PCl5 is 0.05 mol/L.
Therefore, the moles of PCl5 that decomposed is (x – 0.05) mol.
Since the balanced chemical equation for the reaction is PCl5(g) ⇌ PCl₃(g) + Cl₂(g), the moles of PCl₃ and Cl₂ formed are also (x – 0.05) mol each.
The molar concentrations per liter of the reactants and products before the reaction and at the equilibrium point are as follows:
Substance Initial Concentration (mol/L) Equilibrium Concentration (mol/L)
PCl5 x 0.05
PCl₃ 0 x – 0.05
Cl₂ 0 x – 0.05
20. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2.
FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and = 0.80 atm?
Ans :
The reaction quotient (Qp) is greater than the equilibrium constant (Kp), indicating that the reaction is not at equilibrium. To reach equilibrium, the reaction will shift in the reverse direction. This means that the partial pressure of CO₂ will decrease, while the partial pressure of CO will increase.
Denote the decrease in the partial pressure of CO₂ by p atm. Since the reaction has a 1:1 stoichiometric ratio between CO₂ and CO, the partial pressure of CO will increase by the same magnitude, p atm.
21. Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
Ans : The given reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Determine reaction quotient (Qc) by using formula
Qc = [NH₃]² / ([N₂][H₂]³)
Substituting the given concentrations, we get:
Qc = (0.50)² / ((3.0)(2.0)³)
= 0.25 / 24
= 0.0104
Since the value of Qc (0.0104) is less than the equilibrium constant Kc (0.061), the reaction is not in a state of equilibrium. It will proceed in the forward direction until Qc becomes equal to Kc.
22 .Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl (g) Br2 (g) + Cl2 (g) for which Kc = 32 at 500 K.
If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?
Ans : Let x moles of BrCl decompose to reach equilibrium. The initial molar concentration and the equilibrium molar concentration of the different species can be represented as follows::
The given reaction is 2BrCl(g) ⇌ Br₂(g) + Cl₂(g).
The starting concentration of BrCl is 0.0033 moles per liter
The moles of BrCl that decompose at equilibrium will be represented by x.
The equilibrium concentrations of BrCl, Br₂, and Cl₂ are:
BrCl: (0.0033 – x) mol/L
Br₂: x/2 mol/L
Cl₂: x/2 mol/L
equilibrium constant (Kc) for the reaction is defined
Kc = ([Br₂][Cl₂]) / [BrCl]²
Putting the equilibrium concentrations into the expression for Kc,
Kc = (x/2)(x/2) / (0.0033 – x)²
Given that Kc = 32, we can set up the equation:
32 = (x/2)(x/2) / (0.0033 – x)²
Solving for x, we get:
x = 0.003 mol/L
Therefore, the equilibrium concentration of BrCl is 0.0033 – 0.003
= 0.0003 mol/L.
23 .At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass C (s) + CO2 (g) 2CO (g). Calculate Kc for this reaction at the above temperature.
Ans : The given reaction is C(s) + CO₂(g) ⇌ 2CO(g).
Step I: Calculation of Kp for the reaction
The total mass of the gaseous mixture is 100 g.
The mass of CO in the mixture is 90.55 g, and the mass of CO₂ is 9.45 g.
The number of moles of CO and CO₂ can be calculated using their molar masses:
nCO = 90.55 g / 28 g/mol = 3.234 mol
nCO₂ = 9.45 g / 44 g/mol = 0.215 mol
The partial pressures of CO and CO₂ in the mixture can be calculated using the mole fraction and the total pressure (1 atm):
pCO = (3.234 mol / 3.449 mol) x 1 atm
= 0.938 atm
pCO₂ = (0.215 mol / 3.449 mol) x 1 atm
= 0.062 atm
Step II: find Kc
The equilibrium constant Kc is related to Kp by the equation Kc = Kp / (RT)^Δng, where Δng is the change in the number of moles of gas.
For the given reaction, Δng = 2 – 1 = 1.
Substituting the values of Kp, R, and T, we get:
Kc = 14.19 atm / (0.0821 L atm K⁻¹ mol⁻¹) x (1127 K)
= 6.46
Therefore, the equilibrium constants Kp and Kc for the given reaction are 14.19 atm and 6.46, respectively.
24. Calculate a) ∆G and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K
NO (g) + ½ O2 (g) NO2 (g) where ∆f G (NO2) = 52.0 kJ/mol ∆f G (NO) = 87.0 kJ/mol ∆f G (O2) = 0 kJ/mol
Ans : To calculate ΔG and the equilibrium constant for the formation of NO2 from NO and O2 at 298K, we can use the following equations:
ΔG° = ΣnΔfG°(products) – ΣnΔfG°(reactants)
ΔG°= -RT ln K
where:
ΔG° is the standard Gibbs free energy change
ΔfG° is the standard free energy of formation
n is the stoichiometric coefficient
R is the gas constant (8.314 J/mol K)
T is the temperature (298 K)
K is the equilibrium constant
a) Calculate ΔG°:
ΔG° = ΔfG°(NO2) – [ΔfG°(NO) + 1/2 * ΔfG°(O2)]
ΔG° = 52.0 kJ/mol – [87.0 kJ/mol + 1/2 * 0 kJ/mol]
ΔG° = -35.0 kJ/mol
b) Calculate the equilibrium constant (K):
ΔG° = -RT ln K
ln K = -ΔG° / RT
K = e^(-ΔG° / RT)
Substituting the values:
K = e^(-(-35.0 kJ/mol) / (8.314 J/mol K * 298 K))
K ≈ 1.36 x 10^6
Therefore:
ΔG° = -35.0 kJ/mol
K ≈ 1.36 x 10^6
When ΔG is negative, the reaction is said to be spontaneous, meaning it will occur on its own without requiring additional energy input.
The large value of the equilibrium constant K indicates that the reaction strongly favors the formation of NO2 at equilibrium.
25. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl5 (g) PCl3 (g) + Cl2 (g)
(b) CaO (s) + CO2 (g) CaCO3 (s)
(c) 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g)
Ans : To determine the effect of decreasing pressure (increasing volume) on the number of moles of reaction products, we can use Le Chatelier’s principle, Equilibrium systems resist changes and strive to maintain their balance.
a) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
This reaction involves all gaseous species.
Decreasing pressure (increasing volume) will favor the side with more moles of gas to increase the total pressure.
In this case, the product side has more moles of gas (2 moles) than the reactant side (1 mole).
Therefore, the number of moles of reaction products will increase when the pressure is decreased.
b) CaO (s) + CO2 (g) ⇌ CaCO3 (s)
This reaction involves a solid and a gas.
Decreasing pressure (increasing volume) will favor the side with more moles of gas.
In this case, the reactant side has 1 mole of gas, while the product side has no gas.
Therefore, the number of moles of reaction products will decrease when the pressure is decreased.
c) 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) + 4H2 (g)
This reaction involves solids and gases.
Decreasing pressure (increasing volume) will favor the side with more moles of gas.
In this case, both sides have the same number of moles of gas (4 moles).
Therefore, the number of moles of reaction products will remain the same when the pressure is decreased.
26 .Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl2 (g) CO (g) + Cl2 (g)
(ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g)
(iii) CO2 (g) + C (s) 2CO (g)
(iv) 2H2 (g) + CO (g) CH3OH (g)
(v) CaCO3 (s) CaO (s) + CO2 (g)
(vi) 4 NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g)
Ans : The effect of increasing pressure on a reaction depends on the difference in the number of moles of gaseous reactants and products (Δng). If Δng ≠ 0, the reaction will be affected by pressure. If Δng = 0, the reaction will not be affected by pressure.
Among the given reactions, only reaction (1) has Δng = 0. Therefore, increasing pressure will not affect reaction (1). The remaining five reactions will be affected by increasing pressure.
In general, increasing pressure favors reactions that produce fewer moles of gas. Conversely, decreasing pressure favors reactions that produce more moles of gas. This is because increasing pressure increases the concentration of the gases, which can shift the equilibrium in the direction that reduces the total number of gas molecules.
The reaction will shift to the left if the number of moles of products (np) is greater than the number of moles of reactants (nr).
Conversely, the reaction will shift to the right if the number of moles of reactants (nr) is greater than the number of moles of products (np).
(i) The equilibrium is unaffected by an increase in pressure as np equals nr, both being 3.
(ii) Due to np being greater than nr, an increase in pressure will shift the reaction backward.
(iii)The reaction will shift towards the reactants when pressure increases because there are more moles of gas on the products side (np = 10) than on the reactants side (nr = 9).
(iv) Since np (1) is less than nr (2), an increase in pressure will shift the reaction forward.
(v) An increase in pressure will drive the reaction towards the reactants because np is greater than nr.
(vi) The inequality np (1) > nr (0) indicates that an increase in pressure will drive the reaction towards the reactants.
27. The equilibrium constant for the following reaction is 1.6 ×105 at 1024K H2(g) + Br2(g) 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Ans : To find the equilibrium pressures of all gases, we can set up an ICE (Initial, Change, Equilibrium) table:
Gas Initial Pressure (bar) Change (bar) Equilibrium Pressure (bar)
H2 0 +x x
Br2 0 +x x
HBr 10.0 -2x 10.0 – 2x
The equilibrium constant (Kp) is given by:
Kp = (PHBr)^2 / (PH2 * PBr2)
find the equilibrium pressures from the ICE table:
1.6 x 10^5 = (10.0 – 2x)^2 / (x * x)
Solving for x:
x ≈ 0.0313 bar
Therefore, the equilibrium pressures are:
PH2 = 0.0313 bar
PBr2 = 0.0313 bar
PHBr = 10.0 – 2(0.0313) ≈ 9.94 bar
28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH4 (g) + H2O (g) CO (g) + 3H2 (g) (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst?
Ans : (a) The expression for Kp :
Kp =
(PCO * PH2^3)
———————————–
(PCH4 * PH2O)
where:
Kp is the equilibrium constant
PCO, PH2, PCH4, and PH2O are the partial pressures of CO, H2, CH4, and H2O, respectively.
(b) Let’s analyze how the values of Kp and the composition of the equilibrium mixture will be affected by the given changes:
(i) Increasing the pressure:
Le Chatelier’s principle states that an equilibrium will shift to counteract changes in pressure.. In this case, the reactant side has 2 moles of gas, while the product side has 4 moles of gas. Therefore, increasing the pressure will shift the equilibrium towards the left, favoring the formation of CH4 and H2O.
The value of Kp becomes smaller. As the equilibrium shifts to the left, the concentration of CO and H2 will decrease, while the concentrations of CH4 and H2O will increase. This will decrease the value of the numerator and increase the value of the denominator in the Kp expression, resulting in a lower Kp value.
(ii) Increasing the temperature:
Heat is absorbed in the reaction.. According to Le Chatelier’s principle, increasing the temperature will shift the equilibrium towards the endothermic side to counteract the increase in temperature. In this case, that means shifting the equilibrium towards the right, favoring the formation of CO and H2.
The value of the equilibrium constant will rise. As the equilibrium shifts to the right, the concentration of CO and H2 will increase, while the concentrations of CH4 and H2O will decrease. This will increase the value of the numerator and decrease the value of the denominator in the Kp expression, resulting in a higher Kp value.
(iii) Using a catalyst:
Catalysts accelerate both forward and reverse reactions. It does not affect the position of the equilibrium, so the value of Kp will remain unchanged. However, the catalyst will help the reaction reach equilibrium more quickly.
29. Describe the effect of: a) addition of H2
b) addition of CH3OH
c) removal of CO
d) removal of CH3OH on the equilibrium of the reaction:
2H2(g) + CO (g) CH3OH (g)
Ans : The given reaction is 2H₂(g) + CO(g) ⇌ CH₃OH(g).
According to Le Chatelier’s principle, if a system at equilibrium is subjected to a change in concentration, temperature, pressure, or volume, the system will shift in a direction to counteract the applied change.
Let’s analyze the effect of each change on the equilibrium:
(a) Addition of H₂:
Increasing the concentration of H₂ will shift the equilibrium towards the right to consume the excess H₂ and produce more CH₃OH.
(b) Addition of CH₃OH:
Increasing the concentration of CH₃OH will shift the equilibrium towards the left to consume the excess CH₃OH and produce more H₂ and CO.
(c) Removal of CO:
Decreasing the concentration of CO will shift the equilibrium towards the left to produce more CO.
(d) Removal of CH₃OH:
Decreasing the concentration of CH₃OH will shift the equilibrium towards the right to produce more CH₃OH.
30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition is depicted as, 2024-25 212 chemistry PCl5 (g) PCl3 (g) + Cl2 (g) ∆r H = 124.0 kJ mol–1
a) write an expression for Kc for the reaction.
b) what is the value of Kc for the reverse reaction at the same temperature?
c) what would be the effect on Kc if-(i) more PCl5 is added
(ii) pressure is increased
(iii) the temperature is increased ?
Ans : (a) Expression for Kc:
For the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g), the expression for Kc is:
Kc = [PCl3] * [Cl2] / [PCl5]
where:
Kc is the equilibrium constant
[PCl3], [Cl2], and [PCl5] represent the concentrations of PCl3, Cl2, and PCl5 when the reaction has reached equilibrium.
(b)The reverse reaction’s Kc is the inverse of the forward reaction’s Kc.. Therefore, the value of Kc for the reverse reaction at the same temperature is:
Kc(reverse) = 1 / Kc(forward)
Kc(reverse) = 1 / (8.3 x 10^-3)
Kc(reverse) ≈ 120.48
(c) Effect on Kc:
(i) Adding more PCl5:
Increasing the concentration of PCl5 will shift the equilibrium to the right, favoring the formation of PCl3 and Cl2. This will rise the value of Kc.
(ii) Increasing pressure:
Increasing the pressure will have no effect on the value of Kc for this reaction because there are equal numbers of moles of gas on both sides of the equation.
(iii) Increasing the temperature:
The reaction is endothermic (ΔrH = 124.0 kJ/mol), meaning it absorbs heat from the surroundings. When the temperature is increased, the equilibrium shifts towards the right side of the reaction, producing more PCl3 and Cl2. This results in a higher value for the equilibrium constant
31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO (g) + H2O (g) CO2 (g) + H2 (g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pco = pH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C
Ans : The given reaction is CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g).
Let p bar be the partial pressure of hydrogen (H₂) at equilibrium.
The initial and equilibrium partial pressures of the gases are:
Substance Initial Pressure (bar) Equilibrium Pressure (bar)
CO 4.0 4.0 – p
H₂O 4.0 4.0 – p
CO₂ 0 p
H₂ 0 p
The equilibrium constant (Kp) is given by:
Kp = (pCO₂ x pH₂) / (pCO x pH₂O)
Substituting the equilibrium pressures, we get:
Kp = (p x p) / ((4.0 – p) x (4.0 – p)) = p² / (4.0 – p)²
Given that Kp = 0.1, we can set up the equation:
0.1 = p² / (4.0 – p)²
Solving for p, we get:
p = 0.96 bar
Therefore, the equilibrium partial pressure of hydrogen (H₂) is 0.96 bar.
32 .Predict which of the following reaction will have appreciable concentration of reactants and products:
a) Cl2 (g) 2Cl (g) Kc = 5 ×10–39
b) Cl2 (g) + 2NO (g) 2NOCl (g) Kc = 3.7 × 108
c) Cl2 (g) + 2NO2 (g) 2NO2Cl (g) Kc = 1.8
Ans : The equilibrium constant Kc quantifies the ratio of product concentrations to reactant concentrations at equilibrium.. A high Kc value suggests that the reaction proceeds significantly towards product formation, while a low Kc value indicates that the reaction remains predominantly in the reactant state
Based on the given Kc values:
Cl2 (g) ⇌ 2Cl (g) (Kc = 5 × 10^-39):
The very small Kc value indicates that this reaction strongly favors the formation of Cl2. At equilibrium, there will be a very high concentration of Cl2 and a very low concentration of Cl.
Cl2 (g) + 2NO (g) ⇌ 2NOCl (g) (Kc = 3.7 × 10^8):
The large equilibrium constant indicates a significant shift towards the formation of NOCl, resulting in a much higher concentration of NOCl compared to Cl2 and NO at equilibrium
Cl2 (g) + 2NO2 (g) ⇌ 2NO2Cl (g) (Kc = 1.8):
The moderate Kc value indicates that there will be appreciable concentrations of both reactants and products at equilibrium.
Therefore, the reaction with appreciable concentrations of reactants and products is:
c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2Cl (g)
33. The value of Kc for the reaction 3O2 (g) 2O3 (g) is 2.0 ×10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3?
Ans : To find the equilibrium concentration of O3, we can use the equilibrium constant expression:
Kc = [O3]^2 / [O2]^3
where:
Kc is the equilibrium constant
[O3] is the equilibrium concentration of O3
[O2] is the equilibrium concentration of O2
We are given the values of Kc and [O2]. Substituting these values into the equation and solving for [O3]:
2.0 × 10^-50 = [O3]^2 / (1.6 × 10^-2)^3
[O3]^2 =
2.0 × 10^-50 * (1.6 × 10^-2)^3
[O3]^2 ≈ 8.192 × 10^-56
[O3] ≈ √(8.192 × 10^-56)
[O3] ≈ 2.86 × 10^-28 M
Therefore, the equilibrium concentration of O3 in air at 25°C is approximately 2.86 × 10^-28 M.
34 .The reaction, CO(g) + 3H2(g) CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Ans : The given reaction is CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g).
Given, reaction of equilibrium constant (Kc)
Kc = ([CH₄][H₂O]) / ([CO][H₂]³)
Substituting the given equilibrium concentrations, we get:
Kc = (x * 0.02) / (0.30 * 0.1³) = 3.90
Solving for x, we get:
x = (3.90 * 0.30 * 0.1³) / 0.02
= 5.85 x 10⁻² M
Therefore, the equilibrium concentration of CH₄ is 5.85 x 10⁻² M.
35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN– , HClO4, F – , OH– , CO , and S2–
Ans : A conjugate acid-base pair is a set of two chemical species that differ by only one proton (H⁺). For example, HCN and CN⁻ are a conjugate acid-base pair.
Conjugate acid examples: HCN, H₂O, HCO₃⁻, HS⁻
Conjugate base examples: NO₂⁻, ClO₄⁻, O₂⁻
These pairs are related by the acid-base reaction: HA ⇌ A⁻ + H⁺, where HA is the acid and A⁻ is its conjugate base.
36. Which of the followings are Lewis acids? H2O, BF3, H+, and NH4 +
Ans : A Lewis acid is a substance that can form a covalent bond by accepting an electron pair from a donor molecule
Here’s the analysis of the given species:
H2O: While H2O can act as both a Lewis acid and a Lewis base, it is more commonly considered a Lewis base due to its ability to donate a lone pair of electrons.
BF3: Boron in BF3 has an incomplete octet, making it electron-deficient. Therefore, BF3 can accept an electron pair and is a Lewis acid.
H+: A proton (H+) lacks an electron, so it can accept an electron pair. Thus, H+ is a Lewis acid.
NH4+: The nitrogen atom in NH4+ has a full octet. However, due to the positive charge, it can attract electron-rich species. Therefore, NH4+ can accept an electron pair and is a Lewis acid.
In conclusion, the Lewis acids among the given species are BF3, H+, and NH4+.
37. What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO– 3?
Ans : Here are the conjugate bases for the given Brönsted acids:
HF: F- is the conjugate base of HF..
H2SO4: The conjugate base of H2SO4 is HSO4-.
HCO3-: The conjugate base of HCO3- is CO32-.
Remember, a conjugate base is the species that remains after a Brönsted acid has donated a proton.
38.Write the conjugate acids for the following Brönsted bases: NH2 – , NH3 and HCOO– .
Ans : Here are the conjugate acids for the given Brönsted bases:
NH2-: The conjugate acid is NH3.
NH3: The conjugate acid is NH4+.
HCOO-: The conjugate acid is HCOOH.
Remember, a conjugate acid is formed when a Brönsted base accepts a proton.
39 .The species: H2O, HCO3 – , HSO4 – and NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.
Ans : Here are the conjugate acid and base pairs for the given species:
H2O:
Conjugate acid: H3O+
Conjugate base: OH-
HCO3-:
Conjugate acid: H2CO3
Conjugate base: CO32-
HSO4-:
Conjugate acid: H2SO4
Conjugate base: SO42-
NH3:
Conjugate acid: NH4+
Conjugate base: NH2-
These species are called amphiprotic because they can both donate and accept protons, depending on the reaction conditions
40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH– (b) F– (c) H+ (d) BCl3 .
Ans : Lewis Acids and Bases:
Lewis acids are electron acceptors.
Lewis bases are electron donors.
Analysis of the given species:
(a) OH-:
OH- possesses a non-bonding pair of electrons on the oxygen atom..
It can donate this lone pair to another species.
Therefore, OH- is a Lewis base.
(b) F-:
Similar to OH-, F- also has a lone pair of electrons on the fluorine atom.
It can donate this lone pair to another species.
Therefore, F- is a Lewis base.
(c) H+:
H+ is a proton, which lacks an electron.
It can form a covalent bond by attracting and binding to an electron-rich species.
Therefore, H+ is a Lewis acid.
(d) BCl3:
Boron in BCl3 has an incomplete octet.
It can accept an electron pair from another species to complete its octet.
Therefore, BCl3 is a Lewis acid.
In summary:
Lewis acids: H+ and BCl3
Lewis bases: OH- and F-
41. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH?
Ans : The pH of a solution is a numerical scale that indicates its acidity or alkalinity, with a lower pH indicating a higher concentration of hydrogen ions.The pH of a solution is a logarithmic scale that indicates the acidity or alkalinity based on the concentration of hydrogen ions present
pH =
-log[H+]
where:
[H+] represents the number of hydrogen ions present in one liter of solution
In this case, the concentration of hydrogen ions in the soft drink is 3.8 × 10^-3 M. Substituting this value into the pH formula:
pH = -log(3.8 × 10^-3)
Using a calculator to evaluate the logarithm:
pH ≈ 2.42
Hence, the pH of the soft drink is approximately 2.42.
This indicates that the soft drink is acidic, with a pH lower than 7.
42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Ans : To find the concentration of hydrogen ions in the vinegar sample, we can use the formula:
[H+] = 10^-pH
where:
[H+] is the concentration of hydrogen ions
pH is the pH of the solution
Substituting the given pH value:
[H+] = 10^-3.76
Using a calculator to evaluate the exponent:
[H+] ≈ 1.74 × 10^-4 M
Therefore, the concentration of hydrogen ions in the vinegar sample is approximately 1.74 × 10^-4 M.
43. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Ans : The ionization constants of an acid and its conjugate base are connected by a mathematical relationship
Kw = Ka * Kb
where Kw is the ion product constant of water, which has a value of 1.0 × 10^-14 at 298 K.
Therefore, to find the ionization constant of the conjugate base, we can rearrange the equation and substitute the given Ka value:
Kb = Kw / Ka
For HF:
Kb = (1.0 × 10^-14) / (6.8 × 10^-4)
Kb ≈ 1.47 × 10^-11
For HCOOH:
Kb = (1.0 × 10^-14) / (1.8 × 10^-4)
Kb ≈ 5.56 × 10^-11
For HCN:
Kb = (1.0 × 10^-14) / (4.8 × 10^-9)
Kb ≈ 2.08 × 10^-6
Therefore, the ionization constants of the corresponding conjugate bases are:
F-: 1.47 × 10^-11
HCOO-: 5.56 × 10^-11
CN-: 2.08 × 10^-6
44. The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?
Ans : The given reaction is C₆H₅OH ⇌ C₆H₅O⁻ + H⁺.
The initial concentration of phenol (C₆H₅OH) is 0.05 M. Let x be the amount of phenol that dissociates at equilibrium.
The equilibrium concentrations of C₆H₅OH, C₆H₅O⁻, and H⁺ are:
[C₆H₅OH] = 0.05 – x M
[C₆H₅O⁻] = x M
[H⁺] = x M
The acid dissociation constant (Ka) for phenol is given by:
Ka = ([C₆H₅O⁻][H⁺]) / [C₆H₅OH]
Putting the equilibrium concentrations into the expression for Ka,
Ka = (x * x) / (0.05 – x) = 1.0 x 10⁻¹⁰
Solving for x, we get:
x = 2.2 x 10⁻⁶ M
Therefore, the equilibrium concentration of H⁺ ions is 2.2 x 10⁻⁶ M.
In the presence of 0.01 M sodium phenoxide (C₆H₅ONa), the initial concentration of C₆H₅O⁻ is 0.01 M. Let y be the amount of phenol that dissociates in this case.
The equilibrium concentrations of C₆H₅OH, C₆H₅O⁻, and H⁺ are:
[C₆H₅OH] = 0.05 – y M ≈ 0.05 M (since y is expected to be very small compared to 0.05 M)
[C₆H₅O⁻] = 0.01 + y M ≈ 0.01 M
[H⁺] = y M
Substituting these concentrations into the expression for Ka, we get:
Ka = (0.01 * y) / 0.05
= 1.0 x 10⁻¹⁰
Solving for y, we get:
y = 5 x 10⁻¹⁰ M
Therefore, the degree of dissociation (α) of phenol in the presence of 0.01 M sodium phenoxide is:
α = y / c = 5 x 10⁻¹⁰ / 5 x 10⁻²
= 10⁻⁸
45. The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.
Ans : The given reaction is H₂S ⇌ H⁺ + HS⁻.
The initial concentration of H₂S is 0.1 M. Let x be the amount of H₂S that dissociates at equilibrium.
The equilibrium concentrations of H₂S, H⁺, and HS⁻ are:
[H₂S] = 0.1 – x M
[H⁺] = x M
[HS⁻] = x M
The acid dissociation constant (Ka) for H₂S is given by:
Ka = ([H⁺][HS⁻]) / [H₂S]
putting the equilibrium concentrations into the expression for Ka,
Ka = (x * x) / (0.1 – x) = 9.1 x 10⁻⁸
Solving for x, we get:
x = 9.54 x 10⁻⁵ M
Therefore, the equilibrium concentration of HS⁻ is 9.54 x 10⁻⁵ M.
In the presence of 0.1 M HCl, the initial concentration of H⁺ is 0.1 M. Let y be the amount of H₂S that dissociates in this case.
The equilibrium concentrations of H₂S, H⁺, and HS⁻ are:
[H₂S] = 0.1 – y M ≈ 0.1 M
[H⁺] = 0.1 + y M ≈ 0.1 M
[HS⁻] = y M
Putting these concentrations into the expression for Ka,
Ka = (0.1 * y) / 0.1 = 9.1 x 10⁻⁸
Solving for y, we get:
y = 9.1 x 10⁻⁸ M
Therefore, the equilibrium concentration of HS⁻ in the presence of 0.1 M HCl is 9.1 x 10⁻⁸ M.
To calculate the concentration of S²⁻, we need to consider the second ionization step of H₂S:
HS⁻ ⇌ H⁺ + S²⁻
The acid dissociation constant for this step is denoted as Ka₂. The overall acid dissociation constant for the complete ionization of H₂S is given by Ka = Ka₁ * Ka₂.
find the given values,
Ka = 9.1 x 10⁻⁸ * 1.2 x 10⁻¹³ = 1.092 x 10⁻²⁰
The equilibrium expression for the second ionization step is:
Ka₂ = ([H⁺][S²⁻]) / [HS⁻]
In the absence of 0.1 M HCl, [H⁺] = 2[S²⁻]. Substituting this into the expression for Ka₂, we get:
Ka₂ = (4x³) / 0.1 = 1.092 x 10⁻²⁰
Solving for x, we get:
x = 6.497 x 10⁻⁸ M
Therefore, the equilibrium concentration of S²⁻ in the absence of 0.1 M HCl is 6.497 x 10⁻⁸ M.
In the presence of 0.1 M HCl, [H⁺] is approximately equal to 0.1 M. Substituting this into the expression for Ka₂, we get:
Ka₂ = (0.1 * y) / 0.1 = 1.09 x 10⁻²⁰
Solving for y, we get:
y = 1.09 x 10⁻¹⁹ M
Therefore, the equilibrium concentration of S²⁻ in the presence of 0.1 M HCl is 1.09 x 10⁻¹⁹ M.
46. The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Ans : We’re dealing with the ionization of acetic acid (CH₃COOH) in a 0.05 M solution. We need to find:
The degree of dissociation of acetic acid.
The concentration of the acetate ion (CH₃COO-).
The pH of the solution.
Step 1: Write the ionization equation
CH₃COOH(aq) ⇌ CH₃COO-(aq) + H+(aq)
Step 2: Set up an ICE table
Species Initial (M) Change (M) Equilibrium (M)
CH₃COOH 0.05 -x 0.05 – x
CH₃COO- 0 +x x
H+ 0 +x x
Step 3: Write the Ka expression
Ka = [CH₃COO-][H+] / [CH₃COOH]
Replace the equilibrium concentrations in the equation with their values from the ICE table
Ka = (x * x) / (0.05 – x)
Step 4: Solve for x (assuming x is negligible compared to 0.05)
Since Ka is very small, we can assume that x is much smaller than 0.05. Therefore:
Ka ≈ x^2 / 0.05
Solving for x:
x = √(Ka * 0.05) = √(1.74 × 10^-5 * 0.05) ≈ 9.33 × 10^-4 M
Step 5: Calculate the required values:
Degree of dissociation (α): α = x / [CH₃COOH]₀ = 9.33 × 10^-4 M / 0.05 M ≈ 0.01865
Concentration of acetate ion: [CH₃COO-] = x ≈ 9.33 × 10^-4 M
pH: pH = -log[H+] = -log(9.33 × 10^-4) ≈ 3.03
Therefore:
The degree of dissociation of acetic acid is approximately 1.865%.
The concentration of acetate ion in the solution is 9.33 × 10^-4 M.
The pH of the solution is 3.03.
47 .It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Ans : We’re dealing with an organic acid in a 0.01 M solution. Given the pH, we need to find:
The concentration of the anion (conjugate base)
The ionization constant (Ka) of the acid
The pKa of the acid
Step 1: Find the concentration of H+ ions
Since pH = -log[H+], we can find [H+] as follows:
[H+] = 10^(-pH) = 10^(-4.15) ≈ 7.08 × 10^-5 M
Step 2: Find the concentration of the anion
For a weak acid like this organic acid, the concentration of the anion (A-) is equal to the concentration of H+ ions at equilibrium. Therefore:
[A-] = [H+] ≈ 7.08 × 10^-5 M
Step 3: Calculate the ionization constant (Ka)
The ionization constant of an acid is given by:
Ka = [H+][A-] / [HA]
where HA is the undissociated acid. Assuming that the concentration of the undissociated acid remains approximately equal to the initial concentration (0.01 M) due to the weak nature of the acid:
Ka ≈ [H+][A-] / 0.01
Substituting the values:
Ka ≈ (7.08 × 10^-5)^2 / 0.01
Ka ≈ 5.01 × 10^-7
Step 4: Calculate the pKa
pKa = -log Ka
pKa = -log(5.01 × 10^-7) ≈ 6.3
Therefore:
The concentration of the anion is approximately 7.08 × 10^-5 M.
The ionization constant (Ka) of the acid is approximately 5.01 × 10^-7.
The pKa of the acid is approximately 6.3.
48.Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH
Ans : To calculate the pH of the solutions, we’ll use the following formula:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions.
Assuming complete dissociation:
For HCl and HBr, the concentration of H+ ions will be equal to the concentration of the acid.
For NaOH and KOH, the concentration of OH- ions will be equal to the concentration of the base. We can then calculate [H+] using the relationship Kw = [H+][OH-], where Kw is the ion product of water (1.0 × 10^-14).
Calculations:
(a) 0.003 M HCl:
[H+] = 0.003 M
pH = -log(0.003) ≈ 2.52
(b) 0.005 M NaOH:
[OH-] = 0.005 M
[H+] = Kw / [OH-] = 1.0 × 10^-14 / 0.005 ≈ 2.0 × 10^-12 M
pH = -log(2.0 × 10^-12) ≈ 11.70
(c) 0.002 M HBr:
[H+] = 0.002 M
pH = -log(0.002) ≈ 2.70
(d) 0.002 M KOH:
[OH-] = 0.002 M
[H+] = Kw / [OH-] = 1.0 × 10^-14 / 0.002 ≈ 5.0 × 10^-12 M
pH = -log(5.0 × 10^-12) ≈ 11.30
Hence, the pH values of the solutions are as follows
(a) 0.003 M HCl: 2.52
(b) 0.005 M NaOH: 11.70
(c) 0.002 M HBr: 2.70
(d) 0.002 M KOH: 11.30
49. Calculate the pH of the following solutions:
a) 2 g of TlOH dissolved in water to give 2 litre of solution.
b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.
c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
Ans : a) 2 g of TlOH dissolved in 2 L of solution
Molar mass of TlOH: 221 g/mol
Moles of TlOH: 2 g / 221 g/mol = 0.00905 mol
Concentration of TlOH: 0.00905 mol / 2 L = 0.00453 M
pH: -log[OH⁻] = -log(0.00453) ≈ 2.34
b) 0.3 g of Ca(OH)₂ dissolved in 500 mL of solution
Molar mass of Ca(OH)₂: 74 g/mol
Moles of Ca(OH)₂: 0.3 g / 74 g/mol = 0.00405 mol
Concentration of OH⁻: 2 * 0.00405 mol / 0.5 L = 0.0162 M
pH: 14 – (-log[OH⁻]) = 14 – (-log(0.0162)) ≈ 12.21
c) 0.3 g of NaOH dissolved in 200 mL of solution
Molar mass of NaOH: 40 g/mol
Moles of NaOH: 0.3 grams of NaOH corresponds to 0.0075 moles.
Concentration of OH⁻: 0.0075 mol / 0.2 L = 0.0375 M
pH: 14 – (-log[OH⁻]) = 14 – (-log(0.0375)) ≈ 12.58
d) 1 mL of 13.6 M HCl is diluted to 1 L of solution
Molarity after dilution: (1 mL / 1000 mL) * 13.6 M = 0.0136 M
pH: -log[H⁺] = -log(0.0136) ≈ 1.87
Therefore, the pH values of the solutions are:
a) 2.34
b) 12.21
c) 12.58
d) 1.87
50. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Ans : Calculating pH and pKa
Given:
Concentration of bromoacetic acid (HA) = 0.1 M
Degree of ionization (α) = 0.132
Step 1: Calculate the concentration of H⁺ ions
[H⁺] = Cα = 0.1 M * 0.132 = 0.0132 M
Step 2: Calculate pH
pH = -log[H⁺] = -log(0.0132) ≈ 1.88
Step 3: Calculate the acid dissociation constant (Ka)
Ka = Cα² / (1 – α) = 0.1 * (0.132)² / (1 – 0.132) ≈ 0.00201
Step 4: Calculate pKa
pKa = -log Ka = -log(0.00201) ≈ 2.7
Therefore, the pH of the solution is approximately 1.88, and the pKa of bromoacetic acid is approximately 2.7.
51. The pH of 0.005M codeine (C18H21N03) solution is 9.95. Calculate its ionization constant and pKb.
Ans : Codeine (C18H21N03) is a weak base. When it dissolves in water, it accepts a proton to form its conjugate acid.
Step 1: Calculate concentration of hydroxide ions (OH-)
Since pH = -log[H+], we can find [H+] as follows:
[H+] = 10^(-pH) = 10^(-9.95) ≈ 1.12 × 10^-10 M
Here, use the ion product of water (Kw) to find [OH-]:
Kw = [H+][OH-]
[OH-] = Kw / [H+]
[OH-] = 1.0 × 10^-14 / 1.12 × 10^-10
[OH-] ≈ 8.93 × 10^-5 M
Step 2: Find the ionization constant (Kb) of codeine
Codeine acts as a base in water, accepting a proton to form its conjugate acid. The ionization reaction can be represented as:
C18H21NO3 + H2O ⇌ C18H21NO3H+ + OH-
The base ionization constant is Kb
Kb = [C18H21NO3H+][OH-] / [C18H21NO3]
Since the initial concentration of codeine (0.005 M) is much larger than the concentration of hydroxide ions (8.93 × 10^-5 M), we can assume that the concentration of codeine at equilibrium remains approximately 0.005 M.
Therefore:
Kb ≈ (8.93 × 10^-5)^2 / 0.005
Kb ≈ 1.59 × 10^-6
Step 3: Find pKb
pKb = -log Kb
pKb = -log(1.59 × 10^-6) ≈ 5.8
In summary:
The concentration of hydroxide ions in the solution is 8.93 × 10^-5 M.
The ionization constant (Kb) of codeine is 1.59 × 10^-6.
The pKb of codeine is 5.8.
52. What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Ans : The ionization constant of aniline (C₆H₅NH₂) is Kb = 4.3 × 10⁻¹⁰.
Ionization of Aniline:
C₆H₅NH₂(aq) + H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq)
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
C₆H₅NH₂ 0.001 -x 0.001 – x
C₆H₅NH₃⁺ 0 +x x
OH⁻ 0 +x x
Kb Expression:
Kb = [C₆H₅NH₃⁺] * [OH⁻] / [C₆H₅NH₂]
Substituting equilibrium concentrations:
4.3 × 10⁻¹⁰ = (x * x) / (0.001 – x)
Assuming x is very small compared to 0.001 M, we can simplify the equation:
4.3 × 10⁻¹⁰ ≈ x² / 0.001
Solving for x (concentration of OH⁻):
x ≈ √(4.3 × 10⁻¹³) ≈ 6.56 × 10⁻⁷ M
Calculating pH:
pH = 14 – pOH = 14 – (-log[OH⁻]) = 14 – (-log(6.56 × 10⁻⁷)) ≈ 7.18
Calculating Degree of Ionization (α):
α = [OH⁻] / [C₆H₅NH₂] = (6.56 × 10⁻⁷) / (0.001) ≈ 6.56 × 10⁻⁴
Calculating Ka for the conjugate acid (C₆H₅NH₃⁺):
Ka = Kw / Kb = 1.0 × 10⁻¹⁴ / 4.3 × 10⁻¹⁰ ≈ 2.33 × 10⁻⁵
Therefore:
The pH of the 0.001 M aniline solution is approximately 7.18.
The degree of ionization of aniline in the solution is approximately 6.56 × 10⁻⁴.
The ionization constant of the conjugate acid of aniline (C₆H₅NH₃⁺) is approximately 2.33 × 10⁻⁵.
53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains
(a) 0.01M
(b) 0.1M in HCl ?
Ans : To calculate the degree of ionization (α) of 0.05 M acetic acid, we can use the following formula:
α = √(Ka / C)
where:
Ka is the acid dissociation constant
C is the concentration of the acid
Step 1: Calculate Ka from pKa:
Ka = 10^(-pKa) = 10^(-4.74) ≈ 1.82 × 10⁻⁵
Step 2: Calculate α:
α = √(1.82 × 10⁻⁵ / 0.05) ≈ 0.0191
Hence, the degree of ionization of 0.05 M acetic acid is approximately 0.0191 or 1.91%.
Effect of adding HCl:
Adding HCl to the solution will increase the concentration of H⁺ ions. This will suppress the ionization of acetic acid, according to Le Châtelier’s principle. As a result, the degree of ionization of acetic acid will decrease.
The effect of adding HCl will be more pronounced at higher concentrations of HCl. For example, adding 0.1 M HCl will significantly reduce the degree of ionization of acetic acid compared to adding 0.01 M HCl.
54. The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
Ans : Calculating Degree of Ionization and Effect of NaOH
Part 1: Degree of Ionization in 0.02 M Dimethylamine Solution
Reaction: (CH₃)₂NH(aq) + H₂O(l) ⇌ (CH₃)₂NH₂⁺(aq) + OH⁻(aq)
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
(CH₃)₂NH 0.02 -x 0.02 – x
(CH₃)₂NH₂⁺ 0 +x x
OH⁻ 0 +x x
Kb Expression:
Kb = [(CH₃)₂NH₂⁺] * [OH⁻] / [(CH₃)₂NH]
Substituting equilibrium concentrations:
5.4 × 10⁻⁴ = (x * x) / (0.02 – x)
Assuming x is small compared to 0.02 M, we can simplify the equation:
5.4 × 10⁻⁴ ≈ x² / 0.02
Solving for x:
x ≈ √(5.4 × 10⁻⁴ * 0.02) ≈ 0.0033
Degree of Ionization (α):
α = x / C = 0.0033 / 0.02 ≈ 0.165
Percentage Ionization:
(0.165) * 100% ≈ 16.5%
Part 2: Effect of 0.1 M NaOH
Adding 0.1 M NaOH to the solution will significantly increase the concentration of OH⁻ ions. This will suppress the ionization of dimethylamine according to Le Châtelier’s principle. As a result, the degree of ionization of dimethylamine will decrease significantly.
Therefore, the degree of ionization of 0.02 M dimethylamine is approximately 16.5%, and the addition of 0.1 M NaOH will significantly decrease the degree of ionization.
55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83
(b) Human stomach fluid, 1.2
(c) Human blood, 7.38
(d) Human saliva, 6.4.
Ans : To calculate the hydrogen ion concentration ([H+]) from the pH, we use the formula:
[H+] = 10^(-pH)
(a) Human muscle-fluid, pH = 6.83
[H+] = 10^(-6.83) ≈ 1.48 × 10⁻⁷ M
(b) Human stomach fluid, pH = 1.2
[H+] = 10^(-1.2) ≈ 6.31 × 10⁻² M
(c) Human blood, pH = 7.38
[H+] = 10^(-7.38) ≈ 4.17 × 10⁻⁸ M
(d) Human saliva, pH = 6.4
[H+] = 10^(-6.4) ≈ 3.98 × 10⁻⁷ M
Therefore, the hydrogen ion concentrations in the given biological fluids are:
(a) Human muscle-fluid: 1.48 × 10⁻⁷ M
(b) Human stomach fluid: 6.31 × 10⁻² M
(c) Human blood: 4.17 × 10⁻⁸ M
(d) Human saliva: 3.98 × 10⁻⁷ M
56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Ans : To calculate the hydrogen ion concentration ([H+]) from the pH, we use the formula:
[H+] = 10^(-pH)
Milk: pH = 6.8
[H+] = 10^(-6.8) ≈ 1.58 × 10⁻⁷ M
Black coffee: pH = 5.0
[H+] = 10^(-5.0) = 1 × 10⁻⁵ M
Tomato juice: pH = 4.2
[H+] = 10^(-4.2) ≈ 6.31 × 10⁻⁵ M
Lemon juice: pH = 2.2
[H+] = 10^(-2.2) ≈ 6.31 × 10⁻³ M
Egg white: pH = 7.8
[H+] = 10^(-7.8) ≈ 1.58 × 10⁻⁸ M
Therefore, the hydrogen ion concentrations in the given substances are:
Milk: 1.58 × 10⁻⁷ M
Black coffee: 1 × 10⁻⁵ M
Tomato juice: 6.31 × 10⁻⁵ M
Lemon juice: 6.31 × 10⁻³ M
Egg white: 1.58 × 10⁻⁸ M
57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Ans : To calculate the concentrations of potassium, hydrogen, and hydroxyl ions, and the pH of the solution, we need to first determine the concentration of KOH.
Step 1: Calculate the moles of KOH:
Moles of KOH =
mass of KOH
—————————–
molar mass of KOH
Moles of KOH = 0.561 g / 56.11 g/mol ≈ 0.01 mol
Step 2: Calculate the concentration of KOH:
Concentration of KOH = moles of KOH / volume of solution in liters
Concentration of KOH = 0.01 mol / 0.2 L = 0.05 M
Step 3: Determine the concentrations of ions:
KOH undergoes complete ionization when dissolved in water.
KOH(aq) → K⁺(aq) + OH⁻(aq)
Therefore, the concentration of potassium ions (K⁺) and hydroxyl ions (OH⁻) will be equal to the concentration of KOH.
[K⁺] = 0.05 M
[OH⁻] = 0.05 M
Step 4: Calculate the concentration of hydrogen ions (H⁺):
Since the solution is aqueous, we can use the ion product of water (Kw) to calculate the concentration of hydrogen ions:
Kw = [H⁺] * [OH⁻] = 1.0 × 10⁻¹⁴
[H⁺] = Kw / [OH⁻] = (1.0 × 10⁻¹⁴) / (0.05) = 2.0 × 10⁻¹³ M
Step 5: Calculate the pH:
pH = -log[H⁺] = -log(2.0 × 10⁻¹³) ≈ 12.70
Therefore:
The concentration of potassium ions (K⁺) is 0.05 M.
The concentration of hydrogen ions (H⁺) is 2.0 × 10⁻¹³ M.
The concentration of hydroxyl ions (OH⁻) is 0.05 M.
The pH of the solution is 12.70.
58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Ans : To calculate the concentrations of strontium and hydroxyl ions and the pH of the solution, we need to first determine the molar solubility of Sr(OH)₂.
Step 1: Calculate the molar mass of Sr(OH)₂:
Molar mass of Sr(OH)₂ = 87.62 g/mol + 2 * 17.01 g/mol = 121.64 g/mol
Step 2: Calculate the molar solubility:
Molar solubility = (19.23 g/L) / (121.64 g/mol) = 0.1581 mol/L
Step 3: Find the molar concentrations of Sr²⁺ and OH⁻.
Sr(OH)₂ dissociates completely in water according to the equation:
Sr(OH)₂(s) → Sr²⁺(aq) + 2OH⁻(aq)
Therefore, for every mole of Sr(OH)₂ that dissolves, we get 1 mole of Sr²⁺ ions and 2 moles of OH⁻ ions.
[Sr²⁺] = 0.1581 M
[OH⁻] = 2 * 0.1581 M = 0.3162 M
Step 4: Calculate the pH:
pH = 14 – pOH = 14 – (-log[OH⁻]) = 14 – (-log(0.3162)) ≈ 13.5
Therefore, the concentrations of strontium and hydroxyl ions are 0.1581 M and 0.3162 M, respectively, and the pH of the solution is approximately 13.5.
59. The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Ans : To calculate the degree of ionization (α) of 0.05 M propanoic acid and its pH, we can use the following steps:
Step 1: Calculate the hydronium ion concentration ([H+])
Propanoic acid (CH₃CH₂COOH) is a weak acid that ionizes in water according to the equation:
CH₃CH₂COOH(aq) + H₂O(l) ⇌ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)
The equilibrium constant expression defines the relationship between the concentrations of reactants and products at equilibrium.
Ka = [CH₃CH₂COO⁻][H₃O⁺] / [CH₃CH₂COOH]
where Ka is the acid ionization constant, and the square brackets denote molar concentrations.
We can assume that the degree of ionization (α) is small compared to the initial concentration of propanoic acid (0.05 M).
Ka ≈ [H₃O⁺]² / 0.05
Solving for [H₃O+]:
[H₃O⁺] = √(Ka * 0.05) = √(1.32 × 10⁻⁵ * 0.05) ≈ 8.13 × 10⁻⁴ M
Step 2: Calculate the pH:
pH =
-log[H₃O⁺] =
-log(8.13 × 10⁻⁴) ≈ 3.09
Step 3: Find the degree of ionization (α):
α = [H₃O⁺] / [CH₃CH₂COOH] = (8.13 × 10⁻⁴) / 0.05 ≈ 0.0163
Therefore, the degree of ionization of 0.05 M propanoic acid is approximately 0.0163 or 1.63%, and the pH of the solution is 3.09.
Effect of adding 0.01 M HCl:
Adding 0.01 M HCl to the solution will increase the concentration of H₃O⁺ ions. This will suppress the ionization of propanoic acid according to Le Châtelier’s principle. As a result, the degree of ionization of propanoic acid will decrease.
To calculate the new degree of ionization, we can use the same approach as before, but with the initial concentration of H₃O⁺ being 0.01 M (from the added HCl).
Step 1: Calculate the new [H₃O⁺]:
[H₃O⁺] = 0.01 M (from HCl) + x (from propanoic acid ionization)
Step 2: Substitute into the Ka expression:
1.32 × 10⁻⁵ = (x * (0.01 + x)) / (0.05 – x)
Find x
We use quadratic equation
x ≈ 1.32 × 10⁻³ M
Step 3: Calculate the new degree of ionization:
α = x / 0.05 ≈ 0.0264
Therefore, the degree of ionization of propanoic acid in the presence of 0.01 M HCl is approximately 0.0264 or 2.64%.
As expected, the degree of ionization of propanoic acid decreases in the presence of HCl, due to the common ion effect.
60. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Ans : To calculate the ionization constant (Ka) and the degree of ionization (α) of cyanic acid (HCNO), we can use the following steps:
Step 1: Calculate the hydronium ion concentration ([H+]) from the pH:
[H+] = 10^(-pH) = 10^(-2.34) ≈ 4.57 × 10⁻³ M
Step 2: Write the ionization equation for cyanic acid:
HCNO(aq) ⇌ H⁺(aq) + CNO⁻(aq)
Step 3: Set up an ICE (Initial, Change, Equilibrium) table:
Species Initial (M) Change (M) Equilibrium (M)
HCNO 0.1 -x 0.1 – x
H⁺ 0 +x x
CNO⁻ 0 +x x
Step 4: Find the equilibrium concentrations into the Ka expression:
Ka = [H⁺] * [CNO⁻] / [HCNO]
Ka = (x * x) / (0.1 – x)
Step 5: Since the pH is relatively low, we can assume that x is small compared to 0.1 M. This allows us to simplify the equation:
Ka ≈ x² / 0.1
Step 6: Solve for x (the concentration of H⁺):
x² ≈ Ka * 0.1
x = √(Ka * 0.1)
Step 7: Calculate Ka:
Ka = (x²) / 0.1 = (4.57 × 10⁻³) / 0.1 ≈ 2.09 × 10⁻⁴
Step 8: Calculate the degree of ionization (α):
α = x / C = (4.57 × 10⁻³) / 0.1 ≈ 0.0457
Therefore:
The ionization constant (Ka) of cyanic acid is approximately 2.09 × 10⁻⁴.
The degree of ionization of cyanic acid in the 0.1 M solution is approximately 0.0457 or 4.57%.
61. The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Ans : Analyzing Sodium Nitrite Solution
Given:
Ionization constant of nitrous acid (HNO₂): Ka = 4.5 × 10⁻⁴
Concentration of sodium nitrite (NaNO₂) = 0.04 M
Understanding the Solution:
Sodium nitrite (NaNO₂) is the salt of a weak acid (HNO₂) and a strong base (NaOH).
When dissolved in water, NaNO₂ completely dissociates into Na⁺ and NO₂⁻ ions.
The NO₂⁻ ion can react with water to form HNO₂ and OH⁻ ions.
Hydrolysis Reaction:
NO₂⁻(aq) + H₂O(l) ⇌ HNO₂(aq) + OH⁻(aq)
Kb for NO₂⁻:
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (4.5 × 10⁻⁴) ≈ 2.22 × 10⁻¹¹
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
NO₂⁻ 0.04 -x 0.04 – x
HNO₂ 0 +x x
OH⁻ 0 +x x
Kb Expression:
Kb = [HNO₂] * [OH⁻] / [NO₂⁻]
Substituting equilibrium concentrations:
2.22 × 10⁻¹¹ = (x * x) / (0.04 – x)
Assuming x is small compared to 0.04 M, we can simplify the equation:
2.22 × 10⁻¹¹ ≈ x² / 0.04
Solving for x (concentration of OH⁻):
x ≈ √(2.22 × 10⁻¹¹ * 0.04) ≈ 3.0 × 10⁻⁶ M
Calculating pH:
pH = 14 – pOH = 14 – (-log[OH⁻]) = 14 – (-log(3.0 × 10⁻⁶)) ≈ 8.48
Calculating Degree of Hydrolysis (α):
α = [HNO₂] / [NO₂⁻] = x / 0.04 ≈ 3.0 × 10⁻⁶ / 0.04 ≈ 7.5 × 10⁻⁵
Therefore:
The pH of the 0.04 M sodium nitrite solution is approximately 8.48.
The degree of hydrolysis of the nitrite ion (NO₂⁻) is approximately 7.5 × 10⁻⁵.
62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Ans : Pyridinium hydrochloride (C₅H₅NHCl) is the salt of a weak base (pyridine, C₅H₅N) and a strong acid (HCl). In solution, it dissociates completely into pyridinium ions (C₅H₅NH⁺) and chloride ions (Cl⁻).
The pyridinium ion can then undergo hydrolysis to produce pyridine and hydronium ions:
C₅H₅NH⁺(aq) + H₂O(l) ⇌ C₅H₅N(aq) + H₃O⁺(aq)
Given:
pH of 0.02 M pyridinium hydrochloride solution = 3.44
Concentration of pyridinium hydrochloride = 0.02 M
Step 1: Calculate the hydronium ion concentration:
[H₃O⁺] = 10^(-pH) = 10^(-3.44) ≈ 3.63 × 10⁻⁴ M
Step 2: Calculate the concentration of pyridine (C₅H₅N):
Since pyridinium hydrochloride dissociates completely, the concentration of pyridinium ions (C₅H₅NH⁺) is equal to the initial concentration of pyridinium hydrochloride, which is 0.02 M.
[C₅H₅NH⁺] = 0.02 M
Step 3: Calculate the ionization constant of pyridine (Kb):
Kb = [C₅H₅N] * [H₃O⁺] / [C₅H₅NH⁺]
Kb = (x * 3.63 × 10⁻⁴) / 0.02
where x is the equilibrium concentration of pyridine.
Step 4: Since the pH is relatively low, we can assume that x is small compared to 0.02 M. This allows us to simplify the equation:
Kb ≈ (x * 3.63 × 10⁻⁴) / 0.02
Step 5: Solve for x (concentration of pyridine):
x ≈ Kb * 0.02 / 3.63 × 10⁻⁴
Step 6: Substitute the given Kb value and solve for x:
x ≈ (1.52 × 10⁻⁹) * 0.02 / 3.63 × 10⁻⁴ ≈ 8.37 × 10⁻⁸ M
Step 7: Calculate the ionization constant of pyridine (Kb):
Kb = (x * 3.63 × 10⁻⁴) / 0.02 ≈ (8.37 × 10⁻⁸ * 3.63 × 10⁻⁴) / 0.02 ≈ 1.52 × 10⁻⁹
Therefore, the ionization constant of pyridine is approximately 1.52 × 10⁻⁹.
63. Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF
Ans : To predict the acidity or basicity of a salt, we need to consider the nature of its constituent ions.
1. Salts of Strong Acid and Strong Base:
These salts are neutral. Examples: NaCl, KBr
2. Salts of Strong Acid and Weak Base:
These salts are acidic. Examples: NH₄NO₃
3. Salts of Weak Acid and Strong Base:
These salts are basic. Examples: NaCN, NaNO₂
4. Salts of Weak Acid and Weak Base:
The acidity or basicity of these salts depends on the relative strengths of the acid and base.
Based on this information, we can predict the nature of the given salts:
NaCl and KBr: Neutral salts (strong acid + strong base)
NaCN: Basic salt (weak acid + strong base)
NH₄NO₃: Acidic salt (strong acid + weak base)
NaNO₂: Basic salt (weak acid + strong base)
KF: Basic salt (weak acid + strong base)
Therefore, the solutions of NaCl and KBr will be neutral, while the solutions of NaCN, NH₄NO₃, NaNO₂, and KF will be basic.
64. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Ans : Analyzing Chloroacetic Acid and Its Sodium Salt
Chloroacetic Acid (CH₂ClCOOH)
Ionization: CH₂ClCOOH(aq) ⇌ H⁺(aq) + CH₂ClCOO⁻(aq)
Ka: 1.35 × 10⁻³
Sodium Chloroacetate (CH₂ClCOONa)
Complete Dissociation: CH₂ClCOONa(aq) → CH₂ClCOO⁻(aq) + Na⁺(aq)
pH of 0.1 M Chloroacetic Acid Solution
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
CH₂ClCOOH 0.1 -x 0.1 – x
H⁺ 0 +x x
CH₂ClCOO⁻ 0 +x x
Ka Expression:
Ka = [H⁺] * [CH₂ClCOO⁻] / [CH₂ClCOOH]
Substituting equilibrium concentrations:
1.35 × 10⁻³ = x² / (0.1 – x)
Assuming x is small compared to 0.1 M, we can simplify the equation:
1.35 × 10⁻³ ≈ x² / 0.1
Solving for x (concentration of H⁺):
x ≈ √(1.35 × 10⁻³ * 0.1) ≈ 0.0116 M
pH:
pH = -log[H⁺] = -log(0.0116) ≈ 1.94
pH of 0.1 M Sodium Chloroacetate Solution
Since sodium chloroacetate is the salt of a weak acid (chloroacetic acid) and a strong base (NaOH), the solution will be basic. The hydrolysis of the CH₂ClCOO⁻ ion will produce OH⁻ ions, increasing the pH.
Hydrolysis Reaction:
CH₂ClCOO⁻(aq) + H₂O(l) ⇌ CH₂ClCOOH(aq) + OH⁻(aq)
Kb for CH₂ClCOO⁻:
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.35 × 10⁻³) ≈ 7.41 × 10⁻¹²
ICE Table:
Species Initial (M) Change (M) Equilibrium (M)
CH₂ClCOO⁻ 0.1 -x 0.1 – x
CH₂ClCOOH 0 +x x
OH⁻ 0 +x x
Kb Expression:
Kb = [CH₂ClCOOH] * [OH⁻] / [CH₂ClCOO⁻]
Substituting equilibrium concentrations and simplifying:
7.41 × 10⁻¹² ≈ x² / 0.1
Solving for x (concentration of OH⁻):
x ≈ √(7.41 × 10⁻¹² * 0.1) ≈ 8.62 × 10⁻⁷ M
Calculating pH:
pH = 14 – pOH = 14 – (-log[OH⁻]) = 14 – (-log(8.62 × 10⁻⁷)) ≈ 6.07
Therefore:
The pH of 0.1 M chloroacetic acid solution is approximately 1.94.
The pH of 0.1 M sodium chloroacetate solution is approximately 6.07.
65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Ans : The product of the molar concentrations of H⁺ and OH⁻ ions in water is known as Kw.. In a neutral solution, the concentrations of hydronium ions and hydroxide ions are identical.
Hence, we follow :
Kw =
[H⁺] * [OH⁻] = [H⁺]²
Taking the negative logarithm of both sides:
-log(Kw) = -log([H⁺]²) = -2 * log[H⁺]
Since pH = -log[H⁺], we can rewrite the equation as:
pKw = 2 * pH
where pKw represents the negative logarithm of the ion product of water.
Now, we can solve for pH:
pH = pKw / 2 =
-log(2.7 × 10⁻¹⁴) / 2 ≈ 6.8
Hence, the pH of neutral water at 310 K is approximately 6.8.
66. Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
Ans : Calculating pH of Resultant Mixtures
a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl
Moles of Ca(OH)₂: 0.2 M * 0.01 L = 0.002 moles
Moles of HCl: 0.1 M * 0.025 L = 0.0025 moles
Net moles of HCl: 0.0025 – 0.002 = 0.0005 moles
Total volume: 10 mL + 25 mL = 35 mL = 0.035 L
Concentration of HCl: 0.0005 moles / 0.035 L ≈ 0.0143 M
pH: -log[H⁺] = -log(0.0143) ≈ 1.85
b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
Moles of H₂SO₄: 0.01 M * 0.01 L = 0.0001 moles
Moles of Ca(OH)₂: 0.01 M * 0.01 L = 0.0001 moles
Neutralization: H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O
Since the moles of acid and base are equal, the solution is neutral.
pH: 7
c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH
Moles of H₂SO₄: 0.1 M * 0.01 L = 0.001 moles
Moles of KOH=
0.1 M * 0.01 L =
0.001 moles
Neutralization: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Since the moles of acid and base are equal, the solution is neutral.
pH: 7
Therefore, the pH of the resultant mixtures are:
a) 1.85
b) 7
c) 7
67. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.
Ans : The solubilities and molarities of individual ions for the given salts can be calculated using the solubility product constants (Ksp) provided in the table.
1. Silver Chromate (Ag₂CrO₄):
Ksp = 1.1 × 10⁻¹²
Dissolution equation: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)
Ksp = [Ag⁺]² * [CrO₄²⁻]
Let s be the solubility of Ag₂CrO₄.
I.E [Ag⁺]
= 2s and [CrO₄²⁻]
= s.
Ksp = (2s)² * s = 4s³
Solving for s: s = (Ksp / 4)^(1/3) = (1.1 × 10⁻¹²) / 4)^(1/3) ≈ 6.5 × 10⁻⁵ M
Solubility of Ag₂CrO₄: 6.5 × 10⁻⁵ M
[Ag⁺]: 2 * 6.5 × 10⁻⁵ M = 1.3 × 10⁻⁴ M
[CrO₄²⁻]: 6.5 × 10⁻⁵ M
2. Barium Chromate (BaCrO₄):
Ksp = 1.2 × 10⁻¹⁰
Dissolution equation: BaCrO₄(s) ⇌ Ba²⁺(aq) + CrO₄²⁻(aq)
Ksp = [Ba²⁺] * [CrO₄²⁻]
Let s be the solubility of BaCrO₄. Then, [Ba²⁺] = s and [CrO₄²⁻] = s.
Ksp = s²
Solving for s: s = √(Ksp) = √(1.2 × 10⁻¹⁰) ≈ 3.46 × 10⁻⁵ M
Solubility of BaCrO₄: 3.46 × 10⁻⁵ M
[Ba²⁺]: 3.46 × 10⁻⁵ M
[CrO₄²⁻]: 3.46 × 10⁻⁵ M
3. Ferric Hydroxide (Fe(OH)₃):
Ksp = 6.3 × 10⁻³⁸
Dissolution equation: Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH⁻(aq)
Ksp = [Fe³⁺] * [OH⁻]³
Let s be the solubility of Fe(OH)₃. Then, [Fe³⁺] = s and [OH⁻] = 3s.
Ksp = s * (3s)³ = 27s⁴
Solving for s: s = (Ksp / 27)^(1/4) = (6.3 × 10⁻³⁸ / 27)^(1/4) ≈ 1.3 × 10⁻¹⁰ M
Solubility of Fe(OH)₃: 1.3 × 10⁻¹⁰ M
[Fe³⁺]: 1.3 × 10⁻¹⁰ M
[OH⁻]: 3 * 1.3 × 10⁻¹⁰ M = 3.9 × 10⁻¹⁰ M
4. Lead Chloride (PbCl₂):
Ksp = 1.6 × 10⁻⁵
Dissolution equation: PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)
Ksp = [Pb²⁺] * [Cl⁻]²
Let s be the solubility of PbCl₂. Then, [Pb²⁺] = s and [Cl⁻] = 2s.
Ksp = s * (2s)² = 4s³
Solving for s: s = (Ksp / 4)^(1/3) = (1.6 × 10⁻⁵ / 4)^(1/3) ≈ 1.54 × 10⁻² M
Solubility of PbCl₂: 1.54 × 10⁻² M
[Pb²⁺]: 1.54 × 10⁻² M
[Cl⁻]: 2 * 1.54 × 10⁻² M = 3.08 × 10⁻² M
5. Mercurous Iodide (Hg₂I₂):
Ksp = 5.2 × 10⁻²⁹
Dissolution equation: Hg₂I₂(s) ⇌ Hg₂²⁺(aq) + 2I⁻(aq)
Ksp = [Hg₂²⁺] * [I⁻]²
Let s be the solubility of Hg₂I₂. Then, [Hg₂²⁺] = s and [I⁻] = 2s.
Ksp = s * (2s)² = 4s³
Solving for s: s = (Ksp / 4)^(1/3) = (5.2 × 10⁻²⁹ / 4)^(1/3) ≈ 2.25 × 10⁻¹⁰ M
Solubility of Hg₂I₂: 2.25 × 10⁻¹⁰ M
[Hg₂²⁺]: 2.25 × 10⁻¹⁰ M
[I⁻]: 2 * 2.25 × 10⁻¹⁰ M = 4.50 × 10⁻¹⁰ M
68. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Ans : To calculate the ratio of the molarities of saturated solutions of Ag₂CrO₄ and AgBr, we first need to find their individual solubilities using their Ksp values.
Solubility of Ag₂CrO₄:
Ksp(Ag₂CrO₄) = [Ag⁺]² * [CrO₄²⁻]
Let s be the solubility of Ag₂CrO₄.
Hence,
[Ag⁺]
= 2s and [CrO₄²⁻]
= s.
Ksp = (2s)² * s = 4s³
Solving for s:
s = (Ksp / 4)^(1/3) = (1.1 × 10⁻¹²) / 4)^(1/3) ≈ 6.50 × 10⁻⁵ M
Solubility of AgBr:
Ksp(AgBr) = [Ag⁺] * [Br⁻]
Let s be the solubility of AgBr. Then, [Ag⁺] = s and [Br⁻] = s.
Ksp = s²
Solving for s:
s = √(Ksp) = √(5.0 × 10⁻¹³) ≈ 7.07 × 10⁻⁷ M
Ratio of solubilities:
(Solubility of Ag₂CrO₄) / (Solubility of AgBr) = (6.50 × 10⁻⁵ M) / (7.07 × 10⁻⁷ M) ≈ 91.9
Therefore, the ratio of the molarities of saturated solutions of Ag₂CrO₄ and AgBr is approximately 91.9.
69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate?
(For cupric iodate Ksp = 7.4 × 10–8 ). 6.70 The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13.
How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Ans : Part 1: Precipitation of Copper Iodate
Reaction: Cu²⁺(aq) + 2IO₃⁻(aq) ⇌ Cu(IO₃)₂(s)
Ksp for Cu(IO₃)₂: 7.4 × 10⁻⁸
Initial Concentrations after Mixing:
[Cu²⁺] = 0.002 M / 2 = 0.001 M (due to dilution)
[IO₃⁻] = 0.002 M / 2 = 0.001 M (due to dilution)
Ionic Product (Q):
Q = [Cu²⁺] * [IO₃⁻]² = (0.001 M) * (0.001 M)² = 1 × 10⁻⁹
Comparison of Q and Ksp:
Since Q (1 × 10⁻⁹) is less than Ksp (7.4 × 10⁻⁸), no precipitation of copper iodate will occur.
Part 2: Solubility of Silver Benzoate in Buffer vs. Pure Water
Reaction: Ag⁺(aq) + C₆H₅COO⁻(aq) ⇌ AgC₆H₅COO(s)
Ksp for AgC₆H₅COO: 2.5 × 10⁻¹³
Solubility in Pure Water:
Let s be the solubility of AgC₆H₅COO in pure water.
Ksp = [Ag⁺] * [C₆H₅COO⁻] = s²
s = √(Ksp) = √(2.5 × 10⁻¹³) ≈ 5.0 × 10⁻⁷ M
Solubility in Buffer (pH 3.19):
Since the pH of the buffer is 3.19, [H₃O⁺] = 10⁻³.¹⁹ ≈ 6.46 × 10⁻⁴ M
Considering the benzoic acid equilibrium:
C₆H₅COOH(aq) ⇌ H⁺(aq) + C₆H₅COO⁻(aq)
Ka = 6.46 × 10⁻⁵ = [H⁺] * [C₆H₅COO⁻] / [C₆H₅COOH]
Solving for [C₆H₅COO⁻]:
[C₆H₅COO⁻] = Ka * [C₆H₅COOH] / [H₃O⁺]
Since the buffer solution contains a significant amount of benzoic acid, we can assume that [C₆H₅COOH] ≈ [C₆H₅COO⁻].
Therefore, [C₆H₅COO⁻] ≈ Ka = 6.46 × 10⁻⁵ M
Substituting into the Ksp expression for AgC₆H₅COO:
Ksp = [Ag⁺] * [C₆H₅COO⁻]
[Ag⁺] = Ksp / [C₆H₅COO⁻] = (2.5 × 10⁻¹³) / (6.46 × 10⁻⁵) ≈ 3.87 × 10⁻⁹ M
Solubility in buffer:
s’ = [Ag⁺] = 3.87 × 10⁻⁹ M
Ratio of solubilities:
s’ / s = (3.87 × 10⁻⁹ M) / (5.0 × 10⁻⁷ M) ≈ 7.74 × 10⁻³
Therefore, silver benzoate is approximately 7.74 × 10⁻³ times more soluble in a buffer of pH 3.19 compared to its solubility in pure water.
71. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18).
Ans : To prevent precipitation of iron sulfide (FeS), the ionic product (Qsp) must be less than the solubility product constant (Ksp).
Reaction: FeSO₄(aq) + Na₂S(aq) → FeS(s) + Na₂SO₄(aq)
Ksp expression:
Ksp = [Fe²⁺] * [S²⁻] = 6.3 × 10⁻¹⁸
Let’s assume the concentration of both FeSO₄ and Na₂S solutions is x M.
After mixing equal volumes, the concentration of each salt will be halved:
[Fe²⁺] = x/2 M
[S²⁻] = x/2 M
Ionic product (Qsp):
Qsp = [Fe²⁺] * [S²⁻] = (x/2)² = x²/4
To prevent precipitation, Qsp must be less than Ksp:
x²/4 < 6.3 × 10⁻¹⁸
Solving for x:
x² < 2.52 × 10⁻¹⁷
x < √(2.52 × 10⁻¹⁷) ≈ 5.02 × 10⁻⁹ M
Therefore, the maximum concentration of equimolar solutions of ferrous sulfate and sodium sulfide that can be mixed without precipitation is approximately 5.02 × 10⁻⁹ M.
72. What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).
Ans : To determine the minimum volume of water required to dissolve 1 g of calcium sulfate (CaSO₄) at 298 K, we need to calculate its solubility product (Ksp) and then use that value to find the solubility in moles per liter.
Step 1: Calculate the molar mass of CaSO₄:
Molar mass of CaSO₄ = 40.08 g/mol (Ca) + 32.07 g/mol (S) + 4 * 16.00 g/mol (O) = 136.14 g/mol
Step 2: find the molar solubility (s) of CaSO₄ using the Ksp value:
Ksp =
[Ca²⁺] * [SO₄²⁻] =
(s)(s) =
s²
s = √(Ksp) = √(9.1 × 10⁻⁶) ≈ 3.02 × 10⁻³ mol/L
Step 3: Find the minimum volume of water necessary to dissolve 1 gram of CaSO₄.
Moles of CaSO₄ = 1 g / 136.14 g/mol ≈ 0.00735 mol
Volume of water = moles of CaSO₄ / solubility of CaSO₄
Volume of water = 0.00735 mol / 3.02 × 10⁻³ mol/L ≈ 2.43 L
hence, the minimum volume of water required to dissolve 1 g of calcium sulfate at 298 K is approximately 2.43 liters
73. The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?
Ans : To predict whether precipitation will form in each solution, we must calculate the ionic product (Qsp) for each metal sulfide and compare it to its solubility product constant (Ksp). A precipitate will form if the reaction quotient (Qsp) is greater than the solubility product (Ksp).
First, calculatefind the final concentration of sulfide ions (S²⁻) after mixing:
Total volume = 10 mL + 5 mL = 15 mL = 0.015 L
Moles of S²⁻ = 1.0 × 10⁻¹⁹ M * 0.01 L = 1.0 × 10⁻²¹ moles
Final concentration of S²⁻ = 1.0 × 10⁻²¹ moles / 0.015 L ≈ 6.67 × 10⁻²⁰ M
Next, calculate the initial concentration of the metal ions in each solution:
FeSO₄: 0.04 M
MnCl₂: 0.04 M
ZnCl₂: 0.04 M
CdCl₂: 0.04 M
Finally, calculate the ionic product (Qsp) for each metal sulfide:
FeS: Qsp =
[Fe²⁺] * [S²⁻] = (0.04 M) * (6.67 × 10⁻²⁰ M) ≈ 2.67 × 10⁻²¹
MnS: Qsp =
[Mn²⁺] * [S²⁻] = (0.04 M) * (6.67 × 10⁻²⁰ M) ≈ 2.67 × 10⁻²¹
ZnS: Qsp =
[Zn²⁺] * [S²⁻] = (0.04 M) * (6.67 × 10⁻²⁰ M) ≈ 2.67 × 10⁻²¹
CdS: Qsp =
[Cd²⁺] * [S²⁻] = (0.04 M) * (6.67 × 10⁻²⁰ M) ≈ 2.67 × 10⁻²¹
Compare Qsp to Ksp for each metal sulfide:
Ksp(FeS) = 6.3 × 10⁻¹⁸
Ksp(MnS) = 2.5 × 10⁻¹³
Ksp(ZnS) = 1.6 × 10⁻²⁴
Ksp(CdS) = 8.0 × 10⁻²⁷
Precipitation will occur if Qsp > Ksp.
An analysis of Qsp and Ksp values reveals that precipitation takes place in ZnCl₂ and CdCl₂ solutions. This is because the ionic product (Qsp) for ZnS and CdS is greater than their respective solubility product constants (Ksp).
