Answer :
Sure, let's work through the solution step by step.
1. Calculate the hydrogen ion concentration [H⁺]:
The pH of the solution is given as 4.328. The pH is related to the hydrogen ion concentration by the following equation:
[tex]\[ \text{[H⁺]} = 10^{-\text{pH}} \][/tex]
Substituting the given pH value:
[tex]\[ \text{[H⁺]} = 10^{-4.328} \][/tex]
2. Compute [H⁺]:
[tex]\[ \text{[H⁺]} = 4.7 \times 10^{-5} \, \text{M} \][/tex]
3. Determine the equilibrium concentrations:
For a weak acid [tex]\(HX\)[/tex] dissociating in water, the reaction is:
[tex]\[ HX \rightleftharpoons H^+ + X^- \][/tex]
Initially, the concentration of [tex]\(HX\)[/tex] is 0.170 M and we assume [tex]\(x\)[/tex] is the amount that dissociates. At equilibrium:
[tex]\[ [H^+] = x = 4.7 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ [X^-] = x = 4.7 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ [HX] \approx 0.170 \, \text{M} - x \approx 0.170 \, \text{M} \][/tex]
4. Calculate the acid dissociation constant [tex]\(K_a\)[/tex]:
The expression for the equilibrium constant [tex]\(K_a\)[/tex] for this reaction is:
[tex]\[ K_a = \frac{[H^+][X^-]}{[HX]} \][/tex]
Substituting the equilibrium concentrations:
[tex]\[ K_a = \frac{(4.7 \times 10^{-5})(4.7 \times 10^{-5})}{0.170} \][/tex]
5. Calculate [tex]\(K_a\)[/tex]:
[tex]\[ K_a \approx \frac{(4.7 \times 10^{-5})^2}{0.170} = \frac{2.209 \times 10^{-9}}{0.170} \approx 1.3 \times 10^{-8} \][/tex]
6. Calculate the standard Gibbs free energy change ([tex]\(\Delta G^\circ\)[/tex]):
The relation between [tex]\(K_a\)[/tex] and [tex]\(\Delta G^\circ\)[/tex] is given by:
[tex]\[ \Delta G^\circ = -RT \ln K_a \][/tex]
where:
- [tex]\( R \)[/tex] is the universal gas constant [tex]\( 8.314 \, \text{J} / (\text{mol} \cdot \text{K}) \)[/tex]
- [tex]\( T \)[/tex] is the temperature in Kelvin. The temperature given is [tex]\(25^\circ C\)[/tex] which converts to Kelvin as:
[tex]\[ T = 25 + 273.15 = 298.15 \, \text{K} \][/tex]
- [tex]\( K_a \)[/tex] is the acid dissociation constant [tex]\(1.3 \times 10^{-8}\)[/tex].
7. Compute [tex]\(\Delta G^\circ\)[/tex]:
[tex]\[ \Delta G^\circ = - (8.314 \, \text{J} / (\text{mol} \cdot \text{K})) (298.15 \, \text{K}) \ln(1.3 \times 10^{-8}) \][/tex]
Calculate the natural logarithm:
[tex]\[ \ln(1.3 \times 10^{-8}) \approx \ln(10^{-8}) + \ln(1.3) \approx -18.14 \][/tex]
Putting it all together:
[tex]\[ \Delta G^\circ \approx - (8.314 \times 298.15 \times (-18.14)) \, \text{J} / \text{mol} \][/tex]
Simplify:
[tex]\[ \Delta G^\circ \approx (8.314 \times 298.15 \times 18.14) \approx 44841 \, \text{J} / \text{mol} \approx 44.84 \, \text{kJ} / \text{mol} \][/tex]
Therefore, the standard Gibbs free energy change [tex]\(\Delta G^\circ\)[/tex] for the dissociation reaction of the weak acid [tex]\(HX\)[/tex] at [tex]\(25^\circ C\)[/tex] is approximately [tex]\(44.84 \, \text{kJ} / \text{mol}\)[/tex].
1. Calculate the hydrogen ion concentration [H⁺]:
The pH of the solution is given as 4.328. The pH is related to the hydrogen ion concentration by the following equation:
[tex]\[ \text{[H⁺]} = 10^{-\text{pH}} \][/tex]
Substituting the given pH value:
[tex]\[ \text{[H⁺]} = 10^{-4.328} \][/tex]
2. Compute [H⁺]:
[tex]\[ \text{[H⁺]} = 4.7 \times 10^{-5} \, \text{M} \][/tex]
3. Determine the equilibrium concentrations:
For a weak acid [tex]\(HX\)[/tex] dissociating in water, the reaction is:
[tex]\[ HX \rightleftharpoons H^+ + X^- \][/tex]
Initially, the concentration of [tex]\(HX\)[/tex] is 0.170 M and we assume [tex]\(x\)[/tex] is the amount that dissociates. At equilibrium:
[tex]\[ [H^+] = x = 4.7 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ [X^-] = x = 4.7 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ [HX] \approx 0.170 \, \text{M} - x \approx 0.170 \, \text{M} \][/tex]
4. Calculate the acid dissociation constant [tex]\(K_a\)[/tex]:
The expression for the equilibrium constant [tex]\(K_a\)[/tex] for this reaction is:
[tex]\[ K_a = \frac{[H^+][X^-]}{[HX]} \][/tex]
Substituting the equilibrium concentrations:
[tex]\[ K_a = \frac{(4.7 \times 10^{-5})(4.7 \times 10^{-5})}{0.170} \][/tex]
5. Calculate [tex]\(K_a\)[/tex]:
[tex]\[ K_a \approx \frac{(4.7 \times 10^{-5})^2}{0.170} = \frac{2.209 \times 10^{-9}}{0.170} \approx 1.3 \times 10^{-8} \][/tex]
6. Calculate the standard Gibbs free energy change ([tex]\(\Delta G^\circ\)[/tex]):
The relation between [tex]\(K_a\)[/tex] and [tex]\(\Delta G^\circ\)[/tex] is given by:
[tex]\[ \Delta G^\circ = -RT \ln K_a \][/tex]
where:
- [tex]\( R \)[/tex] is the universal gas constant [tex]\( 8.314 \, \text{J} / (\text{mol} \cdot \text{K}) \)[/tex]
- [tex]\( T \)[/tex] is the temperature in Kelvin. The temperature given is [tex]\(25^\circ C\)[/tex] which converts to Kelvin as:
[tex]\[ T = 25 + 273.15 = 298.15 \, \text{K} \][/tex]
- [tex]\( K_a \)[/tex] is the acid dissociation constant [tex]\(1.3 \times 10^{-8}\)[/tex].
7. Compute [tex]\(\Delta G^\circ\)[/tex]:
[tex]\[ \Delta G^\circ = - (8.314 \, \text{J} / (\text{mol} \cdot \text{K})) (298.15 \, \text{K}) \ln(1.3 \times 10^{-8}) \][/tex]
Calculate the natural logarithm:
[tex]\[ \ln(1.3 \times 10^{-8}) \approx \ln(10^{-8}) + \ln(1.3) \approx -18.14 \][/tex]
Putting it all together:
[tex]\[ \Delta G^\circ \approx - (8.314 \times 298.15 \times (-18.14)) \, \text{J} / \text{mol} \][/tex]
Simplify:
[tex]\[ \Delta G^\circ \approx (8.314 \times 298.15 \times 18.14) \approx 44841 \, \text{J} / \text{mol} \approx 44.84 \, \text{kJ} / \text{mol} \][/tex]
Therefore, the standard Gibbs free energy change [tex]\(\Delta G^\circ\)[/tex] for the dissociation reaction of the weak acid [tex]\(HX\)[/tex] at [tex]\(25^\circ C\)[/tex] is approximately [tex]\(44.84 \, \text{kJ} / \text{mol}\)[/tex].
