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Sagot :
To determine the pH of a 0.56 M sulfurous acid ([tex]\(H_2SO_3\)[/tex]) solution, knowing that it is a diprotic acid, we need to follow these steps:
1. First Dissociation:
[tex]\[ H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \][/tex]
The equilibrium constant for the first dissociation [tex]\(K_{a1}\)[/tex] is given as [tex]\(1.2 \times 10^{-2}\)[/tex].
2. We assume that the initial concentration of [tex]\(H_2SO_3\)[/tex] is 0.56 M, and let x be the concentration of [tex]\(H^+\)[/tex] ions produced by the first dissociation.
[tex]\[ Ka_1 = \frac{[H^+][HSO_3^-]}{[H_2SO_3]} \][/tex]
Given [tex]\(Ka_1 = 1.2 \times 10^{-2}\)[/tex] and initial concentration of [tex]\(H_2SO_3 = 0.56\)[/tex]:
[tex]\[ 1.2 \times 10^{-2} = \frac{x \cdot x}{0.56 - x} \approx \frac{x^2}{0.56} \][/tex]
Solving for x:
[tex]\[ x = \sqrt{1.2 \times 10^{-2} \times 0.56} = 0.0819756061276768 \][/tex]
This represents the hydrogen ion concentration ([tex]\([H^+]\)[/tex]) from the first dissociation.
3. Second Dissociation:
[tex]\[ HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \][/tex]
The equilibrium constant for the second dissociation [tex]\(K_{a2}\)[/tex] is given as [tex]\(6.2 \times 10^{-8}\)[/tex].
4. We now consider the contribution of H+ from the second dissociation. Let y be the concentration of [tex]\(H^+\)[/tex] ions produced by this step.
[tex]\[ Ka_2 = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]} \][/tex]
Given [tex]\(Ka_2 = 6.2 \times 10^{-8}\)[/tex] and the initial concentration of [tex]\(HSO_3^- = 0.0819756061276768\)[/tex]:
[tex]\[ 6.2 \times 10^{-8} = \frac{y \cdot y}{0.0819756061276768 + y} \approx \frac{y \cdot y}{0.0819756061276768} \][/tex]
Solving for y:
[tex]\[ y = \frac{6.2 \times 10^{-8} \times 0.0819756061276768}{0.0819756061276768 + 0.0819756061276768} = 5.40830518614672 \times 10^{-8} \][/tex]
5. Total Hydrogen Ion Concentration:
The total [tex]\( [H^+] \)[/tex] is the sum of the hydrogen ion concentrations from both dissociations.
[tex]\[ [H^+]_{total} = 0.0819756061276768 + 5.40830518614672 \times 10^{-8} = 0.0819756602107286 \][/tex]
6. Calculate the pH:
The pH is calculated using the formula:
[tex]\[ pH = -\log_{10} [H^+]_{total} \][/tex]
[tex]\[ pH = -\log_{10}(0.0819756602107286) = 2.50133290247808 / \log(10) = 2.50133290247808 \][/tex]
So, the pH of the 0.56 M sulfurous acid solution is approximately 2.50.
1. First Dissociation:
[tex]\[ H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \][/tex]
The equilibrium constant for the first dissociation [tex]\(K_{a1}\)[/tex] is given as [tex]\(1.2 \times 10^{-2}\)[/tex].
2. We assume that the initial concentration of [tex]\(H_2SO_3\)[/tex] is 0.56 M, and let x be the concentration of [tex]\(H^+\)[/tex] ions produced by the first dissociation.
[tex]\[ Ka_1 = \frac{[H^+][HSO_3^-]}{[H_2SO_3]} \][/tex]
Given [tex]\(Ka_1 = 1.2 \times 10^{-2}\)[/tex] and initial concentration of [tex]\(H_2SO_3 = 0.56\)[/tex]:
[tex]\[ 1.2 \times 10^{-2} = \frac{x \cdot x}{0.56 - x} \approx \frac{x^2}{0.56} \][/tex]
Solving for x:
[tex]\[ x = \sqrt{1.2 \times 10^{-2} \times 0.56} = 0.0819756061276768 \][/tex]
This represents the hydrogen ion concentration ([tex]\([H^+]\)[/tex]) from the first dissociation.
3. Second Dissociation:
[tex]\[ HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \][/tex]
The equilibrium constant for the second dissociation [tex]\(K_{a2}\)[/tex] is given as [tex]\(6.2 \times 10^{-8}\)[/tex].
4. We now consider the contribution of H+ from the second dissociation. Let y be the concentration of [tex]\(H^+\)[/tex] ions produced by this step.
[tex]\[ Ka_2 = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]} \][/tex]
Given [tex]\(Ka_2 = 6.2 \times 10^{-8}\)[/tex] and the initial concentration of [tex]\(HSO_3^- = 0.0819756061276768\)[/tex]:
[tex]\[ 6.2 \times 10^{-8} = \frac{y \cdot y}{0.0819756061276768 + y} \approx \frac{y \cdot y}{0.0819756061276768} \][/tex]
Solving for y:
[tex]\[ y = \frac{6.2 \times 10^{-8} \times 0.0819756061276768}{0.0819756061276768 + 0.0819756061276768} = 5.40830518614672 \times 10^{-8} \][/tex]
5. Total Hydrogen Ion Concentration:
The total [tex]\( [H^+] \)[/tex] is the sum of the hydrogen ion concentrations from both dissociations.
[tex]\[ [H^+]_{total} = 0.0819756061276768 + 5.40830518614672 \times 10^{-8} = 0.0819756602107286 \][/tex]
6. Calculate the pH:
The pH is calculated using the formula:
[tex]\[ pH = -\log_{10} [H^+]_{total} \][/tex]
[tex]\[ pH = -\log_{10}(0.0819756602107286) = 2.50133290247808 / \log(10) = 2.50133290247808 \][/tex]
So, the pH of the 0.56 M sulfurous acid solution is approximately 2.50.
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