acid-base equilibrium

HO₂C-CH₂-COOH ⇄ H⁺ + HO₂C-CH₂-CO₂⁻　　　　Ka1 = 1.4 X 10⁻³

HO₂C-CH₂-CO₂⁻ ⇄ H⁺ + ⁻O₂C-CH₂-CO₂⁻　　　　　Ka2 = 2.0 X 10⁻⁶

Estimate the pH value for a 1:1 mixture of malonic acid and sodium malonate

Ka1 = [H⁺][HO₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-COOH]

Ka2 = [H⁺][⁻O₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-CO₂⁻]

log(Ka1) = log[H⁺] + log{[HO₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-COOH]}

-log(Ka1) = -log[H⁺] - log{[HO₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-COOH]}

pH = pKa1 + log{[HO₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-COOH]}

similarly :

pH = pKa2 + log{[⁻O₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-CO₂⁻]}

For a 1:1 mixture of malonic acid and sodium malonate

log{[HO₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-COOH]} = 0

pH = pKa1 = -logKa1 = -log(1.4 X 10⁻³) = 2.85

In a solution of pH = 2.85

pH = pKa2 + log{[⁻O₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-CO₂⁻]}

2.85 = 5.70 + log{[⁻O₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-CO₂⁻]}

log{[⁻O₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-CO₂⁻]} = 2.85 - 5.70 = -2.85

[⁻O₂C-CH₂-CO₂⁻] / [HO₂C-CH₂-CO₂⁻] = 1.4X10⁻³

Ionization of the second step only gives 0.14% of [H⁺] to the whole mixture.
The ionization of the second step can be ignored.



2014-03-18 14:45:12 補充：
It is very kind of you to give me 5 points

2014-03-19 22:44:01 補充：
HO₂C-CH₂-CO₂⁻ + H⁺ → HO₂C-CH₂-COOH

HO₂C-CH₂-COOH + OH⁻ → HO₂C-CH₂-CO₂⁻ + H₂O