Ascorbic acid (vitamin C) is a diprotic acid, . What is the pH of a solution? The acid ionization co — Ionic Equilibrium Chemistry Question
Question
Ascorbic acid (vitamin C) is a diprotic acid, $\text{H}_2\text{C}_6\text{H}_6\text{O}_6$. What is the pH of a $0.10\text{ M}$ solution? The acid ionization constants are $K_{a1} = 9.0 \times 10^{-5}$ and $K_{a2} = 1.6 \times 10^{-12}$. ($\log 2 = 0.3, \log 3 = 0.48$)
Answer: B
💡 Solution & Explanation
Because $K_{a1} \gg K_{a2}$, the first dissociation dictates the pH. $[\text{H}^+] \approx \sqrt{K_{a1} C} = \sqrt{9.0 \times 10^{-5} \times 0.10} = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3}\text{ M}$. $\text{pH} = -\log(3.0 \times 10^{-3}) = 3 - \log 3 = 3 - 0.48 = 2.52$.
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