Ionic EquilibriumMedium
Question
Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? The acid ionization constants are Ka1 = 9.0 × 10–5 and Ka2 = 1.6 × 10–12. (log 2 = 0.3, log 3 = 0.48).
Options
A.3.52
B.2.52
C.1.52
D.2.48
Solution
For. PH, 2nd dissociation may be neglected.
$\left\lbrack H^{+} \right\rbrack = \sqrt{K_{a_{1}}.C} = \sqrt{9 \times 10^{- 5} \times 0.1} = 3 \times 10^{- 3}\text{ M}$
∴ PH = –log(3 × 10–3) = 2.52
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