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)

Question

Prashant Kotian · Accepted Answer

Correct answer: B. Since Ka1 >> Ka2, [H+] = √(Ka1*C) = √(9.0*10^-5 * 0.10) = √(9.0*10^-6) = 3.0*10^-3 M. pH = 3 - log3 = 3 - 0.48 = 2.52.

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