What is the pH of a 0.32 M solution of HC6H6O6- ? a. 4.10b. 11.80c. 7.95d. 3.85

Answer 1

If, in fact, HC₆H₆O₆^- is the ascorbate anion, we might look at the hydrolysis where it will act as a base with H₂O being the acid.

HC₆H₆O₆^- + H₂O ==> H₂C₆H₆O₆ + OH^-

Kb = [H₂C₆H₆O₆][OH^-] / [HC₆H₆O₆^-]

Since Kb wasn't provided, we can calculate it from the tabular value of Ka(II) for ascorbate anion, which is 1.6x10^-12. Thus Kb = 1x10^-14/1.6x10^-12 = 6.3x10^-3

(NOTE: Since we are taking the ascorbate anion to act as a base, the most likely answer will be 11.80).

6.3x10^-3 = (x)(x) / 0.32-x (we will ignore x in denominator to avoid using the quadratic)

6.3x10^-3 = x² / 0.32

x² = 2.0x10³

x = [OH-] = 0.045 M

pOH = -log 0.045 = 1.35

pH = 12.7 Close enough to 11.8 since didn't use the quadratic and used one of several possible values for Kb

Answer 2

A pK_a (or K_a) is need to solve this. Based on the formula HC₆H₆O₆^–, I'm assuming that this is ascorbate, formed by loss of a proton from ascorbic acid (vitamin C). Using pK_a = 11.8 for the second deprotonation of ascorbic acid leads to

x²/(0.32 – x) = 10^–11.8 as the equation to solve, where x = [H+], and pH = –log [H+] = –log x.

Solving for x, then calculating –log x leads to a pH value a little bit greater that 6. This makes sense because ascorbate is such a weak acid.

Even before calculating anything, its clear that none of the multiple choice answers are possible for an acid with a pKa of 11.

1. choices b and c are greater than 7 which cannot be the result of dissolving an acid.
2. choices a and d are correctly below 7, but the values are too low for an acid with such a high pK_a.

(It's an odd coincidence that the value for choice b is the same as the pKa of the acid ascorbate.)