The moles of the reagents are calculated:
n KOH = M * V = 2,657 M * 0.3678 L = 0.9772 mol
n AA = 1.77 M * 0.231 L = 0.4089 mol
The ratio mol OH- / mol AA is calculated:
Ratio = 0.9772 / 0.4089 = 2.39
Since it is greater than 2, the KOH neutralizes all the Acid (both protons), the remaining OH-moles and their concentration are calculated:
n OH-rem = 0.9772 - (2 * 0.4089) = 0.1594 mol
[OH-] = n / V = 0.1594 / 0.5988 = 0.2662 M
POH and pH are calculated:
pOH = - log 0.2662 = 0.57
pH = 14 - 0.57 = 13.43
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