Glycolic acid which is a monoprotic acid and a constituent in sugar has a pka of 3.9. A 25 ml solution of glycolic acid is tritrated to the stoichiometric point with 35.8 ml of 0.020 M sodium hydroxide solution. What is the ph of the resulting solution at the stoichiometric point?
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Expert's answer
2012-05-09T09:25:50-0400
pKa = -logKa Ka = 10^-3.9 = 1.26x10^-4 Ka = [H+][An-]/[HAn] = [H+]^2 /[HAn] "An" is HOCH2COO-
NaOH + HAn = NaAn + H2O
n of NaOH is: 0.020 mol is in 1000ml x mol is in 35.8 x = 0.000716 mol
so n of HAn is the same,and concentration is 0.000716 mol in 25ml,or 0.029 mol in 1000 ml or 0.029M
Now, acetate ions are a base and will react with water by the reaction:
An- + H2O --> HAn + OH-
Kb for this reaction can be calculated from Ka x Kb = Kw to be 7.94x10^-11
By that equation, [HAn] = [OH-], so Kb = x^2/0.029 You can solve this to show that [OH-] = 1,52x10^-6 M. pOH, then = 5.82 and pH = 14 - pOH = 8.18
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