What is the pH of the resulting solution?

Question

Glycolic acid, which is a monoprotic acid, has a pKa of 3.9. A 25.0 mL solution of glycolic acid is titrated to the stoichiometric point with 35.8 mL of .020 M sodium hydroxide solution. What is the pH of the resulting solution at the stoichiometric point?

Please help!

Answer 1

What is the stoichiometric point anyway? It's the point at which the solution behaves exactly like an equivalent molar solution of the sodium glycolate salt. This is important, since the equilibrium starts there. The reaction is:
GlyO- + H2O = GlyOH + OH- .
This is "logical" inasmuch as sodium salts of weak acids give basic solutions. The stoichiometry of the titration tells us that we have 0.00358 L x 0.2 M
or 7.16x10-4 moles of sodium added, which is equivalent to the moles of glycolate. This is in 60.8 mL of solution, which is 7.16x10-4 moles / 60.8x10-3 L = 0.17M appx in glycolate ion. The above equilibrium yields
[GlyOH][OH-]/ [GlyO-] = Kb
where Kb is a BASIC equilibrium constant we will work on later. From the equilibrium, for each (X) moles/L of glycolate hydrolyzed, x moles each of the acid and hydroxide are formed. So the math becomes
(X)(X)/[0.17-X]= Kb
Kb is expressable in terms of Ka and Kw as:
Kw/Ka, so now we have all the math we need.
Ka = 10^(-pKa) = 1.1x10-4 appx.
So.....
(X)(X)/[0.17-X]= 10-14/1.1x10-4 = 9x10-11
we see that X is much smaller than 0.17, so we can drop it out of that part of the expression.
Then (X)(X) = 1.5x10-11 or X= 3.8x10-6.
we have the OH- conc, so we can compute pOH, and then pH. pOH= 5.4 and pH= 8.6.