Question

Calculate pH of a solution by mixing 15.0 mL of .50 M NaOH and 30.0 mL of .50 benzoic acid solution (benzoic acid is monoprotic, its dissociation constant is 6.5 x 10^ -5).

I am not sure whether I am supposed to do it by acid-base titration way or to use the formula -log (Ka)+ log [acid/base]

Because both give me totally different answers. So i wasn't sure which way I am supposed to do this.

Answer 1

Let HA = benzoic acid
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moles NaOH(initial) = (0.015L)(0.50 mol/L) = 0.0075 mole
moles HA(initial) = (0.0300L)(0.50 mol/L) = 0.015 mol
OH^- + HA --> H2O + A-
moles of A- formed = 0.0075 (same as initial OH-)
moles HA remaining = 0.0150- 0.0075 =0.0075 mol
[A-] = moles/liters = 0.0075mol/(0.045L) = 0.167 M at equilibrium
[HA] = 0.167 M (same math as above) at equilibrium
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In the end, you have a solution with equal concentrations of weak acid, HA, and weak base, A-. Use the Henderson-Hasselbalch equation to get your pH:
pH = pKa + log(base/acid)