ok. im new with this forum so thanks for that advice
a) I believe the equation is HF + NaF --> HF + NaF. i calculated that there is 0.06 mol of HF initially and 0.01 mol of NaF initially. so, 0.06-0.01 = 0.05 mol H+ in excess. the total volume is 40mL + 20 mL = 60 mL. so, 0.05/0.060 = 0.83 M = [H+]. however, im not sure this is correct.
b) HA+ OH- --> A-. i set up an equilibrium table to solve for .125 M HC2H3O2 initially and .375 M NaOH initially.
then Ka= [H+][A-]/[HA]. 1.8x10^-5 = x(.375+x)/(.125-x). x=[H+]=5.996x10^-6. pH= 5.22
final answer: [H+]=6.00X10^-6, [C2H3O2-]=.375, pH = 5.22
again, im not sure if i am correct.