A $58\mathrm{.}7$ mL sample of a $0\mathrm{.}320$ M solution of NaCN is titrated by $0\mathrm{.}400$ M HCl$.$ Kb for CN$-$ is $2\mathrm{.}0$ x $10^-5.$ Calculate the pH of the solution:

a$.$ prior to the start of the titration

pH $=$

b$.$ after the addition of $32\mathrm{.}9$ mL of $0\mathrm{.}400$ M

pH $=$

c$.$ at the equivalence point

pH $=$

d$.$ after the addition of $56\mathrm{.}4$ mL of $0\mathrm{.}400$ M

pH $=$