Calculate the pH at the equivalence point for the following

Calculate the pH at the equivalence point for the following titration: 0.10 M HCN versus 0.10 MNaOH. The Kb of CN– = 2.0 x 10–5. Please show work!

Solution

at equivalence point all HCN changes to NaCN

[NaCN] = 0.10 /2 = 0.05 M

pH = 1/2 [pKw + pKa + logC]

Kb = 2.0 x 10^-5

pKb = - log Kb = - log [2.0 x 10^-5] = 4.70

pKa = 14 - 4.70 = 9.3

pH = 1/2 [14 + 9.3 + log 0.05]

pH = 11.0