Question

The reaction of the weak acid HCNHCN with the strong base KOHKOH is

HCN(aq)+KOH(aq)⟶HOH(l)+KCN(aq)HCN(aq)+KOH(aq)⟶HOH(l)+KCN(aq)

To compute the pH of the resulting solution if 61 mL61 mL of 0.49 M0.49 M HCNHCN is mixed with 4.0×101 mL4.0×101 mL of 0.32 KOH0.32 KOH we need to start with the stoichiometry. Let's do just the stoich in steps:

What are the moles of the pH active product after reaction? (remember the spectator ion is not pH active)

Answer 1

moles of HCN = 61 x 0.49 1000 = 0.02989

moles of KOH = 40 x 0.32 / 1000 = 0.0128

HCN (aq) +   KOH (aq)   -------------> KCN (aq) + H2O (l)

0.02989        0.0128                               0                   0

0.01709            0                                   0.0128

moles of CN- = 0.0128

moles of pH active product = 1.28 x 10^-2 mol