Question

The pH of a 100.0 mL buffer solution of 1.25M HCN and 1.25M NaCN is 9.21. Because the concentrations of the HCN and NaCN are equal, pH = pKa. Calculate the pH after 10.0 mL of 0.200M KOH have been added.

Answer 1

Moles of HCN = M x V ( in L) = 1.25 x 0.1 = 0.125

Moles of NaCN = 1.25 x 0.1 = 0.125 = moles of CN-

KOH moles added = 0.2 x 0.01 = 0.002 = moles of OH-

now OH- reacts with HCN to convert it to CN-

HCN (aq) + OH- (aq) <---> CN- (aq) + H2O (l)

hence after adding 0.002 moles of OH- , HCN moles = 0.125-0.002 = 0.123

CN- moles = 0.125+0.002 = 0.127

now pH = pka + log ( CN- moles / HCN moles)

       = 9.21 + log ( 0.127/0.123)

        = 9.224