What is the pH of a solution obtained by mixing 20.0mL of .200M HCN(aq) and 20mL of .250M NaOH(aq)?

Question

What is the pH of a solution obtained by mixing 20.0mL of .200M HCN(aq) and 20mL of .250M NaOH(aq)?

Ka for HCN is 4.5*10^-10

Answer 1

Moles HCN = 0.0200 L x 0.200 M =  0.00400

Moles NaOH = 0.0200 L x 0.250 M = 0.00500

HCN + OH- >> CN- + H2O

The solution is essentially all Na+ and CN- , with trace of HCN and OH- due to hydrolysis.

There is so little OH- present compared to what we are adding that we can forget about it and simply regards this problem as the addition of

0.00500 - 0.00400 = 0.00100 mole OH- to a volume of 20.0 + 20.0 = 40.0 mL = 0.0400 L

[OH-] = 0.00100 / 0.0400 =0.0250 M

pOH = - log 0.0250 = 1.60

pH = 14 - 1.60 = 12.4