Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘

c.

CN^-1 is the conjugate base of the weak acid HCN (Ka 3 X 10^-9) so CN^-1 is a strong base and will hydrolyze, by the reaction:

CN^-1 + H2O - - > HCN + OH^-1

The amount is based on the hydrolysis constant which is

Kw / KaHCN = 1 X 10^-14 / 3 X 10^-9

= 3.333 X 10 ^-6

0.2 M CN^-1 will ionize to a small extent (A) to produce A amounts of HCN and A amounts of OH^-1

soo Kh = 3.333 X 10^-6 = [A][A] / 0.2 - A

but A <<< 0.2 (much lesser than) so it will be omitted in the denominator

A^2 = 0.2 x 3.333 X 10^-6

A^2 = 6.67 X 10^-7

A = 8.17 X 10^-4

pOH = - log (8.17 X 10^-4) = 3.09

then ph is

pH = 14 - pOH

so

pH = 14 - 3.09

pH = 10.91