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

There are no metaphors.