Question

1) The Ka for acetic acid is 1.8e-5. What is the pH of a 3.18M solution of this acid?

I did 1.8e-5 = x^2/3.18

x=sqrt 1.8e-5 X 3.18 = 7.56e-3

pH=-log(7.56e-3) = 2.12

The pH is 2.12

2) At 25 degrees celsius, the pH of a 1.75M NaCN solution (the Ka for HCN is 4.0e-10) is ?

4.0e-10 = x^2/1.75

x=sqrt 4.0e-10 X 1.75 = 2.64e-5

pH=-log(2.64e-5) = 4.58

Did I solve these problems correctly? Thank you!
1 is right.
2 is not.
The pH of salts is determined by the hydrolysis of the salt.
CN^- + HOH ==> HCN + OH^-
Kb for CN^- is (Kw/Ka for HCN)
is Kw 1.0e-14?
yes
so kb = 1.0e-14 / 4.0e-10 = 2.5e-5
is 2.5e-5 my answer or do I have to do another step?

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