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## Answers (1)

Ka for HCN = 7.2x10^-10

Kb = Kw/Ka = 1x10^-14 / 7.2x10^-10 = 1.39x10^-5

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

Kb = [HCN][OH-] / [CN-]

[HCN]=[OH-] = X

Kb = 1.39x10^-5 = X^2 / (10^-4 - X) = X^2 / 10^-4 (if X << 10^-4)

1.39x10^-5 x10^-4 = X^2 = 1.39x10^-9

X = 3.7x10^-5 and pOH = 4.4

pH = 14 - pOH = 9.6

NOTE: the approximation that X-10^-4 = X is not very good.

You could solve the following equation using the quadratic formula, or you could substitute the

estimated value of X into the following equation and recalculate the value of X-

Kb = 1.39x10^-5 = X^2 / (10^-4 - X) = X^2 / 10^-4

1.39x10^-5 = X^2 / (10^-4 - 3.7x10^-5)

1.39x10^-5 = X^2 / 6.3x10^-5

X = 3.0x10^-5

pOH = 4.52 and pH = 9.5

So 9.5 is a better estimate of the pH

Further iterations would not cause a meaningful change in the pH.