Ka = 4.9 x 10^-10
NaCN ----> Na+ + CN-
0.65 0 0 (Initial)
0.65 - y y y ( Final)
CN- + H2O <-----> HCN + OH-
Kb = Kw/Ka
Kb = 1 x 10^-14 / 4.9 x 10^-10 = 2 x 10^-5
Kb = [Na+] [CN-] / [NaCN]
2 x 10^-5 = y^2 / ( 0.65 - y)
y^2 - 1.3 x 10^-5 + 2 x 10^-5 y = 0
y = 0.0036 M
[OH-]=0.0036 M
pOH = - log(0.0036) = 2.44
pH = 14 - 2.44 =11.56
[HCN] = 0.0036 M
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