Calculate the pH of a 0.021 M NaCN solution. ( Ka(HCN)=4.9*10 to the -10. [Solved]

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NaCN is a strong salt : NaCN => Na CN-

CN- H2O <-----> HCN OH-
the constant of this equilibrium is Kb = Kw/Ka
Kb = 1.0 x 10^-14 / 4.9 x 10^-10=2.0 x 10^-5 = x^2/ 0.021-x

x = [OH-]=0.00065 M
pOH =3.2
pH = 14 3.2 =10.8

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What is the ph of 0.25M of NaCN

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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