Question

If a 0.10 M Solution of KCN has a PH-11.60, what is the value of ka for HON?

I'm not sure if i need an ice rable for both the KCN and HCN? please solve step by step thanks:)

Answer 1

Given

pH = 11.6

[H+] = 10^-pH

= 10^-11.6

= 2.51 * 10^-12M

So

[OH-] = 10^-14/[H+]

= 10^-14/(2.51 * 10^-12M)

= 3.98 * 10^-3 M

Now

---------------------------KCN(aq) + H2O(l) ==> HCN(aq) + KOH(aq)

Initial --------------------0.1-------------------------0---------------0

Change-------------------(-x)------------------------( +x)---------- ( +x)

Equilibrium------------(0.1-x)-----------------------(x)-----------(x)

Kb = [HCN][KOH]/[KCN]

0.1-x approx to 0.1

Here x = [OH-] =3.98 * 10^-3 M

Kb = (3.98 * 10^-3 )( 3.98 * 10^-3 ) / (0.1)

Kb = 1.58 * 10^-4

So

Ka =10^-14/kb

=10^-14/1.58 * 10^-4

= 6.3 * 10^-11

Ka of HCN = 6.3 * 10^-11