Question

What is the `pH` of a `0.10*mol*L^-1` solution of ammonium ion....for which `pK_a=9.26`?

Answer 1

And the `pK_a` of ammonium ion is..............?

Explanation:

We interrogate the equilibrium,

`NH_4^+ + H_2O(l) rightleftharpoons NH_3 + H_3O^(+)`

For which, `K_a=([NH_3][H_3O^+])/([NH_4^+])=10^(-9.25)`

So `10^(-9.25)=x^2/(0.10-x)`, a quadratic in `x`, that we can solve by successive approximation, i.e. assuming that `0.1>x`:

Thus `x~=sqrt(10^(-9.25)xx0.10)`

`x_1=7.5xx10^-6`

`x_2=7.5xx10^-6`

Since `x=[H_3O^+]`, `pH=-log_10(7.5xx10^-6)=5.12`