Question

What is the `pH` of a solution whose concentration is `0.450*mol*L^-1` in ammonia; `K_b=1.80xx10^-5`?

Answer 1

`pH=11.5`

Explanation:

We interrogate the equilibrium:

`NH_3(aq) + H_2O(l) rightleftharpoons NH_4^(+) + HO^-`.

We set up the equilibrium in the usual way:

`K_b=1.80xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])`

And we call the concentration of ammonia that hydrolyzes `x`.

And thus: `K_b=1.80xx10^-5=x^2/(0.450-x)`

If, we assume that `0.450>x`, then `x_1=sqrt(1.80xx10^-5xx0.450)`

`=2.85xx10^-3`, which is indeed small compared to `0.450`, but since we have an estimate for `x_1`, we can put it thru the washing again, and come up with:

`x_2=sqrt(1.80xx10^-5xx(0.450-2.85xx10^-3))=2.84xx10^-3`

`x_3=sqrt(1.80xx10^-5xx(0.450-2.84xx10^-3))=2.84xx10^-3`

Clearly I am making a meal of the successive approximations, but I think you can get the gist of the operation. Approximate, then substitute, then resubstitute, till you reach a consistent answer. Alternatively, you could have used the quadratic equation, but here there are more terms to include, and more mistakes (potentially) to make.

And thus at equilibrium,

`[NH_4^+]=[HO^-]=2.84xx10^-3*mol*L^-1`

`pOH=-log_10[HO^-]=-log_10(2.84xx10^-3)=2.55`

And, since in aqueous solution, `pH+pOH=14`, `pH=11.45`.

If I were you, I would apply this general method to other `pH` problems. You can get very efficient at their solutions, and you must practise how to manipulate your calculator.