Calculate the pH of a solution that contains 1.00 M HF `(K_(a)=7.2xx10^(-4))` and 5.00 M HClO `(K_(a)=3.5xx10^(-8))`.

Explanation

As `K_(a)` for `HClO lt lt K_(a)` for HF, `[H_(3)O^(+)]` obtained from HClO can be neglected in comparison to `[H_(3)O^(+)]` obtained from HF. Hence
`[H_(3)O^(+)]=sqrt(K_(a) (HF)xxC(HF))=sqrt((7.2xx10^(-4))(1.0))=2.68xx10^(-2)M`
`pH = - log (2.68 xx 10^(-2))=2-0.4265 = 1.57` (approx)