Question

What is the `pH` of a solution of `"sulfuric acid"` of `2xx10^-2*mol*L^-1` concentration?

Answer 1

`pH=1.70...........`

Explanation:

By definition, `pH=-log_10[H_3O^+]`

Now for `H_2SO_4` with a concentration of `10^-2*mol*L^-1`, clearly, `[H_3O^+]=2xx10^-2*mol*L^-1`.

That is to a first approx. `H_2SO_4` is considered to be a diacid in aqueous solution and gives 2 equiv of `H_3O^+`. (In the absence of a value for `pK_(a2)` I am perfectly justified in making this assumption!)

And so `pH=-log_10(2xx10^-2)=-log_10(0.02)=1.70`