Question

What is the ratio of pH of 0.1M H2SO4 and 0.1N H2SO4?

Answer 1

Well...`pH=-log_10[H_3O^+]`

Explanation:

And sulfuric acid is diprotic, and we could represent its reaction with water by the equation....

`H_2SO_4(l) + 2H_2O(l) rarr2H_3O^+ +SO_4^(2-)`

And so `0.1*mol*L^-1` is `0.2*mol*L^-1` with respect to `H_3O^+`.

And thus `pH=-log_10(0.2)=-(-0.699)=0.699`.

And on the other hand, `0.1*N` sulfuric acid is in fact `0.05*mol*L^-1` with respect to sulfuric acid....the acid STILL requires TWO equivs of base for stoichiometric equivalence. Capisce?

And so here...`pH=-log_10(0.1)=1.0`