Question

What is the `pH` of a `0.055*mol*L^-1` solution of sulfuric acid in water?

Answer 1

Assuming quantitative dissociation, `pH=0.96`

Explanation:

We assume (reasonably!) that dissociation is quantitative:

`H_2SO_4(aq) + 2H_2O rarr SO_4^(2-) + 2H_3O^+`

Thus `[H_3O^+]` `=` `0.110*mol*L^-1`.

Now `pH` `=` `-log_10[H_3O^+]` `=` `-log_10(0.110)` `=` `0.96`