Question

What is `pH` if `[H_3O^+]=0.001*mol*L^-1`?

Answer 1

`pH=3............`

Explanation:

`pH`, `"pouvoir hydrogene"`, `=-log_10[H_3O^+]`

Now it is a fact that in water the following autoprotolysis takes place:

`2H_2O(l) rightleftharpoonsHO^(-) + H_3O^+`

The ionic product at `298*K` `=` `K_w=[H_3O^+][HO^-]=10^-14`.

If we take `-log_10` of both sides we get:

`pK_w =14=-log_10[H_3O^+]-log_10[HO^-]`

But by definition, `pH =-log_10[H_3O^+]` and `pOH =-log_10[HO^-]`

Thus `pK_w =14=pH+pOH`

But nitric acid should be stoichiometric in `H_3O^+` and `NO_3^-` ion...........

`HNO_3(aq) + H_2O(l) rarr H_3O^+ + NO_3^-`

And so `pH` `=` `-log_10(10^-3)` `=` `3`, and `pOH=??`

And `pH=3`.

Capisce?