Question

Given a solution of nitric acid whose `pH=4`, what are `[H_3O^+]`, and `[NO_3^(-)]`?

Answer 1

`[HNO_3]=0.0001*mol*L^-1`

Explanation:

`pH=-log_10[H_3O^+]`

We are given that `pH=4`

And thus `[H_3O^+]=0.0001*mol*L^-1` (since `log_(10)0.0001=log_(10)10^-4=-(-4)=4`.

In water, nitric acid would undergo complete dissociation, according to the following rxn:

`HNO_3(aq) + H_2O(l)rarrH_3O^+ + NO_3^-`

And thus the formal concentration of nitric acid is equivalent to `[H_3O^+]` and `[NO_3^-]`.