# What is the pH of a 6.2 x 10-2 mol/L solution of nitric acid?

Jun 5, 2017

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = 1.21$

#### Explanation:

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

We know that nitric acid is a moderately strong acid for which protonolysis of the water solvent is relatively complete, i.e. the given equilibrium lies to the right...........

$H N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + N {O}_{3}^{-}$

That is the solution is stoichiometric in ${H}_{3} {O}^{+}$.

And we simply take ${\log}_{10}$............

$- {\log}_{10} \left(6.2 \times {10}^{-} 2\right) = - \left(- 1.207\right) = 1.21$.

For more details, see this old answer.