# The pH of a solution changes fro 300 to 6.00. By what factor does the [H_3O^+] change?

Feb 26, 2017

By a factor of $1000$.

#### Explanation:

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

If $p {H}_{1} = 3.00$, then $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1.$

If $p {H}_{2} = 6.00$, then $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1.$

$\frac{p {H}_{1}}{p {H}_{2}} = \frac{{10}^{-} 3 \cdot m o l \cdot {L}^{-} 1}{{10}^{-} 6 \cdot m o l \cdot {L}^{-} 1} = {10}^{3}$

The latter concentration is less concentrated with respect to ${H}_{3} {O}^{+}$ than the former by a factor of $1000$.