# What is the [H3O+] of a solution with the PH of 4.98?

Apr 23, 2018

$\left[{H}_{3} {O}^{+}\right] = 1.05 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$...

#### Explanation:

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

And given that if ${\log}_{a} y = z$ if follows that ${a}^{z} = y$

And so if $p H = 4.98$, [H_3O^+]=10^(-4.98)*mol*L^-1=??*mol*L^-1..

Apr 23, 2018

${10}^{-} 4.98 = 1.05 \times {10}^{-} 5 m o l {\mathrm{dm}}^{-} 3$

#### Explanation:

As

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

Hence, using log laws:

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

Therefore if $p H = 4.98$

Then

${10}^{- 4.98} = \left[{H}_{3} {O}^{+}\right] = 1.05 \times {10}^{- 5} m o l {\mathrm{dm}}^{-} 3$