# The pH of a solution is 8.7, what is the pOH?

Jul 5, 2016

pOH = 5.3

#### Explanation:

You can answer this question in one of two ways:

• Take the anti-log of the pH to obtain the concentration of ${H}^{+}$ ions in solution. After that, use the self-ionization of water formula:

Where ${K}_{w}$ has a value of $1.0 \times {10}^{-} 14$ Then you can rearrange the equation to solve for [$O {H}^{-}$]. Take the -log of that value to obtain the pOH.

• Subtract the pH from 14 to obtain the pOH.

I'll show you both ways using these equations:

THE PROCESS FOR METHOD 1:

$\left[{H}^{+}\right] = {10}^{- 8.7} = 1.995 \times {10}^{-} 9 M$
${K}_{w} = \left[{\text{H"_3"O"^(+)]["OH}}^{-}\right] = 1.0 \times {10}^{- 14}$

Kw / [H+] = [OH-]

(1.0xx10^(-14))/(1.995xx10^(-9) "M") = 5.01xx10^(-6)"
$\left[O {H}^{-}\right] = 5.01 \times {10}^{- 6} M$
$p O H = - \log \left(5.01 \times {10}^{- 6}\right) M$
$p O H = 5.3$

THE PROCESS FOR METHOD 2:

$14 - 8.7 = 5.3$