# If the pH of a solution is 10.7, what would the pOH be?

Oct 6, 2016

3.3

#### Explanation:

Remember a simple thing that pH+pOH=14
so pOH= 14-10.7 = 3.3

Oct 6, 2016

$p O H = 3.3$

#### Explanation:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

At $298 \cdot K$, we can write and quantify this equilibrium reaction:

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$

And taking ${\log}_{10}$ of both sides:

$\log {K}_{w} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = {\log}_{10} {10}^{-} 14$

OR $14 = - \log {K}_{w} = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right]$

But, by definitions, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, and $- \log {K}_{w} = p {K}_{w}$.

And thus, $p {K}_{w} = 14 = p H + p O H$.

Given your question (finally we've got to it!), $p O H = 14 - p H$ $=$ 14-10.7=?.