# When an aqueous solution at room temperature is analyzed, the [H^+] is found to be 2.0*10^-3 M. What is the [OH^-]?

Aug 13, 2016

$\left[H {O}^{-}\right]$ $=$ $5.01 \times {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

As you know water undergoes autoprotolysis:

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

At $298 \cdot K$ this equilibrium has been measured exhaustively and precisely:

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

If we take $- {\log}_{10}$ of both sides:

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

i.e. $14 = p H + p O H$, because $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$ by definition.

So $\left[{H}_{3} {O}^{+}\right]$ $=$ $2.0 \times {10}^{-} 3$, and $p H = - {\log}_{10} \left(2.0 \times {10}^{-} 3\right)$ $=$ $- \left(- 2.70\right) = 2.70$.

Thus $p O H = 14 - 2.70 = 11.30$. $\left[H {O}^{-}\right]$ $=$ ${10}^{- 11.30} \cdot m o l \cdot {L}^{-} 1 = 5.01 \times {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$.

You can check to see if I am right by multiplying $\left[{H}_{3} {O}^{+}\right]$ and $\left[H {O}^{-}\right]$. What should the product equal?