# If the H ion concentration is 0.00075 M, what is the OH ion concentration?

Mar 2, 2017

#### Answer:

$\left[H {O}^{-}\right] = {10}^{- 10.88} = 1.33 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$.

We assume an aqueous solution under standard conditions.

#### Explanation:

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

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

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

But by definition, when I say ${\log}_{a} b = c$, I ask for the power to which I raise the base $a$ to get $b$. So since ${\log}_{a} b = c$, then ${a}^{c} = b$.

And likewise, ${\log}_{10} \left[{10}^{-} 14\right] = - 14$, because clearly the exponent of ${10}^{-} 14 = - 14$.

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

OR,

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

i.e. $14 = p H + p O H$, because that is how we define $p H$ etc., i.e. $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$.

We are given that $\left[{H}_{3} {O}^{+}\right] = 0.00075 \cdot m o l \cdot {L}^{-} 1$

And $p H = - {\log}_{10} \left(0.00075\right) = - \left(- 3.13\right) = 3.13$

So $p O H = 10.88$.

And $\left[H {O}^{-}\right] = {10}^{- 10.88} = 1.33 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$.