# 3.00 * 10^-3 mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [OH^-] in this solution?

Jul 16, 2017

$\left[H {O}^{-}\right] = 5.33 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We know that ${10}^{-} 14 = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right]$ under standard conditions at $298 \cdot K$......... And should we take ${\log}_{10}$ of BOTH SIDES, we gets......

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

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

And so our defining relationship, $p H + p O H = 14$

Now.............

$\left[{H}_{3} {O}^{+}\right] = \frac{3.00 \times {10}^{-} 3 \cdot m o l}{16 \cdot L} = 1.875 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$

$p H = - {\log}_{10} \left(1.875 \times {10}^{-} 4\right) = - \left(- 3.727\right) = 3.727$

And thus $p O H = 10.27$, and......................

$\left[H {O}^{-}\right] = {10}^{- 10.27} = 5.33 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$