# Question #8a714

Dec 22, 2016

$p H = 12$

#### Explanation:

$\left[H {O}^{-}\right] = \frac{4 \cdot g}{39.99 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{10 \cdot L} = 0.01 \cdot m o l \cdot {L}^{-} 1$.

Now $p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(0.01\right) = 2.0$

But $p H + p O H = 14$, and thus $p H = 12$

That $p H + p O H = 14$ is key, and of course it stems from the autoprotolysis equilibrium, $\left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right] = {10}^{-} 14$.