# What is pH of an aqueous solution that is 0.016*mol*L^-1 with respect to NaOH?

Apr 5, 2017

$p H = 12.2$

#### Explanation:

In aqueous solution, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and likewise, $p H = - {\log}_{10} \left[H {O}^{-}\right]$.

Now since we interrogate the equilibrium:

2H_2Orightleftharpoons H_3O^+ + ""^(-)OH where ${K}_{\text{water}} = {10}^{- 14}$, then..........

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$, and we take ${\log}_{10}$ of both sides......

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

On rearrangement,

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

And this is our defining relationship, $p H + p O H = 14$ (and this has to be known and used!)

So, we find the $p O H$ of a solution that is $0.016 \cdot m o l \cdot {L}^{-} 1$ with respect to $N a O H$:

$p O H \equiv - {\log}_{10} \left(0.016\right) = 1.80$

And finally,

$p H = 14 - p O H = 14 - 1.80 = 12.2$

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