# Question #4d832

Mar 16, 2017

$p O H = 11.2$

#### Explanation:

We interrogate the equilibrium:

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

Now this equilibrium has been carefully measured under standard conditions, and at $298 \cdot K$:

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

This is a mathematical relationship, we can divide, multiply or otherwise manipulate. One thing we can do is to take $- {\log}_{10}$ OF BOTH SIDES,

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

But the RHS, $- {\log}_{10} {10}^{-} 14 = - \left(- 14\right) = 14$, by the very definition of the $\text{logarithmic function}$. And thus,

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

And, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$. This is simply by definition of the $p$ function, and thus,

$+ 14 = p H + p O H$

And thus we are given that $\left[{H}_{3} {O}^{+}\right] = 1.55 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$, and so $p H = - \left(- 2.81\right) = 2.81 .$

And so $p O H = 11.19$

Capisce?