# A 5*g mass of sodium hydroxide is dissolved in a volume of water such that the final solution is 1*L...what are the pOH, and pH of this solution, and what is the final concentration of H_3O^+?

Apr 4, 2017

$p H = 13.1$, $\left[{H}_{3} {O}^{+}\right] = {10}^{- 13.1} \cdot m o l \cdot {L}^{-} 1 =$

$\left[{H}_{3} {O}^{+}\right] = 7.94 \times {10}^{-} 14 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We interrogate the equilibrium:

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

(Note that we may treat ${H}_{3} {O}^{+}$ and ${H}^{+}$ equivalently; this really is a matter of personal preference.)

We know, that this autoprotolysis reaction follows the given equilibrium reaction under standard conditions:

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

K_w=[H_3O^+][""^(-)OH]=10^(-14)

And if we take ${\log}_{10}$ of BOTH sides:

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

EQUIVALENTLY, if we multiply each side by $- 1$,

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

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and..........

$- {\log}_{10} \left[H {O}^{-}\right] = p O H$, and so............

$14 = p H + p O H$, under standard conditions, and this is our defining relationship.

And so, (FINALLY after all that background):

$\left[H {O}^{-}\right] = \frac{5 \cdot g}{40 \cdot g \cdot m o {l}^{-} 1} = 0.125 \cdot m o l$.

And $- {\log}_{10} \left(0.125\right) = 0.90 = p O H$.

And thus $p H = 14 - p O H = 13.1$

$\left[{H}_{3} {O}^{+}\right] = 7.94 \times {10}^{-} 14 \cdot m o l \cdot {L}^{-} 1$