# Question #31b0a

Mar 22, 2017

Well, this is a basic solution; because it contains MORE of the acidium ion than the hydroxide ion.........

#### Explanation:

We interrogate the equilibrium:

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

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

Taking ${\log}_{10}$ of both sides:

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

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

i.e. $14 = p H + p O H$.

$\text{Concentration of sodium hydroxide}$

$= \left(\text{Moles of sodium hydroxide")/("Volume of solution}\right)$

$=$ $\frac{\frac{0.800 \cdot g}{39.998 \cdot g \cdot m o {l}^{-} 1}}{2.00 \cdot L} = 1.00 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

Now, clearly, we can take the $p H$ of this solution, given that we know that $p H + p O H = 14$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$
$= - {\log}_{10} \left(1.00 \times {10}^{-} 2\right) = 2.0$

Thus $p H = 14 - 2 = 12$......clearly an alkaline solution,