# Question #ed4f1

Aug 8, 2017

$p H = 11.2$

#### Explanation:

We use the relationship $\text{concentration}$ $=$ $\text{moles of solute"/"volume of solution}$

And thus $\text{moles of solute"="concentration"xx"volume}$

And thus ...................

$200 \cdot \times {10}^{-} 3 \cdot L \times 1 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1 = 2.00 \times {10}^{-} 4 \cdot m o l$ with respect to $N a O H$.

Likewise.....................

$300 \cdot \times {10}^{-} 3 \cdot L \times 2 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1 = 6.00 \times {10}^{-} 4 \cdot m o l$ with respect to $N a O H$.

And so $\left[N a O H\right] = \frac{\left(6.00 + 2.00\right) \times {10}^{-} 4 \cdot m o l}{\left(300 + 200\right) \times {10}^{-} 3 \cdot L}$

$= 1.60 \times {10}^{-} 3 \cdot m o l \cdot L$ with respect to $N a O H$; and of course $\left[N {a}^{+}\right] = \left[H {O}^{-}\right] = 1.60 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$.

Note that I have reasonably assumed that the volumes are additive.

Now it is a fact that $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$......

but given ${K}_{w} = {10}^{-} 14 = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$, we know that......

$p {K}_{w} = 14 = p H + p O H$

From the above.....$p O H = - {\log}_{10} \left(1.60 \times {10}^{-} 3\right) = 2.80$

And $p H = 14 - p O H = 14 - 2.80 = 11.2$.........

If you have followed my reasoning, you should be able to tackle most problems that request values of $p H$ and $p O H$..........