What is pH in the following scenario?

$187.5 \cdot m L$ of $0.60 \cdot m o l \cdot {L}^{-} 1$ $K O H$ is reacted with $250.0 \cdot m L$ of $0.30 \cdot m o l \cdot {L}^{-} 1$ $H N {O}_{3}$.

Aug 4, 2017

Answer:

$p H \cong 13$

Explanation:

We assess the extent of the following reaction......

$K O H \left(a q\right) + H N {O}_{3} \left(a q\right) \rightarrow {H}_{2} O \left(l\right) + K N {O}_{3} \left(a q\right)$....

There is thus 1:1 stoichiometry, but unequal molar quantities of $K O H$ and $H N {O}_{3}$.

$\text{Moles of KOH} = 187.5 \times {10}^{-} 3 \cdot L \times 0.60 \cdot m o l \cdot {L}^{-} 1 = 0.1125 \cdot m o l .$

$\text{Moles of nitric acid} = 250.0 \times {10}^{-} 3 \cdot L \times 0.30 \cdot m o l \cdot {L}^{-} 1 = 0.0750 \cdot m o l$.

And thus with respect to $K O H$ there is a concentration of

$\frac{0.1125 \cdot m o l - 0.0750 \cdot m o l}{437.5 \times {10}^{-} 3 \cdot L} \equiv 0.0857 \cdot m o l \cdot {L}^{-} 1$.

Note that we assume (very reasonably) that the volumes are additive.......and really we must assume this.

And since $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, and since $p O H = - {\log}_{10} \left(0.0857\right) = 1.07$, and further since $p H + p O H = 14$, $p H = 12.93$.

Capisce?