# What is the pH of the final solution...?

## A $25 \cdot m L$ volume of $H B r \left(a q\right)$ at $0.050 \cdot m o l \cdot {L}^{-} 1$ concentration is mixed with a $10 \cdot m L$ volume of $K O H \left(a q\right)$ at $0.020 \cdot m o l \cdot {L}^{-} 1$ concentration. What is the $p H$ of the final solution?

Mar 23, 2017

A stoichiometric equation is required:

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

We get (finally) $p H = 1.52$

#### Explanation:

And thus we need to find the amount of substance of both $\text{hydrogen bromide}$, $\text{hydrobromic acid}$, and $\text{potassium hydroxide}$.

We use the relationship, $\text{Concentration"="Moles of solute"/"Volume of solution}$, OR

$\text{Concentration"xx"Volume"="Moles of solute}$

$\text{Moles of HBr} = 25 \times {10}^{-} 3 \cancel{L} \times 0.050 \cdot m o l \cdot \cancel{{L}^{-} 1} = 1.25 \times {10}^{-} 3 \cdot m o l .$

$\text{Moles of KOH} = 10 \times {10}^{-} 3 \cancel{L} \times 0.020 \cdot m o l \cdot \cancel{{L}^{-} 1} = 0.200 \times {10}^{-} 3 \cdot m o l .$

Note that I converted the $m L$ volume to $L$ by using the relationship: $1 \cdot m L \equiv 1 \times {10}^{-} 3 \cdot L$

Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are $\frac{\left(1.25 \cdot m o l - 0.200 \cdot m o l\right) \times {10}^{-} 3}{35 \times {10}^{-} 3 L}$ $H B r$ remaining.

So $\left[H B r\right] = \frac{\left(1.25 \cdot m o l - 0.200 \cdot m o l\right) \times {10}^{-} 3}{35 \times {10}^{-} 3 L}$

$= \frac{1.05 \cdot m o l \times {10}^{-} 3}{35 \times {10}^{-} 3 L} = 0.030 \cdot m o l \cdot {L}^{-} 1$ with respect to $H B r$.

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(0.030\right) = 1.52$