# Question 109c2

Mar 25, 2017

$\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right) = 1.3$

#### Explanation:

Start by calculating the $\text{p} {K}_{a}$ of hypobromous acid, $\text{HBrO}$, by using the fact that

$\text{p} {K}_{a} = - \log \left({K}_{a}\right)$

You will have

$\text{p} {K}_{a} = - \log \left(2.3 \cdot {10}^{- 9}\right) = 8.64$

Now, notice that the pH of the buffer is slightly higher than the $\text{p} {K}_{a}$ of the weak acid. This should tell you that the buffer contains more conjugate base, which in this case is the hypobromite anion, ${\text{BrO}}^{-}$, than weak acid.

In other words, you should expect to find

$\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right) > 1$

The pH of a buffer that contains a weak acid and its conjugate base can be calculated by using the Henderson - Hasselbalch equation

color(blue)(ul(color(black)("pH" = - "p"K_a + log( (["conjugate base"])/(["weak acid"])))))

"pH" = "p"K_a + log((["BrO"^(-)])/(["HBrO"]))#

which is

$8.75 = 8.64 + \log \left(\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right)\right)$

Rearrange the equation as

$\log \left(\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right)\right) = 8.75 - 8.64$

This will be equivalent to

${10}^{\log} \left(\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right)\right) = {10}^{0.11}$

which gets you

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right) = 1.3}}}$

The answer is rounded to two sig figs, the number of decimal places you have for the pH of the solution.

As predicted, the buffer contains more conjugate base than weak acid.