# How many moles of sodium hypobromite (NaBrO) must be added to a 2.00 L solution of 0.025 M hypobromous acid (HBrO) to form a buffer solution of pH 9.0 (Ka HBrO is 2.5x10^-9)? Assume the volume of solution does not change on addition of NaBrO?

## a) How many moles of sodium hypobromite (NaBrO) must be added to a 2.00 L solution of 0.025 M hypobromous acid (HBrO) to form a buffer solution of pH 9.0 (Ka HBrO is 2.5x10^-9)? Assume the volume of solution does not change on addition of NaBrO? b) If you started with a solution of 0.500 M HBrO and made a pH 9.0 buffer, would the buffer capacity of this system be larger, smaller or the same as the one in part A. Explain in no more than 3 sentences.

Sep 21, 2016

(a)

0.125 mol

(b)

Larger. See below

#### Explanation:

(a)

Hypobromous acid is a weak acid and dissociates:

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

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[O B {r}^{-}\right]}{\left[H O B r\right]} = 2.5 \times {10}^{- 9} \textcolor{w h i t e}{x} \text{mol/l"" } \textcolor{red}{\left(1\right)}}$

Please note that these refer to equilibrium concentrations and not initial concentrations.

The initial number of moles of $\textsf{H O B r}$ is given by:

$\textsf{{n}_{H O B r} = c \times v = 0.025 \times 2 = 0.050}$

Since $\textsf{p H = 9}$ then $\textsf{\left[{H}^{+}\right] = {10}^{- p H} = {10}^{- 9} \textcolor{w h i t e}{x} \text{mol/l}}$

Rearranging sf(color(red)((1)) gives:

sf([H^(+)]=K_axx([HBrO])/([BrO^(-)])" "color(red)((2))

At this point I will make the important assumption that, because the value of $\textsf{{K}_{a}}$ is so small, then the equilibrium concentrations are a good approximation to the initial concentrations.

The fact that you are asked to assume that the volume change is negligible is actually irrelevant. Since $\textsf{\left[\textcolor{w h i t e}{x}\right] = \frac{n}{v}}$ you can see that the volume is common to both acid and co - base so will cancel in sf(color(red)((2)).

This means we can write out sf(color(red)((2)) using molessf(rArr

$\therefore$sf(10^(-9)=2.5xx10^(-9)xx(0.050)/(nBrO^(-))

$\therefore$$\textsf{n B r {O}^{-} = \frac{2.5 \times \cancel{{10}^{- 9}} \times 0.050}{\cancel{{10}^{- 9}}} = 0.125}$

(b)

From sf(color(red)((2)) you can see that the pH of the buffer depends on the ratio of acid to co - base concentrations, whereas the amount of added $\textsf{{H}^{+}}$ or sf(OH^- which the buffer can cope with will depend on the individual concentrations of acid and co - base.

This is referred to as the buffer capacity.

In (b) to maintain a pH of 9 the concentration of $\textsf{H O B r}$ and of $\textsf{O B {r}^{-}}$ must have both been increased, so we would expect a greater buffer capacity.