# What "CH"_3"COO"^(-) concentration is required to prepare a buffer solution with a pH of 5.10 if ["CH"_3"COOH"] is 0.150 M? K_a of "CH"_3"COOH" is 1.8 xx 10^-5?

Jun 21, 2016

$\text{0.344 M}$

#### Explanation:

This is a pretty straightforward application of the Henderson - Hasselbalch equation, which for a buffer that contains a weak acid and its conjugate base looks like this

color(blue)(|bar(ul(color(white)(a/a)"pH" = "p"K_a + log((["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))

Here you have

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{p} {K}_{a} = - \log \left({K}_{a}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where ${K}_{a}$ - the acid dissociation constant for the weak acid

Before using the H - H equation, calculate the $p {K}_{a}$ and compare it with the pH of the buffer. This will help you determine how much conjugate base you need relative to the amount of weak acid present in solution.

You will have

$\text{p} {K}_{a} = - \log \left(1.8 \cdot {10}^{- 5}\right) = 4.74$

The pH of the buffer is set at $5.10$. Since you have

$\text{pH " > " p} {K}_{a}$

you know for a fact that the buffer must contain more conjugate base, which in this case is the acetate anion, ${\text{CH"_3"COO}}^{-}$, usually delivered to the solution by sodium acetate, $\text{CH"_3"COONa}$, than weak acid, which in this case is acetic acid, $\text{CH"_3"COOH}$.

Your goal now is to use the H - H equation to find the concentration of the acetate anion. You will have

$5.10 = 4.74 + \log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right)$

Rearrange to get

$\log \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = 0.36$

This will be equivalent to

${10}^{\log} \left(\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right)\right) = {10}^{0.36}$

$\left(\left[\text{CH"_3"COO"^(-)])/(["CH"_3"COOH}\right]\right) = 2.291$

Therefore, the concentration of the acetate anions must be

$\left[\text{CH"_3"COO"^(-)] = 2.291 xx ["CH"_3"COOH}\right]$

["CH"_3"COO"^(-)] = 2.291 xx "0.150 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.344 M")color(white)(a/a)|)))

The answer is rounded to three sig figs.

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