Question

What is the component concentration ratio, $\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]\mathbf{\text{/}}\left[{\text{CH}}_{\text{3}}\text{COOH}\right]$, of a buffer that has a pH of 4.39 ( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of $\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$)?

Short answer

The component concentration ratio obtained is $\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\text{= 0.45.}$$\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\text{= 0.45.}$

Step-by-step solution

Buffer solution

The pH value of an acidic buffer solution can be determined by using the formula below:

$\mathbf{pH}\mathbf{=}\mathbf{pKa}\mathbf{+}\mathbf{log}\frac{\left[\mathbf{salt}\right]}{\left[\mathbf{acid}\right]}$

Explanation

First, write the ${\text{CH}}_{\text{3}}\text{COOH}$ dissociation process.

${\text{CH}}_{\text{3}}{\text{COOH +H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ CH}}_{\text{3}}{\text{COO}}^{\text{-}}.$

We must alter the Henderson-Hassle Balch equation for solving $\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$. The pH and ${\text{K}}_{\text{a}}$are given. Therefore, first solve for the ${\text{pK}}_{\text{a}}$.

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{= - logK}}_{\text{a}} \\ =-\log (1.8\times {10}^{-5}) \\ =4.74 \end{gathered}$$

Then alter the Henderson-Hassle Balch equation to find the answer for

$\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}.$

role="math" $$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \\ \text{log}\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\text{= pH - pK}}_{\text{a}} \\ \frac{\left[{\text{CH}}_{\text{3}}\text{COO -}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\text{= 10}}^{{\text{pH - pK}}_{\text{a}}} \\ {\text{= 10}}^{\text{4.39 - 4.74}} \\ \text{= 0.45.} \end{gathered}$$

Therefore, the component concentration ratio obtained is $\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\text{= 0.45}$.