Q19.26P - Chemistry: Molecular Nature Of Matter And Change

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 $\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$)?

Step-by-Step Solution

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 1: Buffer solution

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

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

Step 2: 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  

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

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

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