Q19.25P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the component concentration ratio, $\mathbf{\left[}\mathit{B}\mathit{r}{\mathit{O}}^{\mathbf{-}}\mathbf{\right]}\mathbf{/}\mathbf{\left[}\mathit{H}\mathit{B}\mathit{r}\mathit{O}\mathbf{\right]}$, of a buffer that has a pH of  7.95( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of $\mathit{H}\mathit{B}\mathit{r}\mathit{O}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}$ )?

Step-by-Step Solution

Answer

The component concentration ratio obtained is $\frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}}\text{ = 0.20}$

Step 1: Buffer solution

A buffer solution withstands the change in a pH value. An acidic buffer is formed by combining a weak acid and the salt of its conjugate base.

Step 2: Evaluate the component concentration ratio

Write the HBrO dissociation reaction first.

${\text{HBrO + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ + BrO}}^{\text{ - }}.$

We must alter the Henderson-Hasselbalch equation to find $\frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}}$.

The pH and ${\text{K}}_{\text{a}}$ are known.   Therefore, first solve for the ${\text{pK}}_{\text{a}}$.

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(2.3×10}}^{-9}\text{)} \\ =8.64 \end{gathered}$$

Then, solve for $\frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}}$ using the Henderson-Hasselbalch equation as:

$$\begin{gathered} {\text{ pH = pK}}_{\text{a}}\text{ + log}\frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}} \\ \text{log}\frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}}{\text{ = pH - pK}}_{\text{a}} \\ \frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}}{\text{ = 10}}^{{\text{pH - pK}}_{\text{a}}} \\ {\text{ = 10}}^{\text{7.95 - 8.64}} \\ \text{ = 0.20.} \end{gathered}$$

Therefore, the component concentration ratio obtained is $\frac{{\text{[BrO}}^{\text{ - }}\text{]}}{\text{[HBrO]}}\text{ = 0.20}$.