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 of7.95(${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of$\mathit{H}\mathit{B}\mathit{r}\mathit{O}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}$)?

Short answer

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

Step-by-step solution

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.

Evaluate the component concentration ratio

Write theHBrOdissociation 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, thecomponent concentration ratio obtained is$\frac{{\text{[BrO}}^{\text{-}}\text{]}}{\text{[HBrO]}}\text{= 0.20}$.