Question

Consider the acids in Table 7.2. Which acid would be the best choice for preparing a pH 5 7.00 buffer? Explain how to make 1.0 L of this buffer.

Short answer

Answer

The buffer of1 L has 0.35 M OCl- and 1 M HOCl- for pH=7.

Step-by-step solution

Definition of best buffer

The best buffer can be defined as the equal and large quantities of weak acid and base of the conjugate. The Henderson-Hasselbalch equation can be denoted as,

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}\frac{\left[basse\right]}{\left[acid\right]} \\ \mathit{p}\mathit{H}\mathbf{=}\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{+}\mathbf{0} \\ \mathit{p}\mathit{H}\mathbf{\approx }\mathit{p}{\mathit{K}}_{\mathbf{a}} \end{gathered}$$

For the best buffer, it should be $\mathit{p}\mathit{H}\mathbf{\approx }\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{.}$

Best acid for pH=7

The value of pKa should be near 7. The value for suited acid is HOCl. Specifically, ${\mathit{K}}_{\mathbf{a}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{,}\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{46}\mathbf{.}$

By using the Henderson-Hasselbalch equation,

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}\frac{\left[base\right]}{\left[acid\right]} \\ \mathbf{7}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{46}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}\frac{\left[base\right]}{\left[acid\right]} \\ \mathbf{=}\frac{\left[OC{l}^{-}\right]}{\left[HOC{l}^{-}\right]}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{46}} \\ \mathbf{=}\frac{\left[OC{l}^{-}\right]}{\left[HOC{l}^{-}\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35} \end{gathered}$$

The ratio for $\frac{\left[OC{l}^{-}\right]}{\left[HO{C}^{-}\right]}$has pH 7. The buffer of1 L has 0.35 M OCl- and 1 M HOCl- for pH=7.