Q19.38P - Chemistry: Molecular Nature Of Matter And Change

Question

Oxoanions of phosphorus are buffer components in blood. For a ${\mathbf{\text{KH}}}_{\mathbf{\text{2}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}{\mathbf{\text{/Na}}}_{\mathbf{\text{2}}}{\mathbf{\text{HPO}}}_{\mathbf{\text{4}}}$ solution with $\mathbf{\text{pH = 7.40}}$ (pH of normal arterial blood), what is the buffer-component concentration ratio?

Step-by-Step Solution

Answer

The ratio is $\frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}\text{ = 1.58}$.

Step 1: Buffer solution

A buffer solution can withstand pH changes when acidic or basic components are added. It can neutralize little amounts of additional acid or base, allowing the pH of the solution to remain relatively constant.

Step 2: Explanation

To find the buffer components, first create the dissolving equation for each salt.

$$\begin{gathered} {\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\to {\text{K}}^{\text{ + }}{\text{ + H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }} \\ {\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\to {\text{2Na}}^{\text{ + }}{\text{ + HPO}}_{\text{4}}^{\text{2 - }} \end{gathered}$$

As ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ and ${\text{HPO}}_{\text{4}}^{\text{2 - }}$ are the ions that will be involved in the buffer reaction.

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{HPO}}_{\text{4}}^{\text{2 - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }}$

Calculate the from ${\text{pK}}_{\text{a}}$ the ${\text{K}}_{\text{a}}$ of above reaction in Appendix C.

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ =-\log \left(6.3\times {10}^{-8}\right) \\ {\mathrm{pK}}_{\mathrm{a}}=7.20 \end{gathered}$$

Solve for the concentration ratio $\left(\frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}\right)$ using the Henderson-Hassle Balch equation.

$$\begin{gathered} {\text{ pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]} \\ \text{log}\frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}\text{ = pH - pKa} \end{gathered}$$

$$\begin{gathered} \text{log}\frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}\text{ = 7.4 - 7.20} \\ \frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}{\text{ = 10}}^{\text{7.4 - 7.20}} \\ \frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}\text{ = 1.58} \end{gathered}$$

 

Therefore, $\frac{\left[{\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}\right]}{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\right]}\text{ = 1.58}$.