Q19.23P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the component concentration ratio,${\mathbf{\text{[Pr}}}^{\mathbf{\text{ - }}}\mathbf{\text{]/[HPr]}}$, of a buffer that has a pH of  5.44 ( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of $\mathbf{\text{HPr}}\mathbf{ }\mathbf{\text{ = }}\mathbf{ }{\mathbf{\text{1.3×10}}}^{\mathbf{-}\mathbf{5}}$ )?

Step-by-Step Solution

Answer

The component concentration ratio obtained is $\frac{{\text{[Pr}}^{\text{ - }}\text{]}}{\text{[HPr]}}\text{ = 3.54}$.

Step 1: Ionic Equilibrium

AN ionic equilibrium refers to an equilibrium that exists in the solution of weak electrolytes between unionized molecules and ions. The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation, which can be mathematically presented as:

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

Step 2: Calculation

Write the HPr dissociation reaction first.

${\text{HPr + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ + Pr}}^{\text{ - }}.$

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

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

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(1.3×10}}^{-5}\text{)} \\ =4.89 \end{gathered}$$
Then, solve for $\frac{{\text{[Pr}}^{\text{ - }}\text{]}}{\text{[HPr]}}$ using the Henderson-Hasselbalch equation as:

$$\begin{gathered} {\text{ pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{Pr}}^{\text{ - }}\right]}{\text{[HPr]}} \\ \text{log}\frac{\left[{\text{Pr}}^{\text{ - }}\right]}{\text{[HPr]}}{\text{ = pH - pK}}_{\text{a}} \\ \frac{\left[{\text{Pr}}^{\text{ - }}\right]}{\text{[HPr]}}{\text{ = 10}}^{{\text{pH - pK}}_{\text{a}}} \\ {\text{ = 10}}^{\text{5.44 - 4.89}} \\ \text{ = 3.54.} \end{gathered}$$

Therefore, the component concentration ratio obtained is $\frac{{\text{[Pr}}^{\text{ - }}\text{]}}{\text{[HPr]}}\text{ = 3.54}$.