Question

What is the component concentration ratio,${\mathbf{\text{[Pr}}}^{\mathbf{\text{-}}}\mathbf{\text{]/[HPr]}}$, of a buffer that has a pH of5.44 (${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of $\mathbf{\text{HPr}}\mathbf{ }\mathbf{\text{=}}\mathbf{ }{\mathbf{\text{1.3×10}}}^{\mathbf{-}\mathbf{5}}$)?

Short answer

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

Step-by-step solution

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:

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

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