Q19.13P - Chemistry: Molecular Nature Of Matter And Change

Question

What are the  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ and the pH of a buffer that consists of  0.55 M ${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}$ and  0.75 M  ${\mathbf{\text{KNO}}}_{\mathbf{\text{2}}}$ ( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ of ${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}{\mathbf{\text{ = 7.1×10}}}^{\mathbf{-}\mathbf{4}}$ )?

Step-by-Step Solution

Answer

The pH of propanoic acid is 3.28 and the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ obtained is $5.2\times {10}^{-4}$.

Step 1: Ionic Equilibrium

An ionic equilibrium refers to an equilibrium that exists in the solution of weak electrolytes between unionized molecules and ions.

Step 2: What is H 3 O  +  and pH ?

Write the  0.55 M  ${\text{HNO}}_{\text{2}}$ dissociation equation first.

 ${\text{HNO}}_{\text{2}}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons \text{N}{{\text{O}}_{\text{2}}}^{\text{ - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }}.$

Then, write the  0.75 M  ${\text{KNO}}_{\text{2}}$ dissociation equation.

 ${\text{KNO}}_{\text{2}}\to \text{N}{{\text{O}}_{\text{2}}}^{\text{ - }}{\text{ + K}}^{\text{ + }}.$

We now have the initial nitric acid and nitrate.

 $$\begin{gathered} {\text{HNO}}_{\text{2}}\text{ = 0.55M} \\ {\text{NO}}_{\text{2}}^{\text{ - }}\text{ = 0.75 M} \end{gathered}$$

To find the pH, we may utilize the Henderson-Hasselbalch equation. First, solve for the ${\text{pK}}_{\text{a}}$.

  $$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(7.1×10}}^{-4}\text{)} \\ =3.15 \\ \text{pH = pKa + log}\frac{\left[{\text{NO}}_{\text{2}}^{\text{ - }}\right]}{\left[{\text{HNO}}_{\text{2}}\right]} \\ \text{ = 3.15 + log}\frac{\text{0.75 M}}{\text{0.55 M}} \\ \text{ = 3.28.} \end{gathered}$$

Then, solve for  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ from the calculated pH as:

 $$\begin{gathered} \text{pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ = 10}}^{\text{ - pH}} \\ {\text{ = 10}}^{\text{ - 3.28}} \\ {\text{ = 5.2×10}}^{-4} \end{gathered}$$

Therefore, the values obtained are:

 $$\begin{gathered} \text{ pH = 3.28.} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ = 5.2×10}}^{-4} \end{gathered}$$