Question

What are the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$and the pH of a buffer that consists of 0.55M ${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}$and0.75M ${\mathbf{\text{KNO}}}_{\mathbf{\text{2}}}$( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}{\mathbf{\text{= 7.1×10}}}^{\mathbf{-}\mathbf{4}}$ )?

Short 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-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.

What is H3O + 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 ${\text{pK}}_{\text{a}}$.

role="math" localid="1663406489385" $$\begin{gathered} {\text{pK}}_{\text{a}}{\text{= - logK}}_{\text{a}} \\ {\text{= - log(7.1×10}}^{-4}\text{)} \\ =3.15 \\ \text{pH = pKa + log}\frac{\left[{\text{NO}}_{\text{2}}^{\text{-}}\right]}{\left[{\text{HNO}}_{\text{2}}\right]} \\ \text{= 3.15 + log}\frac{\text{0.75 M}}{\text{0.55 M}} \\ \text{= 3.28.} \end{gathered}$$

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

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

Therefore, the values obtained are:

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