Q19.24P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the component concentration ratio, $\mathbf{\left[}\mathit{N}{\mathit{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\right]}\mathbf{/}\mathbf{\left[}\mathit{H}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\right]}$, of a buffer that has a pH of 2.95 ( ${\mathit{K}}_{\mathbf{a}}$of ${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}{\mathbf{\text{ = 7.1×10}}}^{\mathbf{-}\mathbf{4}}$ )?

Step-by-Step Solution

Answer

The component concentration ratio obtained is $\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\text{ = 0.63}$.

Step 1: Henderson-Hesselbalch Equation

The pH of a buffer solution can be determined by using the Henderson-Hesselbalch equation. The formula to calculate the pH of an acidic buffer solution is:

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

Step 2: Calculation

Write the ${\text{HNO}}_{\text{2}}$ dissociation reaction first.

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

We must alter the Henderson-Hasselbalch equation to find $\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}$.

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

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(7.1×10}}^{-4}\text{)} \\ =3.15 \end{gathered}$$

Then, solve for $\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}$ using the Henderson-Hasselbalch equation as:

$$\begin{gathered} {\text{ pH = pK}}_{\text{a}}\text{ + log}\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}} \\ \text{log}\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}{\text{ = pH - pK}}_{\text{a}} \\ \frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}{\text{ = 10}}^{{\text{pH - pK}}_{\text{a}}} \\ {\text{ = 10}}^{\text{2.95 - 3.15}} \\ \text{ = 0.63.} \end{gathered}$$

Therefore, the component concentration ratio obtained is $\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{ - }}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\text{ = 0.63}$.