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 ${\mathbf{\text{HNO}}}_{\mathbf{\text{2}}}{\mathbf{\text{= 7.1×10}}}^{\mathbf{-}\mathbf{4}}$)?

Short 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-by-step solution

Henderson-Hesselbalch Equation

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

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

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:

role="math" localid="1663389158459" $$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}} \\ \text{log}\frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}{\text{= pH - pK}}_{\text{a}} \\ \frac{\text{[N}{{\text{O}}_{\text{2}}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}{\text{= 10}}^{{\text{pH - pK}}_{\text{a}}} \\ {\text{= 10}}^{\text{2.95 - 3.15}} \\ \text{= 0.63.} \end{gathered}$$

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