Question

Question: What volumes of ${\mathbf{\text{0.50MHNO}}}_{\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{50}{\mathbf{MNaNO}}_{\mathbf{2}}$ must be mixed to prepare 1.00 L of a solution buffered at pH = 3.55

Short answer

The volume of ${\mathrm{NaNO}}_2$added will be 0.587 L.

The volume of${\mathrm{HNO}}_2$ added will be 0.413 L.

Step-by-step solution

Definition of buffered solution

A buffered solution is defined as an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa.

Calculations

The pH of a buffer solution can be calculated by the Henderson – Hasselbalch equation.

$$\begin{gathered} \mathrm{pH}={\mathrm{PK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]} \\ {\mathrm{pK}}_{\mathrm{a}}=3.4 \\ \mathrm{pH}={\mathrm{PK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{NaNO}}_2\right]}{\left[{\mathrm{HNO}}_2\right]} \mathrm{or} \end{gathered}$$

$$\begin{gathered} 3.55=3.4+\log \frac{\left[{\mathrm{NaNO}}_2\right]}{\left[{\mathrm{HNO}}_2\right]} \\ \log \frac{\left[{\mathrm{NaNO}}_2\right]}{\left[{\mathrm{HNO}}_2\right]}=0.152 \\ \frac{\left[{\mathrm{NaNO}}_2\right]}{\left[{\mathrm{HNO}}_2\right]}={10}^{0.152} \\ \frac{\left[{\mathrm{NaNO}}_2\right]}{\left[{\mathrm{HNO}}_2\right]}=1.42 \end{gathered}$$

The value ofrole="math" localid="1649052591463" $\left[{\mathrm{NaNO}}_2\right]$will be 0.5a.

The value of role="math" localid="1649052626391" $\left[{\mathrm{HNO}}_2\right]$will be 0.5b.

role="math" localid="1649052172195" $$\begin{gathered} \frac{0.5\mathrm{a}}{0.5\mathrm{b}}=1.42 \\ \frac{\mathrm{a}}{\mathrm{b}}=1.42 \end{gathered}$$

$\text{Thetotalvolumeis1L.}$

$$\begin{gathered} \mathrm{a}+\mathrm{b}=1\mathrm{L} \\ \mathrm{b}=1-\mathrm{a} \\ \frac{\mathrm{a}}{\mathrm{b}}=1.42 \end{gathered}$$

role="math" localid="1649052545556" $$\begin{gathered} \frac{\mathrm{a}}{1-\mathrm{a}}=1.42 \\ \mathrm{a}=1.42-1.42\mathrm{a} \\ 2.42\mathrm{a}=1.42 \\ \mathbf{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{587} \end{gathered}$$

$\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{b}\mathrm{will}\mathrm{be}:$

$$\begin{gathered} \mathrm{b}=1-\mathrm{a} \\ =1-0.587 \\ \mathbf{b}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{413} \end{gathered}$$