Question

You make of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{L}$a buffered solution by mixing acetic acid and sodium acetate. You have$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{M}$ solutions for each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

Short answer

$154 \text{mL}$of sodium acetate and $846 \text{mL}$of acetic acid have to be mixed to make a buffer solution.

Step-by-step solution

Definition of buffer solution

A buffer solution is stated as an aqueous mixture of a weak acid and its corresponding base, or vice versa.

Calculations

The molarity of the solution is$0.18197$.

$$\begin{gathered} \text{Molarity}=\frac{\text{Moles}}{\text{Volume}} \\ \frac{\frac{{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{V}}_{\text{T}}}}{\frac{{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{V}}_{\text{T}}}}=0.18197 \\ {\text{V}}_{\text{1}}=0.18197{\text{V}}_{\text{2}} \end{gathered}$$

The total volume of the solutionis$1.00 \text{L}$.

${\text{V}}_{\text{1}}+{\text{V}}_{\text{2}}=1.00 \text{L}$

Now,

$$\begin{gathered} 0.18197{\text{V}}_{\text{2}}+{\text{V}}_{\text{2}}=1.00 \\ 1.18197 {\text{V}}_{\text{2}}=1.00 \\ {\text{V}}_{\text{2}}=\frac{1.00}{1.18197} \\ {\text{V}}_{\text{2}}=0.846 \text{L} \end{gathered}$$

$$\begin{gathered} {\text{V}}_1=1.00-0.846 \\ {\text{V}}_1=0.154 \text{L} \end{gathered}$$

Abuffer can be madebymixing$154 \text{mL}$of sodium acetate and$\text{846mL}$of acetic acid.