Question

Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodium acetate, CH3COONa, to make 500 mL of solution. Ka for CH3COOH = 1.76×10–5

Answer 1

Answer:

Approximately $4.75$.

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of  $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ are equal.

${\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}$

Since $\rm CH_3COONa$ is a salt soluble in water. Once in water, it would readily ionize to give $\rm CH_3COO^{-}$ and $\rm Na^{+}$ ions.

Assume that the $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ ions in this solution did not disintegrate at all. The solution would contain:

$0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COOH$, and

$0.06\; \rm mol$ of $\rm CH_3COO^{-}$ from $0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COONa$.

Accordingly, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ would be:

$\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$.

$\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$.

In other words, in this buffer solution, the initial concentration of the weak acid $\rm CH_3COOH$ is the same as that of its conjugate base, $\rm CH_3COO^{-}$.

Hence, once in equilibrium, the $\rm pH$ of this buffer solution would be the same as the ${\rm pK}_{a}$ of $\rm CH_3COOH$.

Calculate the ${\rm pK}_{a}$ of $\rm CH_3COOH$ from its ${\rm K}_{a}$:

$\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}$.