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

Approximately
`4.75`.

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 * 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 * 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} \\ &= (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} \\ &= (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 * 10^(-5)) \\ &\approx 4.75\end{aligned}`.