Answer:
Approximately [tex]4.75[/tex].
Step-by-step explanation:
Remark: this approach make use of the fact that in the original solution, the concentration of [tex]\rm CH_3COOH[/tex] and [tex]\rm CH_3COO^{-}[/tex] are equal.
[tex]{\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}[/tex]
Since [tex]\rm CH_3COONa[/tex] is a salt soluble in water. Once in water, it would readily ionize to give [tex]\rm CH_3COO^{-}[/tex] and [tex]\rm Na^{+}[/tex] ions.
Assume that the [tex]\rm CH_3COOH[/tex] and [tex]\rm CH_3COO^{-}[/tex] ions in this solution did not disintegrate at all. The solution would contain:
[tex]0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol[/tex] of [tex]\rm CH_3COOH[/tex], and
[tex]0.06\; \rm mol[/tex] of [tex]\rm CH_3COO^{-}[/tex] from [tex]0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol[/tex] of [tex]\rm CH_3COONa[/tex].
Accordingly, the concentration of [tex]\rm CH_3COOH[/tex] and [tex]\rm CH_3COO^{-}[/tex] would be:
[tex]\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}[/tex].
[tex]\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}[/tex].
In other words, in this buffer solution, the initial concentration of the weak acid [tex]\rm CH_3COOH[/tex] is the same as that of its conjugate base, [tex]\rm CH_3COO^{-}[/tex].
Hence, once in equilibrium, the [tex]\rm pH[/tex] of this buffer solution would be the same as the [tex]{\rm pK}_{a}[/tex] of [tex]\rm CH_3COOH[/tex].
Calculate the [tex]{\rm pK}_{a}[/tex] of [tex]\rm CH_3COOH[/tex] from its [tex]{\rm K}_{a}[/tex]:
[tex]\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}[/tex].
