Exactly of is mixed with of . The pH of the resulting solution would be (Given for ): — Ionic Equilibrium Chemistry Question
Question
Exactly $100 \text{ mL}$ of $0.2 \text{ M } HCl$ is mixed with $100 \text{ mL}$ of $0.2 \text{ M } CH_3COONa$. The pH of the resulting solution would be (Given $pK_a$ for $CH_3COOH = 4.74$):
💡 Solution & Explanation
The reaction is $HCl + CH_3COONa \rightarrow CH_3COOH + NaCl$. Since we mix $100 \text{ mL}$ of $0.2 \text{ M}$ each, there are $20 \text{ mmol}$ of both reactants, meaning they neutralize completely to form $20 \text{ mmol}$ of $CH_3COOH$. The total volume is $200 \text{ mL}$, making the $CH_3COOH$ concentration $= 20/200 = 0.1 \text{ M}$. The pH of a weak acid is $pH = \frac{1}{2}(pK_a - \log C) = \frac{1}{2}(4.74 - \log 0.1) = \frac{1}{2}(4.74 + 1) = 2.87$.