• Solution A with $\text{pH} = 6$ has $[H^+]_A = 10^{-6}\text{ M}$. Solution B with $\text{pH} = 4$ has $[H^+]_B = 10^{-4}\text{ M}$.
• When equal volumes $V$ of both solutions are mixed, the total volume becomes $2V$ and the resultant hydrogen ion concentration is:
  $$[H^+]_{\text{mix}} = \frac{[H^+]_A \times V + [H^+]_B \times V}{2V} = \frac{10^{-6} + 10^{-4}}{2}$$
• Factoring out $10^{-4}$:
  $$[H^+]_{\text{mix}} = \frac{10^{-4}(0.01 + 1)}{2} = \frac{1.01 \times 10^{-4}}{2} \approx 5.05 \times 10^{-5}\text{ M}$$
• Taking negative logarithm:
  $$\text{pH} = -\log(5.05 \times 10^{-5}) = 5 - \log 5.05 \approx 5 - 0.70 = 4.30$$
  Since $4.30$ lies strictly between 4 and 5, option 3 is correct.

The acid dissociation constant ($K_a$) of acetic acid (ethanoic acid) at 298 K (25 °C) is $1.75 \times 10^{-5} \text{ mol L}^{-1}$.

• Mathematical Formula: The $\text{p}K_a$ is defined as the negative logarithm (base 10) of the dissociation constant:
  $$\text{p}K_a = -\log_{10}(K_a)$$
  $$\text{p}K_a = -\log_{10}(1.75 \times 10^{-5}) = 5 - \log_{10}(1.75) \approx 4.76$$
• Buffer Applications: An equimolar buffer solution of acetic acid and sodium acetate ($ ext{CH}_3 ext{COOH} / ext{CH}_3 ext{COONa}$) maintains a pH equal to its $\text{p}K_a$ (4.76) by the Henderson-Hasselbalch equation.
• Acid Strength: Formic acid ($\text{HCOOH}$) is stronger with a lower $\text{p}K_a$ of 3.75.
• Food Use: Acetic acid (4-8% aqueous solution) is commonly used as vinegar for food preservation and pickling.