Question

The dissociation constant of acetic acid is $x \times 10^{-5}$ When $25 \,mL$ of $0.2 \,M \,CH _3 COONa$ solution is mixed with $25 \,mL$ of $0.02\, M \,CH _3 COOH$ solution, the $pH$ of the resultant solution is found to be equal to 5. The value of $x$ is ___

Answer 1

Correct Answer: 10

To determine the value of $x$, we start by analyzing the given information: a solution containing acetic acid $(CH_3COOH)$ and sodium acetate $(CH_3COONa)$. The problem involves understanding the concept of buffer solutions and the Henderson-Hasselbalch equation, which is used to calculate the pH of a buffer solution.
The Henderson-Hasselbalch equation is expressed as:
$$pH = pK_a + \log \frac{[A^-]}{[HA]}$$
Where $[A^-]$ is the concentration of the acetate ion $(CH_3COO^-)$, and $[HA]$ is the concentration of acetic acid $(CH_3COOH)$.
Given: The final solution is formed by mixing equal volumes (25 mL each) of $0.2 \, M$ $CH_3COONa$ and $0.02 \, M$ $CH_3COOH$. Therefore, the concentrations in the mixed solution become:
$[CH_3COONa] = [A^-] = \frac{(0.2 \, M \times 25 \, mL)}{50 \, mL} = 0.1 \, M$
$[CH_3COOH] = [HA] = \frac{(0.02 \, M \times 25 \, mL)}{50 \, mL} = 0.01 \, M$
Plug these values into the Henderson-Hasselbalch equation:
$$5 = pK_a + \log \frac{0.1}{0.01}$$
$$5 = pK_a + \log 10$$
Since $\log 10 = 1$, this simplifies to:
$$5 = pK_a + 1$$
Solving for $pK_a$ gives:
$$pK_a = 4$$
The $pK_a$ is related to $K_a$ by the equation $pK_a = -\log K_a$. Therefore:
$4 = -\log K_a \implies K_a = 10^{-4}$
Given that the dissociation constant $K_a$ of acetic acid is $x \times 10^{-5}$, we equate and solve:
$$x \times 10^{-5} = 10^{-4}$$
$$x = \frac{10^{-4}}{10^{-5}} = 10$$
Thus, the value of $x$ is 10. This solution falls within the specified range of 10,10.
Therefore, the value of $x$ is confirmed to be 10.