Question

An organic monobasic acid has dissociation constant $2.25 \times 10^{-6}$. What is percent dissociation in its $0.01 \text{ M}$ solution?

• A. 1.5%
• B. 15%
• C. 5%
• D. 0.5%

Hint:

For weak acids, the degree of dissociation ($\alpha$) can often be approximated using $\alpha = \sqrt{K_a/C_0}$, where $C_0$ is the initial concentration. Remember to convert $\alpha$ to a percentage for the final answer.

Answer 1

The Correct Option is A

Step 1: Write the dissociation equilibrium and $K_a$ expression. For a monobasic weak acid HA, the dissociation equilibrium in water is: \[ \text{HA} \leftrightharpoons \text{H}^+ + \text{A}^- \] The dissociation constant ($K_a$) is given by: \[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Step 2: Express equilibrium concentrations in terms of initial concentration and degree of dissociation. Let $C_0$ be the initial concentration of the acid ($0.01 \text{ M}$). Let $\alpha$ be the degree of dissociation. At equilibrium: $[\text{HA}] = C_0(1 - \alpha)$ $[\text{H}^+] = C_0\alpha$ $[\text{A}^-] = C_0\alpha$ For a weak acid, $\alpha$ is typically very small ($<<1$), so we can approximate $(1 - \alpha) \approx 1$. Thus, $[\text{HA}] \approx C_0$. 
Step 3: Substitute into the $K_a$ expression and solve for $\alpha$. Substituting the approximate equilibrium concentrations into the $K_a$ expression: \[ K_a = \frac{(C_0\alpha)(C_0\alpha)}{C_0} = \frac{C_0^2\alpha^2}{C_0} = C_0\alpha^2 \] Now, solve for $\alpha$: \[ \alpha^2 = \frac{K_a}{C_0} \] \[ \alpha = \sqrt{\frac{K_a}{C_0 \] 
Step 4: Plug in the given values and calculate $\alpha$. Given $K_a = 2.25 \times 10^{-6}$ and $C_0 = 0.01 \text{ M}$. \[ \alpha = \sqrt{\frac{2.25 \times 10^{-6{0.01 \] \[ \alpha = \sqrt{2.25 \times 10^{-4 \] \[ \alpha = 1.5 \times 10^{-2} \] 
Step 5: Convert the degree of dissociation to percent dissociation. Percent dissociation is $\alpha \times 100%$. \[ \text{Percent dissociation} = (1.5 \times 10^{-2}) \times 100% \] \[ \text{Percent dissociation} = 1.5% \] Thus, the percent dissociation of the acid in its $0.01 \text{ M}$ solution is $1.5%$.