Question:
The pH of a 0.1 M solution of a weak monobasic organic acid is 4.0. What is the dissociation constant of the acid?
The pH of a 0.1 M solution of a weak monobasic organic acid is 4.0. What is the dissociation constant of the acid?
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$K_a = C \alpha^2$ where $\alpha = [H^+]/C$.
Updated On: Jun 10, 2026
- $1.0 \times 10^{-8}$
- $1.0 \times 10^{-7}$
- $1.0 \times 10^{-6}$
- $1.0 \times 10^{-5}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
For a weak acid, $pH = -\log[H^+]$. The dissociation constant $K_a = [H^+]^2 / C$.
Step 2: Analysis
Given $pH = 4.0$, $[H^+] = 10^{-pH} = 10^{-4}$ M. Concentration $C = 0.1$ M. $K_a = (10^{-4})^2 / 0.1 = 10^{-8} / 10^{-1} = 10^{-7}$? Re-calculating: $10^{-8} / 10^{-1} = 10^{-7}$. Checking the options provided in source, $1.0 \times 10^{-7}$ is (B).
Step 3: Conclusion
The dissociation constant is $1.0 \times 10^{-7}$.
Final Answer: (B)
For a weak acid, $pH = -\log[H^+]$. The dissociation constant $K_a = [H^+]^2 / C$.
Step 2: Analysis
Given $pH = 4.0$, $[H^+] = 10^{-pH} = 10^{-4}$ M. Concentration $C = 0.1$ M. $K_a = (10^{-4})^2 / 0.1 = 10^{-8} / 10^{-1} = 10^{-7}$? Re-calculating: $10^{-8} / 10^{-1} = 10^{-7}$. Checking the options provided in source, $1.0 \times 10^{-7}$ is (B).
Step 3: Conclusion
The dissociation constant is $1.0 \times 10^{-7}$.
Final Answer: (B)
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