Q118 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Chemistry
If $K_a$ of weak acid HA is $1 \times 10^{-7}$, calculate the $[\text{H}_3\text{O}^+]$ of $0.1\ \text{M}$ solution at equilibrium.
- A$1 \times 10^{-3}\ \text{M}$
- B$1 \times 10^{-5}\ \text{M}$
- Correct answerC$1 \times 10^{-4}\ \text{M}$
- D$1 \times 10^{-2}\ \text{M}$
Solution
For a weak acid HA, the dissociation reaction is:
$\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$
The dissociation constant is given by $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$.
Since $K_a$ is very small, the degree of dissociation $\alpha$ is much less than $1$, and the concentration of hydronium ions can be approximated as:
$[\text{H}_3\text{O}^+] = \sqrt{K_a \times C}$
Substituting the given values $K_a = 1 \times 10^{-7}$ and $C = 0.1\ \text{M}$:
$[\text{H}_3\text{O}^+] = \sqrt{1 \times 10^{-7} \times 0.1}$
$[\text{H}_3\text{O}^+] = \sqrt{10^{-8}}$
$[\text{H}_3\text{O}^+] = 1 \times 10^{-4}\ \text{M}$
Answer: $1 \times 10^{-4}\ \text{M}$
$\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$
The dissociation constant is given by $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$.
Since $K_a$ is very small, the degree of dissociation $\alpha$ is much less than $1$, and the concentration of hydronium ions can be approximated as:
$[\text{H}_3\text{O}^+] = \sqrt{K_a \times C}$
Substituting the given values $K_a = 1 \times 10^{-7}$ and $C = 0.1\ \text{M}$:
$[\text{H}_3\text{O}^+] = \sqrt{1 \times 10^{-7} \times 0.1}$
$[\text{H}_3\text{O}^+] = \sqrt{10^{-8}}$
$[\text{H}_3\text{O}^+] = 1 \times 10^{-4}\ \text{M}$
Answer: $1 \times 10^{-4}\ \text{M}$
Hence, the correct answer is option: 'C'.