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NEET CHEMISTRYEquilibriumMedium

Question

For a weak acid HA\text{HA}, the percentage of dissociation is nearly 1%1\% at equilibrium. If the concentration of acid is 0.1 mol L−10.1\text{ mol L}^{-1}, then the correct option for its KaK_a at the same temperature will be:

A

1×10−41 \times 10^{-4}

B

1×10−61 \times 10^{-6}

C

1×10−51 \times 10^{-5}

D

1×10−31 \times 10^{-3}

Step-by-Step Solution

For a weak acid HA, the ionization constant (KaK_a) is related to its concentration (cc) and degree of dissociation (α\alpha) by the formula: Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}

Given the values: Concentration of acid, c=0.1 mol L−1=10−1 Mc = 0.1\text{ mol L}^{-1} = 10^{-1}\text{ M} Percentage of dissociation = 1%1\% Degree of dissociation, α=1100=0.01=10−2\alpha = \frac{1}{100} = 0.01 = 10^{-2}

Since α\alpha (0.010.01) is very small compared to 11, the term (1−α)(1 - \alpha) in the denominator can be approximated as 11 . The formula simplifies to: Ka≈cα2K_a \approx c\alpha^2

Substituting the given values: Ka=(10−1)×(10−2)2K_a = (10^{-1}) \times (10^{-2})^2 Ka=10−1×10−4K_a = 10^{-1} \times 10^{-4} Ka=10−5K_a = 10^{-5}

Thus, the ionization constant KaK_a is 1×10−51 \times 10^{-5}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumtexthapercentagedissociationnearlyequilibrium

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