The degree of dissociation of a monobasic acid in its 0.1M solution at 300 K is 1.342 × 10^ -2 . Calculate the degree of dissociation of the same acid in its 0.02M solution.

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Quantrex Academy · Accepted Answer

Correct answer: C. 3.0 × 10^ -2

Using Ostwald's dilution law for a weak monobasic acid:

K_a = C \alpha^2

For the first solution,

C_1 = 0.1

M and

\alpha_1 = 1.342 \times 10^{-2}

:

K_a = 0.1 \times (1.342 \times 10^{-2})^2 = 0.1 \times 1.8 \times 10^{-4} = 1.8 \times 10^{-5}

For the second solution,

C_2 = 0.02

M:

K_a = C_2 \alpha_2^2

1.8 \times 10^{-5} = 0.02 \times \alpha_2^2

\alpha_2^2 = \dfrac{1.8 \times 10^{-5}}{0.02} = 9 \times 10^{-4}

\alpha_2 = 3.0 \times 10^{-2}

Answer:

3.0 \times 10^{-2}

The degree of dissociation of a monobasic acid in its 0.1M solution...

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