A weak monobasic acid, \text{HA}, has an acid dissociation constant (K_a) of 4.5 \times 10… — JAMB UTME

A weak monobasic acid, $\text{HA}$ , has an acid dissociation constant ( $K_a$ ) of $4.5 \times 10^{-5}\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Calculate the percentage ionization of a $0.05\text{ mol dm}^{-3}$ solution of this acid.

Answer: 3 %

Answer

The percentage ionization of the weak monobasic acid solution is 3.0%.

According to Ostwald's dilution law for weak monobasic acids,

K_a = \alpha^2 C

. Rearranging to solve for the degree of ionization gives

\alpha = \sqrt{K_a / C} = \sqrt{(4.5 \times 10^{-5}) / 0.05} = \sqrt{9.0 \times 10^{-4}} = 0.03

. Expressed as a percentage,

0.03 \times 100\% = 3.0\%

Step-by-Step Solution

Relate acid dissociation constant (

K_a

), initial molar concentration (

C

), and degree of ionization (

\alpha

)

K_a = \alpha^2 C

For a weak monobasic acid undergoing partial ionization (

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

), Ostwald's dilution law simplifies to

K_a = \frac{\alpha^2 C}{1 - \alpha} \approx \alpha^2 C

because

\alpha \ll 1

Substitute the given values into the simplified expression and calculate

\alpha

\alpha = \sqrt{\frac{4.5 \times 10^{-5}}{0.05}} = \sqrt{9.0 \times 10^{-4}} = 0.03

Rearranging the equation yields

\alpha = \sqrt{K_a / C}

. Dividing the acid dissociation constant by the molar concentration gives

9.0 \times 10^{-4}

, and taking the square root yields

0.03

Convert the fractional degree of ionization to percentage ionization

3.0%

Multiplying the fractional degree of ionization by 100% gives the percentage of acid molecules ionized in solution.

Key Concept

Ostwald's Dilution Law and Degree of Ionization of Weak Acids

Estimated Time:1m 30s

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