At 25^\circ\text{C}, an aqueous solution of a weak monoacidic base has a concentration of … — JAMB UTME

At $25^\circ\text{C}$ , an aqueous solution of a weak monoacidic base has a concentration of $0.08\text{ mol dm}^{-3}$ and a degree of ionization ( $\alpha$ ) of $0.025$ ( $2.5\%$ ). What is the concentration of hydroxide ions, $[\text{OH}^-]$ , in the solution in $\text{mol dm}^{-3}$ ?

Cevap: 0.002 mol dm^-3

Cevap

The concentration of hydroxide ions,

[\text{OH}^-]

, in the solution is

0.002\text{ mol dm}^{-3}

For a weak monoacidic base in aqueous solution, only a fraction (

\alpha

) of the base molecules ionize to form hydroxide ions. The concentration of hydroxide ions is given by

[\text{OH}^-] = C \alpha

. Substituting the concentration

0.08\text{ mol dm}^{-3}

and degree of ionization

0.025

yields

[\text{OH}^-] = 0.08 \times 0.025 = 0.002\text{ mol dm}^{-3}

Adım Adım Çözüm

Identify the relationship between degree of ionization and hydroxide ion concentration for a weak monoacidic base

[\text{OH}^-] = C \cdot \alpha

A weak monoacidic base

\text{BOH}

ionizes partially according to

\text{BOH}_{(aq)} \rightleftharpoons \text{B}^+_{(aq)} + \text{OH}^-_{(aq)}

, so the concentration of produced hydroxide ions equals the initial concentration multiplied by the degree of ionization.

Substitute the given values into the equation

[\text{OH}^-] = 0.08\text{ mol dm}^{-3} \times 0.025

The initial concentration

C = 0.08\text{ mol dm}^{-3}

and the degree of ionization

\alpha = 2.5\% = 0.025

Perform the multiplication to find the final concentration

[\text{OH}^-] = 0.002\text{ mol dm}^{-3}

Multiplying

0.08

0.025

yields

0.002\text{ mol dm}^{-3}

(or

2.0 \times 10^{-3}\text{ mol dm}^{-3}

Anahtar Kavram

Ionization equilibrium of weak bases and calculation of hydroxide ion concentration

Tahmini Süre:1m 15s

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