An aqueous solution of a weak monoacidic base, \text{BOH}, has a concentration of 0.20\tex… — JAMB UTME

An aqueous solution of a weak monoacidic base, $\text{BOH}$ , has a concentration of $0.20\text{ mol dm}^{-3}$ and a base dissociation constant $K_b = 5.0 \times 10^{-6}\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Calculate the pH of this solution.

Answer: 11

Answer

The pH of the weak monoacidic base solution is 11.0.

For a weak base, partial equilibrium ionization gives

[\text{OH}^-] = \sqrt{K_b \times C} = \sqrt{5.0 \times 10^{-6} \times 0.20} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The pOH is

-\log_{10}(1.0 \times 10^{-3}) = 3.0

. Using

\text{pH} = 14.0 - \text{pOH}

, the pH of the basic solution is

14.0 - 3.0 = 11.0

Step-by-Step Solution

Formulate the dissociation equilibrium equation for the weak base BOH.

The ionization is

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

, giving

K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}

Weak bases ionize incompletely in aqueous media.

Calculate the equilibrium hydroxide ion concentration [OH⁻].

[\text{OH}^-] = \sqrt{K_b \times C} = \sqrt{(5.0 \times 10^{-6})(0.20)} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

Since the base is weak and degree of ionization is small,

[\text{BOH}] \approx 0.20\text{ mol dm}^{-3}

and

[\text{B}^+] = [\text{OH}^-]

Determine the pOH of the solution.

\text{pOH} = -\log_{10}(1.0 \times 10^{-3}) = 3.0

pOH is defined as

-\log_{10}[\text{OH}^-]

Calculate the pH using the relationship between pH and pOH at 25°C.

\text{pH} = 14.0 - \text{pOH} = 14.0 - 3.0 = 11.0

25^\circ\text{C}

\text{pH} + \text{pOH} = 14.0

Key Concept

Weak base dissociation equilibrium and pH determination

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