An aqueous solution of barium hydroxide, \text{Ba(OH)}_2, has a molar concentration of 0.0… — JAMB UTME

An aqueous solution of barium hydroxide, $\text{Ba(OH)}_2$ , has a molar concentration of $0.0005\text{ mol dm}^{-3}$ . Assuming complete dissociation of the base at $25^\circ\text{C}$ , what is the pH of this solution?

$11.0$ Answer
B
$3.0$
C
$10.7$
D
$3.3$

Answer

The pH of the solution is

11.0

Barium hydroxide (

\text{Ba(OH)}_2

) produces two hydroxide ions (

\text{OH}^-

) per formula unit upon complete dissociation. Thus, a

0.0005\text{ mol dm}^{-3}

solution produces

[\text{OH}^-] = 2 \times 0.0005 = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The pOH is

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

. Subtracting pOH from 14.0 yields a pH of

11.0

Step-by-Step Solution

Determine the concentration of hydroxide ions

[\text{OH}^-]

from the mole ratio of

\text{Ba(OH)}_2

Since each mole of

\text{Ba(OH)}_2

dissociates into two moles of

\text{OH}^-

[\text{OH}^-] = 2 \times 0.0005\text{ mol dm}^{-3} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

Barium hydroxide is a strong dibasic base that fully dissociates in aqueous solution.

Calculate the pOH of the solution.

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

The pOH scale is defined as the negative logarithm of the hydroxide ion concentration.

Calculate the pH using the relationship

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

\text{pH} = 14.0 - 3.0 = 11.0

25^\circ\text{C}

, the sum of pH and pOH for any dilute aqueous solution equals 14.0.

Key Concept

pH and pOH calculations for strong basic solutions considering stoichiometry

Estimated Time:1m 0s

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