The solubility product (K_{sp}) of calcium tetraoxosulfate(VI), \text{CaSO}_4, is 2.5 \tim… — JAMB UTME

The solubility product ( $K_{sp}$ ) of calcium tetraoxosulfate(VI), $\text{CaSO}_4$ , is $2.5 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}$ at $25^\circ\text{C}$ . What is the molar solubility of $\text{CaSO}_4$ in water at this temperature?

( $\text{Molar mass of CaSO}_4 = 136\text{ g mol}^{-1}$ )

$5.0 \times 10^{-3}\text{ mol dm}^{-3}$ Answer
B
$0.68\text{ mol dm}^{-3}$
C
$1.25 \times 10^{-5}\text{ mol dm}^{-3}$
D
$2.5 \times 10^{-5}\text{ mol dm}^{-3}$

Answer

The molar solubility of calcium tetraoxosulfate(VI) is

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

For a 1:1 sparingly soluble salt like

\text{CaSO}_4

, dissociation yields equal molar concentrations of

\text{Ca}^{2+}

and

\text{SO}_4^{2-}

ions (

s

). The solubility product expression is

K_{sp} = s^2

. Taking the square root of

2.5 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}

gives

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

Step-by-Step Solution

Write the solubility equilibrium equation and

K_{sp}

expression for

\text{CaSO}_4

\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)

, so

K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}]

Establishing the stoichiometric relationship between dissolved ions and molar solubility.

Substitute molar solubility

s

into the

K_{sp}

expression.

[\text{Ca}^{2+}] = s

and

[\text{SO}_4^{2-}] = s

, then

K_{sp} = s \times s = s^2

Expressing

K_{sp}

in terms of a single variable

s

Calculate the value of

s

by taking the square root of

K_{sp}

s = \sqrt{2.5 \times 10^{-5}} = \sqrt{25 \times 10^{-6}} = 5.0 \times 10^{-3}\text{ mol dm}^{-3}

Solving the algebraic equation for molar solubility.

Key Concept

Solubility product constant (

K_{sp}

) and molar solubility relation for AB-type binary ionic salts.

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