Question

The minimum volume of water (in L) required to dissolve 1.5 g of \(CaSO_{4}\) (molar mass = 136 g mol\(^{-1}\)) at 298 K is given that \(K_{sp} = 9 \times 10^{-6}\).

• A. 6.37
• B. 7.37
• C. 3.67
• D. 3.73

Hint:

For AB-type salts: \(K_{sp} = s^2\). Always confirm dissociation type before solving.

Answer 1

The Correct Option is C

Step 1: Write the dissolving equation.
Calcium sulfate dissolves only a little in water: \[ CaSO_4(s) \rightleftharpoons Ca^{2+} + SO_4^{2-} \] Each formula unit gives one calcium ion and one sulfate ion.

Step 2: Link solubility to $K_{sp}$.
If $s$ is the molar solubility, then both ion concentrations equal $s$. So \[ K_{sp} = [Ca^{2+}][SO_4^{2-}] = s \times s = s^2 \]

Step 3: Solve for solubility.
\[ s = \sqrt{K_{sp}} = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3} \text{ mol/L} \] This is the most $CaSO_4$ that one litre of water can hold.

Step 4: Find the moles to be dissolved.
The mass given is $1.5$ g and the molar mass is $136$ g/mol: \[ n = \frac{1.5}{136} = 0.01103 \text{ mol} \]

Step 5: Find the water volume needed.
Volume equals moles divided by solubility per litre: \[ V = \frac{n}{s} = \frac{0.01103}{3 \times 10^{-3}} \]

Step 6: Get the final value.
\[ V \approx 3.67 \text{ L} \] So about $3.67$ litres of water are needed to just dissolve all the salt.
\[ \boxed{3.67 \text{ L}} \]