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Exercises · 6.72

Q.What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10⁻⁶).

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Use the solubility product constant to find the molar solubility of calcium sulphate, convert to grams per litre, then calculate the volume needed to dissolve 1 g. The minimum volume required is 2.44 L.

The solubility product constant KspK_{sp} tells us the maximum concentration of ions that can coexist in solution at equilibrium. For a sparingly soluble salt like calcium sulphate, this equilibrium is established between the solid and its dissolved ions. When we know KspK_{sp}, we can work backwards to find how much salt dissolves in a given volume of water — or, as this problem asks, how much water we need to dissolve a given mass of salt.

Calcium sulphate dissociates as:

CaSOX4(s)⇌CaX2+(aq)+SOX4X2−(aq)\ce{CaSO4(s) <=> Ca^{2+}(aq) + SO4^{2-}(aq)}

If the molar solubility is ss mol/L, then at equilibrium [CaX2+]=s[\ce{Ca^{2+}}] = s and [SOX4X2−]=s[\ce{SO4^{2-}}] = s.


1. Write the expression for KspK_{sp} and solve for solubility

The solubility product is:

Ksp=[CaX2+][SOX4X2−]=s⋅s=s2K_{sp} = [\ce{Ca^{2+}}][\ce{SO4^{2-}}] = s \cdot s = s^2

Given Ksp=9.1×10−6K_{sp} = 9.1 \times 10^{-6}:

s2=9.1×10−6s^2 = 9.1 \times 10^{-6}

s=9.1×10−6=3.02×10−3 mol/Ls = \sqrt{9.1 \times 10^{-6}} = 3.02 \times 10^{-3} \text{ mol/L}

This is the maximum amount of CaSOX4\ce{CaSO4} that dissolves per litre of water.


2. Convert molar solubility to mass solubility

The molar mass of CaSOX4\ce{CaSO4} is:

M=40+32+4(16)=136 g/molM = 40 + 32 + 4(16) = 136 \text{ g/mol}

So the mass that dissolves in 1 L of water is:

Solubility=3.02×10−3 mol/L×136 g/mol=0.411 g/L\text{Solubility} = 3.02 \times 10^{-3} \text{ mol/L} \times 136 \text{ g/mol} = 0.411 \text{ g/L}


3. Calculate the volume needed to dissolve 1 g …

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