Question 68 of 69intermediate🔧 ApplyNumerical3 marks

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, KspK_{sp} is 9.1×10−69.1 \times 10^{-6}).

Correct Answer

The minimum volume of water required is 2.442.44 L.

Exercise: EXERCISES | Q: 6.72 | (Page 46)
For More Understanding

Explanation

To find the volume of water required to dissolve 1g of calcium sulphate, we first determine its molar solubility (SS) using the given solubility product constant (KspK_{sp}). For calcium sulphate, which dissociates into two ions of equal concentration, Ksp=S2K_{sp} = S^2. Once SS is found in mol L−1^{-1}, it is converted to g L−1^{-1} using the molar mass of CaSO4CaSO_4 (136136 g mol−1^{-1}). Finally, the volume needed to dissolve 1 g is calculated by dividing 1 g by the solubility in g L−1^{-1}.

Solution Steps

  1. Step 1: Write the dissociation equation

    CaSO4(s)⇌Ca2+(aq)+SO42−(aq)CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)

    If molar solubility is SS, then [Ca2+]=S[Ca^{2+}] = S and [SO42−]=S[SO_4^{2-}] = S.

  2. Step 2: Calculate molar solubility (SS)

    Ksp=[Ca2+][SO42−]=(S)(S)=S2K_{sp} = [Ca^{2+}][SO_4^{2-}] = (S)(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=3.016×10−3S = \sqrt{9.1 \times 10^{-6}} = 3.016 \times 10^{-3} mol L−1^{-1}

  3. Step 3: Convert molar solubility to gram solubility Molar mass of CaSO4=40+32+(4×16)=136CaSO_4 = 40 + 32 + (4 \times 16) = 136 g mol−1^{-1} Solubility in g L−1=S×Molar Mass=3.016×10−3×136=0.410^{-1} = S \times \text{Molar Mass} = 3.016 \times 10^{-3} \times 136 = 0.410 g L−1^{-1}

  4. Step 4: Calculate the minimum volume of water required To dissolve 1 g of CaSO4CaSO_4: Volume = MassSolubility (g L−1)=10.410=2.439\frac{\text{Mass}}{\text{Solubility (g L}^{-1})} = \frac{1}{0.410} = 2.439 L Thus, the minimum volume of water required is 2.442.44 L.

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