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NCERT Exemplar · Q35

Q.Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (Ksp of PbCl2 = 3.2 × 10^-8, atomic mass of Pb = 207 u).

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The solubility of PbCl₂ is found from its Ksp, then the volume needed to dissolve 0.1 g is calculated using molar mass and the solubility in mol/L. The required volume is 0.179 L (or 179 mL).

Concept First: Why Solubility Product Works

Lead(II) chloride is a sparingly soluble salt. When it dissolves, it establishes an equilibrium with its ions:

PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)

The solubility product constant, KspK_{sp}, is the equilibrium constant for this dissolution. For a saturated solution, the product of the ion concentrations (each raised to its stoichiometric coefficient) equals KspK_{sp}. This is the key: at saturation, the solution holds the maximum possible amount of dissolved salt, and that maximum is fixed by KspK_{sp}.

If we let the molar solubility of PbCl₂ be ss mol/L, then in a saturated solution:

  • [Pb2+]=s[\text{Pb}^{2+}] = s
  • [Cl−]=2s[\text{Cl}^-] = 2s

So:

Ksp=[Pb2+][Cl−]2=s⋅(2s)2=4s3K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s \cdot (2s)^2 = 4s^3

This relation is the bridge between the equilibrium constant and the actual amount we can dissolve.

Step-by-Step Solution

1. Find the molar solubility ss from KspK_{sp}.

We have Ksp=3.2×10−8K_{sp} = 3.2 \times 10^{-8}. Using Ksp=4s3K_{sp} = 4s^3:

4s3=3.2×10−84s^3 = 3.2 \times 10^{-8}

s3=3.2×10−84=8.0×10−9s^3 = \frac{3.2 \times 10^{-8}}{4} = 8.0 \times 10^{-9}

s=8.0×10−93=2.0×10−3 mol/Ls = \sqrt[3]{8.0 \times 10^{-9}} = 2.0 \times 10^{-3} \text{ mol/L}

So, in one litre of saturated solution, 2.0×10−32.0 \times 10^{-3} moles of PbCl₂ are dissolved.

Tip

The cube root of 8.0×10−98.0 \times 10^{-9} is 2.0×10−32.0 \times 10^{-3} because (2×10−3)3=8×10−9(2 \times 10^{-3})^3 = 8 \times 10^{-9}. Always check the exponent: 10−910^{-9} has a cube root of 10−310^{-3}.

2. Calculate the molar mass of PbCl₂.

Atomic masses: Pb = 207 u, Cl = 35.5 u.

Molar mass of PbCl2=207+2(35.5)=207+71=278 g/mol\text{Molar mass of PbCl}_2 = 207 + 2(35.5) = 207 + 71 = 278 \text{ g/mol}

3. Find the number of moles in 0.1 g of PbCl₂.

Moles=massmolar mass=0.1278≈3.597×10−4 mol\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.1}{278} \approx 3.597 \times 10^{-4} \text{ mol} …

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