Will a precipitate form when 0.01M Pb²⁺ is mixed with 0.02M Cl⁻? (Ksp PbCl₂ = 1.7×10⁻⁵)

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Tags Ionic Equilibrium

Question

Will a precipitate of PbCl2\text{PbCl}_2 form when 0.01 M Pb2+\text{Pb}^{2+} solution is mixed with 0.02 M Cl−\text{Cl}^- solution? Given Ksp(PbCl2)=1.7×10−5K_{sp}(\text{PbCl}_2) = 1.7 \times 10^{-5}.

Solution — Step by Step

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

The solubility product expression is:

Ksp=[Pb2+][Cl−]2K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2

Note the exponent of 2 for Cl−\text{Cl}^- because one formula unit of PbCl2\text{PbCl}_2 releases two chloride ions.

The ionic product (or reaction quotient QQ) is calculated using the actual concentrations present in the solution — before any precipitation occurs.

Using the given concentrations:

  • [Pb2+]=0.01 M=10−2 M[\text{Pb}^{2+}] = 0.01\text{ M} = 10^{-2}\text{ M}
  • [Cl−]=0.02 M=2×10−2 M[\text{Cl}^-] = 0.02\text{ M} = 2 \times 10^{-2}\text{ M}
Q=[Pb2+][Cl−]2=(10−2)×(2×10−2)2Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (10^{-2}) \times (2 \times 10^{-2})^2 Q=(10−2)×(4×10−4)=4×10−6Q = (10^{-2}) \times (4 \times 10^{-4}) = 4 \times 10^{-6}

Ksp=1.7×10−5K_{sp} = 1.7 \times 10^{-5}

Q=4×10−6=0.4×10−5Q = 4 \times 10^{-6} = 0.4 \times 10^{-5}

Since Q=4×10−6<Ksp=1.7×10−5Q = 4 \times 10^{-6} < K_{sp} = 1.7 \times 10^{-5}:

Q<KspQ < K_{sp} — the solution is unsaturated with respect to PbCl2\text{PbCl}_2.

Since Q<KspQ < K_{sp}, the solution is unsaturated — the actual ion concentrations are below the saturation point. No precipitate forms.

For precipitation to occur, we would need Q>KspQ > K_{sp}, meaning the ion concentrations exceed what the solubility product allows in solution.

Why This Works

The KspK_{sp} tells us the maximum concentration of ions that can coexist in a saturated solution at equilibrium. When we mix ions together, we calculate QQ (the current ion product) and compare:

  • Q<KspQ < K_{sp}: Unsaturated — can dissolve more solid; no precipitate
  • Q=KspQ = K_{sp}: Saturated — at equilibrium
  • Q>KspQ > K_{sp}: Supersaturated — precipitate forms until QQ drops back to KspK_{sp}

It’s exactly the same logic as comparing QQ to KcK_c to predict the direction of any equilibrium — just applied to solubility.

Alternative Method

We can also approach this by finding the maximum [Cl−][\text{Cl}^-] that can exist without precipitation when [Pb2+]=0.01 M[\text{Pb}^{2+}] = 0.01\text{ M}:

[Cl−]2=Ksp[Pb2+]=1.7×10−510−2=1.7×10−3[\text{Cl}^-]^2 = \frac{K_{sp}}{[\text{Pb}^{2+}]} = \frac{1.7 \times 10^{-5}}{10^{-2}} = 1.7 \times 10^{-3} [Cl−]max=1.7×10−3≈0.041 M[\text{Cl}^-]_{max} = \sqrt{1.7 \times 10^{-3}} \approx 0.041\text{ M}

Since our actual [Cl−]=0.02 M<0.041 M[\text{Cl}^-] = 0.02\text{ M} < 0.041\text{ M}, no precipitation occurs. Same conclusion.

In JEE and CBSE Class 11 ionic equilibrium problems, always write the KspK_{sp} expression carefully, paying attention to the stoichiometric coefficients as exponents. For Ca3(PO4)2\text{Ca}_3(\text{PO}_4)_2, the expression is Ksp=[Ca2+]3[PO43−]2K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 — missing the exponents is the most common error. The Q vs KspK_{sp} comparison logic is identical to the Q vs K comparison in chemical equilibrium.

Common Mistake

The classic error here is writing Q=[Pb2+][Cl−]Q = [\text{Pb}^{2+}][\text{Cl}^-] (without the exponent of 2). The dissolution of PbCl2\text{PbCl}_2 gives two Cl−\text{Cl}^- ions per formula unit, so [Cl−][\text{Cl}^-] must be squared. Without the exponent: Q=0.01×0.02=2×10−4Q = 0.01 \times 0.02 = 2 \times 10^{-4}, which is greater than KspK_{sp} — a completely wrong (and opposite) conclusion. Always match the KspK_{sp} expression to the balanced dissolution equation.

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