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JEE Chemistry 2023 Question with Solution

At 298 K298 \, \text{K}, the solubility of silver chloride in water is 1.434×10−3 g L−11.434 \times 10^{-3} \, g \, L^{-1}. The value of −log⁡Ksp-\log K_{sp} for silver chloride is:

(Given mass of Ag is 107.9 g mol−1107.9 \, g \, mol^{-1} and mass of Cl is 35.5 g mol−135.5 \, g \, mol^{-1})

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: Solubility of AgCl is 1.434×10−3 g L−11.434 \times 10^{-3} \, \text{g L}^{-1} at 298 K298 \, \text{K}.

Find: The value of −log⁡Ksp-\log K_{sp}.

The dissociation of silver chloride in water is

AgCl(s)⇌Ag+(aq)+Cl−(aq)\text{AgCl(s)} \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)

For a salt of type AB\text{AB}, if molar solubility is SS, then

Ksp=S2K_{\text{sp}} = S^2

Molar mass of AgCl is

107.9+35.5=143.4 g mol−1107.9 + 35.5 = 143.4 \, \text{g mol}^{-1}

Hence molar solubility is

S=1.434×10−3143.4 mol L−1=1×10−5 mol L−1S = \frac{1.434 \times 10^{-3}}{143.4} \, \text{mol L}^{-1} = 1 \times 10^{-5} \, \text{mol L}^{-1}

Now,

Ksp=(1×10−5)2=10−10K_{\text{sp}} = (1 \times 10^{-5})^2 = 10^{-10}

Therefore,

−log⁡Ksp=−log⁡(10−10)=10-\log K_{\text{sp}} = -\log(10^{-10}) = 10

So, the required numerical value is 1010.

Using the extracted solution steps

Given: AgCl\text{AgCl} dissolves as

AgCl(s)→Ag+(aq.)+Cl−(aq.)\text{AgCl(s)} \rightarrow \text{Ag}^+(aq.) + \text{Cl}^-(aq.)

Find: −log⁡Ksp-\log K_{sp}.

From the provided working,

Ksp=S2K_{sp} = S^2

Using the molar mass 143.4 g mol−1143.4 \, \text{g mol}^{-1}, the molar solubility becomes

S=1.434×10−3143.4=1×10−5S = \frac{1.434 \times 10^{-3}}{143.4} = 1 \times 10^{-5}

Therefore,

Ksp=S2=(1×10−5)2=10−10K_{sp} = S^2 = (1 \times 10^{-5})^2 = 10^{-10}

So,

−log⁡Ksp=10-\log K_{sp} = 10

The correct answer is 1010.

Common mistakes

  • Using the given solubility directly in g L−1\text{g L}^{-1} inside KspK_{sp} is incorrect, because KspK_{sp} requires molar concentration. First convert solubility to mol L−1\text{mol L}^{-1} using the molar mass of AgCl.

  • Taking Ksp=SK_{sp} = S is wrong for AgCl. Since AgCl⇌Ag++Cl−\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-, both ion concentrations are SS, so Ksp=[Ag+][Cl−]=S2K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S^2.

  • Forgetting to add the atomic masses of Ag and Cl to get the molar mass of AgCl leads to an incorrect molar solubility. Use 107.9+35.5=143.4 g mol−1107.9 + 35.5 = 143.4 \, \text{g mol}^{-1}.

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