JEE Challenger
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ChemEasyAdv 2013 • Paper 2

Solubility of Silver Chromate in Silver Nitrate Solution

Calculate the solubility (expressed in mol/L\text{mol/L}) of Ag2CrO4\text{Ag}_2\text{CrO}_4 inside a 0.1 M AgNO30.1\text{ M } \text{AgNO}_3 solution, provided that KspK_{\text{sp}} for Ag2CrO4\text{Ag}_2\text{CrO}_4 is 1.1×10−121.1 \times 10^{-12} at 298 K298\text{ K}.

Options

A

1.1×10−111.1 \times 10^{-11}

B

1.1×10−101.1 \times 10^{-10}

Correct
C

1.1×10−121.1 \times 10^{-12}

D

1.1×10−91.1 \times 10^{-9}

Topics & Concepts

EquilibriumSolubility Product

Step-by-Step Solution

To find the solubility of silver chromate (Ag2CrO4\text{Ag}_2\text{CrO}_4) in a 0.1 M AgNO30.1\text{ M } \text{AgNO}_3 solution, we consider the contribution of silver ions (Ag+\text{Ag}^+) from both the fully dissociated strong electrolyte AgNO3\text{AgNO}_3 and the sparingly soluble salt Ag2CrO4\text{Ag}_2\text{CrO}_4.

  1. Dissociation of AgNO3\text{AgNO}_3: AgNO3(aq)⟶Ag+(aq)+NO3−(aq)\text{AgNO}_3(aq) \longrightarrow \text{Ag}^+(aq) + \text{NO}_3^-(aq) Since AgNO3\text{AgNO}_3 is a strong electrolyte and fully dissociates, the concentration of Ag+\text{Ag}^+ provided by AgNO3\text{AgNO}_3 is: [Ag+]initial=0.1 M[\text{Ag}^+]_{\text{initial}} = 0.1\text{ M}

  2. Equilibrium of Ag2CrO4\text{Ag}_2\text{CrO}_4: Let ss be the molar solubility of Ag2CrO4\text{Ag}_2\text{CrO}_4 in mol/L\text{mol/L}. Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq)

    At equilibrium, the concentration of the ions will be: [CrO42−]=s M[\text{CrO}_4^{2-}] = s\text{ M} [Ag+]=(0.1+2s) M[\text{Ag}^+] = (0.1 + 2s)\text{ M}

  3. Solubility Product Expression: The solubility product expression for Ag2CrO4\text{Ag}_2\text{CrO}_4 is: Ksp=[Ag+]2[CrO42−]K_{\text{sp}} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}]

    Substituting the equilibrium concentrations and the given KspK_{\text{sp}} value: 1.1×10−12=(0.1+2s)2×s1.1 \times 10^{-12} = (0.1 + 2s)^2 \times s

  4. Simplification: Since KspK_{\text{sp}} is very small (1.1×10−121.1 \times 10^{-12}), the value of ss will be extremely small compared to 0.1 M0.1\text{ M}. Thus, we can apply the approximation: 0.1+2s≈0.10.1 + 2s \approx 0.1

    Substituting this back into the equation: 1.1×10−12=(0.1)2×s1.1 \times 10^{-12} = (0.1)^2 \times s 1.1×10−12=10−2×s1.1 \times 10^{-12} = 10^{-2} \times s s=1.1×10−1210−2=1.1×10−10 mol/Ls = \frac{1.1 \times 10^{-12}}{10^{-2}} = 1.1 \times 10^{-10}\text{ mol/L}

Thus, the solubility of Ag2CrO4\text{Ag}_2\text{CrO}_4 in 0.1 M AgNO30.1\text{ M } \text{AgNO}_3 solution is 1.1×10−10 mol/L1.1 \times 10^{-10}\text{ mol/L}.

Correct Option: (B)

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