Question 69 of 69advanced🔧 ApplyNumerical5 marks

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0×10−191.0 \times 10^{-19} M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4_4, MnCl2_2, ZnCl2_2 and CdCl2_2, in which of these solutions precipitation will take place?

Correct Answer

Precipitation will take place in the solutions of ZnCl2_2 and CdCl2_2.

Exercise: EXERCISES | Q: 6.73 | (Page 47)
For More Understanding

Explanation

To determine if precipitation occurs, we calculate the ionic product (QspQ_{sp}) of the metal sulphides after mixing and compare it with their respective solubility product constants (KspK_{sp}). If QspQ_{sp} exceeds KspK_{sp}, precipitation takes place. The KspK_{sp} values for FeS, MnS, ZnS, and CdS from the textbook data are 6.3×10−186.3 \times 10^{-18}, 2.5×10−132.5 \times 10^{-13}, 1.6×10−241.6 \times 10^{-24}, and 8.0×10−278.0 \times 10^{-27} respectively.

Solution Steps

  1. Step 1: Calculate the concentration of metal ions (M2+M^{2+}) after mixing. Total volume = 10 mL+5 mL=15 mL10 \text{ mL} + 5 \text{ mL} = 15 \text{ mL}. Initial concentration of metal ion solutions = 0.04 M0.04 \text{ M}. [M2+]=0.04×515=0.0133 M[M^{2+}] = \frac{0.04 \times 5}{15} = 0.0133 \text{ M}.

  2. Step 2: Calculate the concentration of sulphide ions (S2−S^{2-}) after mixing. Initial concentration of S2−=1.0×10−19 MS^{2-} = 1.0 \times 10^{-19} \text{ M}. [S2−]=1.0×10−19×1015=6.67×10−20 M[S^{2-}] = \frac{1.0 \times 10^{-19} \times 10}{15} = 6.67 \times 10^{-20} \text{ M}.

  3. Step 3: Calculate the ionic product (QspQ_{sp}). Qsp=[M2+][S2−]=(0.0133)×(6.67×10−20)=8.87×10−21 M2Q_{sp} = [M^{2+}][S^{2-}] = (0.0133) \times (6.67 \times 10^{-20}) = 8.87 \times 10^{-21} \text{ M}^2.

  4. Step 4: Compare QspQ_{sp} with KspK_{sp} values. For FeS: Ksp=6.3×10−18K_{sp} = 6.3 \times 10^{-18}. Since Qsp<KspQ_{sp} < K_{sp}, no precipitation. For MnS: Ksp=2.5×10−13K_{sp} = 2.5 \times 10^{-13}. Since Qsp<KspQ_{sp} < K_{sp}, no precipitation. For ZnS: Ksp=1.6×10−24K_{sp} = 1.6 \times 10^{-24}. Since Qsp>KspQ_{sp} > K_{sp}, precipitation occurs. For CdS: Ksp=8.0×10−27K_{sp} = 8.0 \times 10^{-27}. Since Qsp>KspQ_{sp} > K_{sp}, precipitation occurs.

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