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Exercises · 6.73

Q.The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10⁻¹⁹ M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?

Telangana TsbieTextbookSubjective· 3mImportance★★★★★est
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The common ion effect from HCl suppresses H₂S dissociation, giving a fixed [S2−]=1.0×10−19 M[S^{2-}] = 1.0 \times 10^{-19}\ \text{M}. After mixing, the diluted metal ion concentration is 0.0133 M0.0133\ \text{M}. Precipitation occurs only when the ionic product [M2+][S2−][M^{2+}][S^{2-}] exceeds the metal sulfide’s KspK_{sp}. Only CdS precipitates here.

The key to this problem is understanding that the sulphide ion concentration is not free to change — it is fixed by the common ion effect. In a 0.1 M HCl solution saturated with H₂S, the high [H+][H^+] from HCl pushes the equilibrium

H2S⇌2H++S2−H_2S \rightleftharpoons 2H^+ + S^{2-}

far to the left. The result is a very low, constant [S2−]=1.0×10−19 M[S^{2-}] = 1.0 \times 10^{-19}\ \text{M} — a value given directly in the problem. This is the concentration before any dilution.

Now, when we mix 10 mL of this sulphide solution with 5 mL of a metal salt solution, both the sulphide ion and the metal ion get diluted. We must calculate the new concentrations after mixing, then compare the ionic product [M2+][S2−][M^{2+}][S^{2-}] with the solubility product KspK_{sp} of each metal sulphide.

Step-by-step

1. Find the diluted concentration of sulphide ion

Total volume after mixing = 10 mL+5 mL=15 mL10\ \text{mL} + 5\ \text{mL} = 15\ \text{mL}.

Using C1V1=C2V2C_1V_1 = C_2V_2:

[S2−]after=(1.0×10−19 M)(10 mL)15 mL=1.0×10−1815=6.67×10−20 M[S^{2-}]_{\text{after}} = \frac{(1.0 \times 10^{-19}\ \text{M})(10\ \text{mL})}{15\ \text{mL}} = \frac{1.0 \times 10^{-18}}{15} = 6.67 \times 10^{-20}\ \text{M}

2. Find the diluted concentration of each metal ion

Each metal salt solution is 0.04 M. After mixing 5 mL of it into 15 mL total:

[M2+]after=(0.04 M)(5 mL)15 mL=0.215=0.0133 M[M^{2+}]_{\text{after}} = \frac{(0.04\ \text{M})(5\ \text{mL})}{15\ \text{mL}} = \frac{0.2}{15} = 0.0133\ \text{M}

This is the same for all four salts.

3. Calculate the ionic product for each case

Ionic product Q=[M2+][S2−]=(0.0133)(6.67×10−20)Q = [M^{2+}][S^{2-}] = (0.0133)(6.67 \times 10^{-20})

Q=8.87×10−22Q = 8.87 \times 10^{-22}

This is the same for all four metal ions because both concentrations are identical before considering KspK_{sp}.

4. Compare with the solubility products

We need the KspK_{sp} values for the metal sulphides. From standard data:

Metal SulphideKspK_{sp}
FeS6×10−186 \times 10^{-18}
MnS2.5×10−102.5 \times 10^{-10}
ZnS1.2×10−231.2 \times 10^{-23}
CdS1.0×10−281.0 \times 10^{-28}
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