Solveeit Logo
Login

Question

Question: The sulphide ion concentration $[S^{2-}]$ in saturated $H_2S$ solution is $1 \times 10^{-22}$. Which...

The sulphide ion concentration [S2−][S^{2-}] in saturated H2SH_2S solution is 1×10−221 \times 10^{-22}. Which of the following sulphides should be quantitatively precipitated by H2SH_2S in the presence of dil. HClHCl

A

1.4×10−161.4 \times 10^{-16}

B

1.2×10−221.2 \times 10^{-22}

C

8.2×10−468.2 \times 10^{-46}

D

5.0×10−345.0 \times 10^{-34}

Answer

(III) and (IV)

Explanation

Solution

For a sulphide MSMS to precipitate, the ionic product Qsp=[M2+][S2−]Q_{sp} = [M^{2+}][S^{2-}] must be greater than its solubility product KspK_{sp}.
The sulphide ion concentration [S2−][S^{2-}] in the saturated H2SH_2S solution is given as 1×10−221 \times 10^{-22}.

The question asks which sulphides should be "quantitatively precipitated". This implies that the concentration of the metal ion in solution should be reduced to a very low level. A common criterion for quantitative precipitation is that the metal ion concentration should be reduced to 10−5M10^{-5} M or 10−6M10^{-6} M or less.

Let's assume that for quantitative precipitation, the concentration of the metal ion [M2+][M^{2+}] in the solution should be reduced to at least 1×10−6M1 \times 10^{-6} M.
At equilibrium, Ksp=[M2+][S2−]K_{sp} = [M^{2+}][S^{2-}].
If we want [M2+][M^{2+}] to be 1×10−6M1 \times 10^{-6} M or less, then the KspK_{sp} must satisfy:
Ksp≤(1×10−6)×(1×10−22)K_{sp} \leq (1 \times 10^{-6}) \times (1 \times 10^{-22})
Ksp≤1×10−28K_{sp} \leq 1 \times 10^{-28}

Now, let's compare the given solubility products with this threshold value:

(I) Ksp=1.4×10−16K_{sp} = 1.4 \times 10^{-16}
Since 1.4×10−16>1×10−281.4 \times 10^{-16} > 1 \times 10^{-28}, sulphide (I) will NOT be quantitatively precipitated.

(II) Ksp=1.2×10−22K_{sp} = 1.2 \times 10^{-22}
Since 1.2×10−22>1×10−281.2 \times 10^{-22} > 1 \times 10^{-28}, sulphide (II) will NOT be quantitatively precipitated.

(III) Ksp=8.2×10−46K_{sp} = 8.2 \times 10^{-46}
Since 8.2×10−46<1×10−288.2 \times 10^{-46} < 1 \times 10^{-28}, sulphide (III) WILL be quantitatively precipitated.
(Specifically, the maximum [M2+][M^{2+}] that can remain in solution for sulphide (III) is Ksp/[S2−]=(8.2×10−46)/(1×10−22)=8.2×10−24MK_{sp}/[S^{2-}] = (8.2 \times 10^{-46}) / (1 \times 10^{-22}) = 8.2 \times 10^{-24} M, which is very low.)

(IV) Ksp=5.0×10−34K_{sp} = 5.0 \times 10^{-34}
Since 5.0×10−34<1×10−285.0 \times 10^{-34} < 1 \times 10^{-28}, sulphide (IV) WILL be quantitatively precipitated.
(Specifically, the maximum [M2+][M^{2+}] that can remain in solution for sulphide (IV) is Ksp/[S2−]=(5.0×10−34)/(1×10−22)=5.0×10−12MK_{sp}/[S^{2-}] = (5.0 \times 10^{-34}) / (1 \times 10^{-22}) = 5.0 \times 10^{-12} M, which is very low.)

Therefore, sulphides (III) and (IV) should be quantitatively precipitated.