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Chemistry Question on Equilibrium

What is the value of KspK_{sp} for bismuth sulphide (Bi2S3)(Bi_2S_3) which has a solubility of 1.0×10−15mol/L1.0 \times 10^{-15} mol / L at 25∘C25^{\circ}C ?

A

1.08×10−731.08\times 10^{-73}

B

1.08×10−741.08\times 10^{-74}

C

1.08×10−721.08\times 10^{-72}

D

1.08×10−751.08\times 10^{-75}

Answer

1.08×10−731.08\times 10^{-73}

Explanation

Solution

The system initially contains H2OH_2O and solid Bi2S3Bi_2S_3,
which dissolves as follows
Bi2S3(s)<=>2Bi3+(aq)+3S2(aq)Bi_2S_3(s) {<=>} 2Bi^{3+} (aq) + 3S^2 (aq)
Therefore, Ksp=[Bi3+]2[S2−]3K_{sp} = [Bi^{3+}]^2 [S^{2-}]^3
Since, no Bi3+Bi^{3+} and S2−S^{2-} were present in solution before the Bi2S3Bi_2S_3 dissolved,
[Bi3+]0=[S2−]0=0[Bi^{3+}]_0 = [S^{2-}]_0 = 0
Thus, the equilibrium concentration of these ions will be determined by the amount of salt that dissolves to reach equilibrium
1.0×10−15 mol/LBi2S3(s)1.0 \times 10^{-15}\,mol/LBi_2S_3(s)
−>2(1.0×10−15mol/L)Bi3+(aq){->} 2(1.0 \times 10^{-15}mol/L) Bi_{3+}(aq)
+3(1.0×10−16mol/L)S2(ag)+ 3(1.0 \times 10^{-16} mol /L)S^2(ag)
The equilibrium reaction are
[Bi3+]=[Bi3+]0+[Bi^{3+}] = [Bi^{3+}]_0 + change =0+2.0×10−15 mol/L= 0 + 2.0 \times 10^{-15}\, mol/L
[S2−]=[S2−]+[S^{2- }] = [S^{2-}] + change =0+3.0×10−15 mol/L= 0 + 3.0 \times 10^{-15}\, mol/ L
Ksp=[Bi3−]2[S2−]3K_{sp} = [Bi^{3-}]^2[S^{2-}]^3
=(2.0×10−15)2(3.0×10−15)2=(2.0 \times 10^{-15})^2 (3.0 \times 10^{-15})^2
=1.1×10−73= 1.1 \times 10^{-73}