Question

What is the value of $ K_{sp} $ for bismuth sulphide $ (Bi_2S_3) $ which has a solubility of $ 1.0 \times 10^{-15} mol / L $ at $ 25^{\circ}C $ ?

• A. $ 1.08\times 10^{-73} $
• B. $ 1.08\times 10^{-74} $
• C. $ 1.08\times 10^{-72} $
• D. $ 1.08\times 10^{-75} $

Answer 1

Answer: (A)

The system initially contains $H_2O$ and solid $Bi_2S_3$,
which dissolves as follows
$Bi_2S_3(s) \rightleftharpoons 2Bi^{3+} (aq) + 3S^2 (aq)$
Therefore, $K_{sp} = [Bi^{3+}]^2 [S^{2-}]^3$
Since, no $Bi^{3+}$ and $S^{2-}$ were present in solution before the $Bi_2S_3$ dissolved,
$[Bi^{3+}]_0 = [S^{2-}]_0 = 0$
Thus, the equilibrium concentration of these ions will be determined by the amount of salt that dissolves to reach equilibrium
$1.0 \times 10^{-15}\,mol/LBi_2S_3(s)$
$\rightarrow 2(1.0 \times 10^{-15}mol/L) Bi_{3+}(aq)$
$+ 3(1.0 \times 10^{-16} mol /L)S^2(ag)$
The equilibrium reaction are
$[Bi^{3+}] = [Bi^{3+}]_0 +$ change $= 0 + 2.0 \times 10^{-15}\, mol/L$
$[S^{2- }] = [S^{2-}] +$ change $= 0 + 3.0 \times 10^{-15}\, mol/ L$
$ K_{sp} = [Bi^{3-}]^2[S^{2-}]^3 $
$=(2.0 \times 10^{-15})^2 (3.0 \times 10^{-15})^2$ $ = 1.1 \times 10^{-73}$