Question

What is the value of $K_{sp}$ for bismuth sulphide $(B{i}_2{S}_3)$ 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 $H_{2}O$ and solid $B{i}_2{S}_3$,
which dissolves as follows
$B{i}_2{S}_3(s)\rightleftharpoons 2B{i}^{3+}(aq)+3S^2(aq)$
Therefore, $K_{sp}=[B{i}^{3+}{]}^2[S^{2-}{]}^3$
Since, no $B{i}^{3+}$ and $S^{2-}$ were present in solution before the $B{i}_2{S}_3$ dissolved,
$[B{i}^{3+}{]}_0=[S^{2-}{]}_0=0$
Thus, the equilibrium concentration of these ions will be determined by the amount of salt that dissolves to reach equilibrium
$1.0\times 10^{-15}mol/LB{i}_2{S}_3(s)$
$\to 2(1.0\times 10^{-15}mol/L)B{i}_{3+}(aq)$
$+3(1.0\times 10^{-16}mol/L)S^2(ag)$
The equilibrium reaction are
$[B{i}^{3+}]=[B{i}^{3+}{]}_0+$ change $=0+2.0\times 10^{-15}mol/L$
$[S^{2-}]=[S^{2-}]+$ change $=0+3.0\times 10^{-15}mol/L$
$K_{sp}=[B{i}^{3-}{]}^2[S^{2-}{]}^3$
$=(2.0\times 10^{-15}{)}^2(3.0\times 10^{-15}{)}^2$ $=1.1\times 10^{-73}$