Question

Which of the following metals are expected to form precipitate if sulphide ion concentration in their 0.01 M solution is made ${10}^{-16}$ M?
(i) $AS(KspofAS={10}^{-16})$
(ii) $BS(KspofBS={10}^{-21})$
(iii) $WS(KspofWS={10}^{-14})$
(iv) $C_{2}S(Kspof{C}_{2}S={10}^{-18})$
(v) $D_{2}S(KspofD2S={10}^{-21})$

• A. $BS,C_{2}S,D_{2}S$
• B. $BS,C_{2}S$
• C. $AS,WS$
• D. $BS,D_{2}S$
• E. $BS,WS$

Answer 1

Answer: (D)

$BS,D_{2}S$

From given values,

$\left[AS\right]=\left[A^{+2}\right[S^{-2}]=0.01\times {10}^{-16}={10}^{-18}<K_{sp}$

$\left[BS\right]=\left[B^{+2}\right[S^{-2}]=0.01\times {10}^{-16}={10}^{-18}>K_{sp}$

$\left[WS\right]=\left[W^{+2}\right[S^{-2}]=0.01\times {10}^{-16}={10}^{-18}<K_{sp}$

$\left[C_{2}S\right]=\left[C^{+}{]}^2\right[S^{-2}]={0.01}^2\times {10}^{-16}={10}^{-20}<K_{sp}$

$\left[D_{2}S\right]=\left[D^{+}{]}^2\right[S^{-2}]={0.01}^2\times {10}^{-16}={10}^{-20}>K_{sp}$

So only $BS$ and $D_{2}S$ will precipitate in solution.