A solution containing 0-1 M Zn 2- and 0-01 M Cu 2- saturated with H2S- The S 2- concentration is 8-1- 10 -31 M- Will ZnS and CuS precipitate- K sp ZnS-3-0- 10 -23 and K sp CuS-8-0- 10 -4

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Standard XII

Chemistry

Common Ion Effect

A solution co...

Question

A solution containing $0.1 M$ ${Z n}^{2 +}$ and $0.01 M$ ${C u}^{2 +}$ saturated with $H_{2} S$ . The $S^{2 -}$ concentration is $8.1 \times 10^{- 31} M$ . Will $Z n S$ and $C u S$ precipitate?
$K_{s p} (Z n S) = 3.0 \times 10^{- 23}$ and
$K_{s p} (C u S) = 8.0 \times 10^{- 4}$

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Solution

For the precipitation of $M S$ the product of concentration of $M^{2 +}$ and $S^{2 -}$ will be greater than their solubility products.
Now, $[Z n^{2 +}] [S^{2 -}] = 0.1 \times 10^{- 31} \times 8.1 = 0.81 \times 10^{- 31} < K_{s p}$ of $Z n S$ and $[C u^{2 +}] [S^{2 -}] = 0.01 \times 8.1 \times 10^{- 31} = 8.1 \times 10^{- 33} < K_{s p}$ of $C u S$ .
Therefore $Z n S$ and $C u S$ will not precipitate.

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A solution containing 0.1 M Zn2+ and 0.01 M Cu2+ saturated with H2S. The S2− concentration is 8.1×10−31 M. Will ZnS and CuS precipitate?Ksp(ZnS)=3.0×10−23 andKsp(CuS)=8.0×10−4

For the precipitation of MS the product of concentration of M2+ and S2− will be greater than their solubility products.Now, [Zn2+][S2−]=0.1×10−31×8.1=0.81×10−31<Ksp of ZnS and [Cu2+][S2−]=0.01×8.1×10−31=8.1×10−33<Ksp of CuS. Therefore ZnS and CuS will not precipitate.

A solution containing $0.1 M$ ${Z n}^{2 +}$ and $0.01 M$ ${C u}^{2 +}$ saturated with $H_{2} S$ . The $S^{2 -}$ concentration is $8.1 \times 10^{- 31} M$ . Will $Z n S$ and $C u S$ precipitate?
$K_{s p} (Z n S) = 3.0 \times 10^{- 23}$ and
$K_{s p} (C u S) = 8.0 \times 10^{- 4}$