Question

A solution which is $10^{ - 3 }$ M each in $Mn^2, \, Fe^{ 2 + }, \, Zn^{ 2 + }$ and $Hg^{ 2 + }$ is treated with $10^{ - 1 6 } M \, sulphide \, ion,$ If $K_{ sp }$ of $MnS, FeS, ZnS$ and $HgS$ are $10^{ - 15 }, \, 10^{ - 13 }, \, 10^{ - 20} \, and \, 10^{ - 34 }$ respectively, which one will precipitate first?

• A. FeS
• B. MgS
• C. HgS
• D. ZnS

Answer 1

Answer: (D)

ZnS

Answer 2

Minimum $S^2$ concentration would be required for precipitation of least soluble HgS.
For HgS, $S^{ 2 - }$ required for precipitation is
$[ S^{ 2 - } ] = \frac{ K}{ [ Hg^{ 2 + } ] } = \frac{ 10^{ - 54 }}{ 10^{ - 3 }} = 10^{ - 51 }$ M