A solution which is $10^{- 3}M\ each\ in\ Mn^{2 +},\ Fe^{2 +},\ Zn^{2 +}\ and\ Hg^{2 +}$ is treated with $10^{- 16}$ M sulphide ion. If $K_{sp}\ of\ MnS,\ FeS,\ ZnS\ \ and\ HgS\ are\ 10^{- 15},\$ $10^{- 23},\ 10^{- 20}\ and\ 10^{- 54}$ respectively, which one will precipitate first?

Question

Dr. Abubacker · Accepted Answer

(c)
On the basis of precipitation concept an electrolyte will be precipitated firstly whose ionic product is exceeded to its solubility product. In similar ionic conc. of cations, an electrolyte is precipitated firstly for which $K_{sp}$ is lower. In these electrolytes $MnS,\ FeS,\ ZnS\ \ and\ HgS,\ HgS$ is precipitated firstly.

Q. A solution which is $10^{- 3}M\ each\ in\ Mn^{2 +},\ Fe^{2 +},\ Zn^{2 +}\ and\ Hg^{2 +}$ is treated with $10^{- 16}$ M sulphide ion. If $K_{sp}\ of\ MnS,\ FeS,\ ZnS\ \ and\ HgS\ are\ 10^{- 15},\$ $10^{- 23},\ 10^{- 20}\ and\ 10^{- 54}$ respectively, which one will precipitate first?

Explanation

Reviewed by Dr. Abubacker

Want more practice?