Question

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. in which of these solutions precipitation will take place?

Answer 1

For precipitation to take place, it is required that the calculated ionic product exceeds the K_sp value.

Before mixing:

`["S"^(2-)] = 1.0 xx 10^(-19) "M" ["M"^(2+)] = 0.04 "M"`

Volume = - 10 mL             volume = 5 mL

After mixing:

`["S"^(2-)] = ?`                           `["M"^(2+)] = ?`

Volume = (10 + 5) = 15mL     Volume  = 15 mL

`["S"^(2-)] = (1.0 xx 10^(-19) xx 10)/15 = 6.67 xx 10^(-20)  "M"`

`["M"^(2+)] = (0.04 xx 5)/15 = 1.33 xx 10^(-2) "M"`

Ionic Product = `["M"^(2+)]["S"^(2-)]`

`= (1.33 xx 10^(-2))(6.67 xx 10^(-20))`

`= 8.87 xx 10^(-22)`

This ionic product exceeds the K_sp of Zns and CdS. Therefore, precipitation will occur in CdCl₂ and ZnCl₂ solutions.