Question

A solution contains 1.44×10^-2 M silver acetate and 6.44×10^-3 M copper(II) nitrate.
Solid potassium sulfide is added slowly to this mixture.

A. What is the formula of the substance that precipitates first?

formula =

B. What is the concentration of sulfide ion when this precipitation first begins?
[sulfide] =

M

Answer 1

In this, concentration of silver acetate is 1.44* 10-2M

So, [Ag+]= [CH3COO-] = 1.44 * 10-2M

Conc. Of copper nitrate , Cu(NO3)2=6.44 *10-3M

[Cu2+] =6.44* 10-3M , [NO3-] = 2* 6.44*10-3M = 12.88*10-3M

Precipitation occurs when ionic product exceeds the solubility product.

When K2S is added in silver acetate , precipitate of silver sulphide, Ag2S will form and in copper nitrate , precipitate of copper sulphide , CuS will form.

Ionic product of Ag2S = [Ag+]2 [S2-]

= (1.44* 10-2)² [S^2-]

= 2.07 * 10^-4[S^2-]

Ionic product of CuS= [Cu2+] [S2-] = 6.44* 10-3 [S^2-]

Using Ksp values: Ksp of Ag2S =6* 10-51

Ksp of CuS = 6* 10-37

concentration of sulphide ion to precipitate Ag2S given by:

Ksp = [Ag+]² [S2-]  

6 * 10-51 = 2.07 * 10-4[S^2-]

[S^2-] = 6*10-51 /2.07 * 10-4

= 2.9 * 10^-47 M

Conc. Of sulphide ion required to precipitate CuS given by:

Ksp = [Cu²⁺] [S^2-]

6* 10^-37 = 6.44 *10^-3 [S^2-]

[S^2-] = 0.93 * 10^-34M

To precipitate Ag₂S less amount of sulphide ion needed so it will precipitate first.

2.9* 10^-47 M of sulphide ion will be required to precipitate Ag₂S.