Question

A solution containing 0.10 M Zn²⁺ and 0.10 M Fe²⁺ is saturated with H₂S (i.e. 0.10 M H₂S). (a) What must the H+ concentration be to separate these ions by selectively precipitating ZnS? Note: ZnS has Ksp=1.2x10^-23; FeS has Ksp=3.7x10^-19. (b) What is the smallest concentration of Zn²⁺ that can be achieved without precipitating any of the Fe²⁺ as FeS?

Answer 1

Given

[Zn2+] = 0.1M

[Fe2+] = 0.1M

[H2S] = 0.1M

Ksp for ZnS = 1.2*10-23

Ksp for FeS = 3.7 * 10-19

a)  In order to precipitate ZnS only not FeS, the ionic product of FeS must be smaller than Ksp of FeS

i.e. For FeS, Kip < Ksp ( for FeS not to be precipitated )

[Fe2+][S2-] < (3.7 *10-19)

(0.1)* [S2-] < (3.7*10-19)

[S2-] <( 3.7 * 10-18)

or the maximum [S2-] = 3.7*10-18 M at which FeS will began to precipitate or upto this concentration no precipitation will occur.

Now, H2S ---------> 2H+ + S2-

From the above equation , it is clear that [H+] is double the [S2-]

So, [H+] for precipitation of ZnS only but not FeS = 2 * [S2-]

= ( 2* 3.7 * 10-18 ) M

= 7.4 * 10-18 M

Thus, Maximum concentration of H+ that can precipitate only Zn2+ ions not Fe2+ is 7.4 * 10-18 M.

b ) As given, Ksp for ZnS is 1.2 * 10-23

i.e. [Zn2+] * [S2-] = 1.2 * 10-23

(0.1) * [S2-] = 1.2 * 10-23

[S2-] = 1.2 * 10-22  M

Thus, minimum concentration of S2- required to precipitate Zn2+ ion as ZnS is 1.2 * 10-22

now, ZnS --------> Zn2+ + S2-

From above equation,

[Zn2+]  = [S2-]

So ,[ Zn2+ ]= 1.2 * 10-22 M

Thus, the smallest concentration of Zn2+ that can be achieved without precipitating any of the Fe2+ ion as FeS is 1.2 * 10-22 M .

Note :

For Subpart (a) :

Maximum concentration of H+ that can precipitate only Zn2+ ions not Fe2+ is 7.4 * 10-18 M.

Minimum concentration of H+ that can precipitate only Zn2+ ions not Fe2+ is ( 2 * 1.2 * 10-22 )M. = 2.4 * 10-22 M.

( As H2S ---------> 2H+ + S2-

and [H+] is double the [S2-] )