Question

What is the equilibrium Zn²⁺ concentration when 1.36 L of a 0.166 M zinc sulfate solution are mixed with 1.43 L of a 0.286 M ammonium sulfide solution?

[Zn²⁺] =  M

Answer 1

The reaction of zinc sulfate (ZnSO4) and ammonium sulfide (NH₄)₂S is

ZnSO4_(aq) + (NH₄)₂S_(aq) ----------> ZnS_(s) + (NH₄)₂SO4_(aq)

This equation indicates that  when we react 1 mole aqueous ZnSO4 adn 1 mole (NH₄)₂S, it produces 1 mole insoluble ZnS and 1 mole aqueous (NH₄)₂SO4

Now the ZnSO4 present in the reactant side , being soluble in water can be converted to Zn+2. But the ZnS present in the product, being insoluble cant be converted into Zn+2

Thus after the complete reaction, the ZnSO4 left will only produce Zn+2

Now given ZnSO4 taken = 1.36 L of a 0.166 M

And (NH₄)₂S taken = 1.43 L of a 0.286 M

Now after mixing the two, the final volume = 1.36 L + 1.43 L = 2.79‬ L

Thus the new initial concentration of ZnSO4 = M1V1/V2 = 0.166 M * 1.36 L / 2.79‬ L = 0.321 M

Similarly new initial concentration of (NH₄)₂S = M1V1/V2 = 0.286 M * 1.43 L / 2.79‬ L = 0.146 M

Thus the concentration at equilibrium is

Reaction	ZnSO4	(NH4)2S	ZnS(s) + (NH4)2SO4(aq)
Initial	0.321 M	0.146 M	0 + 0
Change	-  0.146 M	- 0.146 M	+0.146 M  + 0.146 M
Equilibrium	0.175	0	0.146 M  + 0.146 M

That is after the complition of reaction, amount of ZnSO4 left at equilibrium = 0.175 M

Now this ZnSO4 will dissociate as   ZnSO4 -----------> Zn⁺² + SO₄^2-

It indicates that from 1 mole of ZnSO4, we get 1 mole of Zn⁺²

Thus form 0.175 M of ZnSO4, we get 0.175 M of Zn⁺²

Hence finally at equilibrium, [Zn²⁺] =  0.175 M