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
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 + SO42-
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
What is the equilibrium Zn2+ concentration when 1.36 L of a 0.166 M zinc sulfate solution...
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