Question

A solution contains 1.20×10-2 M zinc acetate and 8.67×10-3 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?
[S2-] =____________ M
  

Answer 1

Zn(CH3COO)2(aq) + K2S(aq) ---> ZnS(s) + 2K(CH3COO)
Zn2+(aq) + S2-(aq) ---> ZnS(s)

2CuNO3(aq) + K2S(aq) =======> Cu2S(s) + 2KNO3(aq)

Now we have to convert this equation into ionic equation. From above equation it is obvious that all compounds are dissolved in water except cu2S which is in the form of precipitates.

2Cu+1(aq) + 2NO3-1(aq) + 2K+1(aq) + S-2(aq) =====> Cu2S(s) + 2K+1(aq) + 2NO3-1(aq)

Now we can write ionic equation by removing spectator ions. Spectators ions are those which are present on both sides i.e they are not taking part in reaction. In above case K and NO3 are spectator ions.

2Cu+1(aq) + S-2 (aq) =====> Cu2S(s)

A)

zinc sulphide   ppts first as its solubility is less

B)

         ZnS   > Zn+2 + S-2

         1 :            1           ; 1   mole ratio

         1.2*10^-2 M 1.2*10^-2 M   1.2*10^-2M

so sulphide ions =   1.2*10^-2 M