please help. Thanks In this step, you start with three drops each of solutions o

Question

please help. Thanks In this step, you start with three drops each of solutions of Group II cations: Pb^2+, Bi^3+, Cu^2+ and Cd^2+. You then add (NH_4)_2 S(aq) to form a precipitate. What precipitate is formed from the reaction between Pb^2+ and (NH_4)_2 S? What precipitate is formed from the reaction between Bi^3+ and (NH_4)_2 S? What precipitate is formed from the reaction between Cu^2+ and (NH_4)_2 S? What precipitate is formed from the reaction between Cd^2+ and (NH_4)_2S? What are the four balanced chemical equations for the four above reactions?

Explanation / Answer

To know the precipitates formed, you need to know the solubility rules. Then, you'll know by that what kind of precipitate will form:

Now according to these rules, let's see the precipitate that form in each of this reactions (I will put that in a reaction, in that way, I'll answer the e part too)

a) Pb2+(aq) + (NH4)2S(aq) = PbS(s) + 2NH4+(aq)
b) 2Bi3+(aq) + 3(NH4)2S(aq) = Bi2S3(s) + 6NH4+(aq)
c) Cu2+(aq) + (NH4)2S(aq) = CuS(s) + 2NH4+(aq)
d) Cd2+(aq) + (NH4)2S(aq) = CdS(s) + 2NH4+(aq)

Hope this helps

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