Question

Write the equations that correspond to the chemical changes that occured on treating each of the carbonate/hydroxide precipitates [1) Fe (III), [2) Cu(II), [3) Zn(II), and 4) Ag (I] with

A) 6M NH3 (if no reaction occurs, write NR)

B) 6M NaOH (If no reaction occurs, write NR)

Answer 1

A) 6M NH3

This reaction occurs because ammonia is a weak base:
NH3(aq) + H2O(l) <--> NH4+(aq) + OH-(aq)
Three OH- then react with the Fe3+ to give a precipitate of Fe(OH)3. So overall, the equation is:
Fe3+(aq) + 3NH3(aq) + 3H2O --> Fe(OH)3(s) + 3NH4+(aq)

Dilute ammonia: you get precipitation:

Cu2+ + 2 NH3 + 2 H2O --> Cu(OH)2 + 2 NH4+

Excess ammonia: complex formation (lovely deep blue):

Cu2+ + 4 NH3 --> Cu(NH3)4 2+

Zn2+ + 4 NH3 --> [Zn(NH3)4](2+)

Ag+(aq) + 2 NH3(aq) ↔ Ag(NH3)2+(aq)

B) 6M NaOH

CuCO3 (aq) + NaOH (aq) -> Cu(OH)2 (s) +Na2CO3 (s)

ZnCO3 + NaOH = Na2[Zn(OH)4] + Na2CO3

AgCO3 + NaOH ----> Na2CO3 + AgOH