Three solutions are mixed together to form a single solution. One contains 0.2 mol \rm mol Pb(CH 3 COO) 2 \rm Pb(CH_3COO)_2, the second contains 0.1 mol \rm mol Na 2 S \rm Na_2S, and the third contains 0.1 mol \rm mol CaCl 2 \rm CaCl_2. Write the precipitation reaction.
Pb^2+(aq) + S^2-(aq) ==> PbS(s)
To determine the precipitation reaction, we need to know if any insoluble salts can form when the three solutions are mixed together. Looking at the combination of the three solutions, we have:
Solution 1: 0.2 mol of Pb(CH3COO)2 (lead(II) acetate)
Solution 2: 0.1 mol of Na2S (sodium sulfide)
Solution 3: 0.1 mol of CaCl2 (calcium chloride)
Now, let's examine the possible combinations of these ions:
Pb2+ (from Pb(CH3COO)2) can react with S2- (from Na2S) to form PbS (lead(II) sulfide). PbS is an insoluble salt and can precipitate.
Ca2+ (from CaCl2) can react with S2- (from Na2S) to form CaS (calcium sulfide). CaS is also an insoluble salt and can precipitate.
Based on this information, the precipitation reactions that can occur are:
Pb(CH3COO)2 (aq) + Na2S (aq) → PbS (s) + 2CH3COONa (aq)
CaCl2 (aq) + Na2S (aq) → CaS (s) + 2NaCl (aq)
So, when the three solutions are mixed together, lead(II) sulfide (PbS) and calcium sulfide (CaS) will precipitate out of the solution.