1. Solid ammonium sulfide is slowly added to 175 mL of a 0.334 M lead(II) nitrat
ID: 1035873 • Letter: 1
Question
1. Solid ammonium sulfide is slowly added to 175 mL of a 0.334 M lead(II) nitrate solution until the concentration of sulfide ion is 0.0621 M. The percent of lead ion remaining in solution is __%.
2. Solid zinc bromide is slowly added to 150 mL of a 0.255 M potassium phosphate solution until the concentration of zinc ion is 0.0491 M. The percent of phosphate ion remaining in solution is ___ %.
3. Consider the insoluble compound copper(II) carbonate , CuCO3. The copper(II) ion also forms a complex with ammonia . Write a balanced net ionic equation to show why the solubility of CuCO3(s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction.
For Cu(NH3)42+, Kf = 6.8×1012. Use the pull-down boxes to specify states such as (aq) or (s).
+
+
K =
4. Consider the insoluble compound iron(II) carbonate , FeCO3. The iron(II) ion also forms a complex with cyanide ions . Write a balanced net ionic equation to show why the solubility ofFeCO3(s) increases in the presence of cyanide ions and calculate the equilibrium constant for this reaction.
For Fe(CN)64-, Kf = 7.7×1036. Use the pull-down boxes to specify states such as (aq) or (s).
+
+
K =
Table of Solubility Product Constants (Ksp at 25 oC)
(aq)(s)(l)(g)+
(aq)(s)(l)(g) (aq)(s)(l)(g)+
(aq)(s)(l)(g)Explanation / Answer
1) The dissociation of lead sulfide, PbS is given as
PbS (s) <=====> Pb2+ (aq) + S2- (aq)
The solubility product constant of PbS is given as
Ksp = [Pb2+][S2-]
Given [S2-] = 0.0621 M and Ksp = 3.2*10-28, plug in values and obtain
3.2*10-28 = [Pb2+]*(0.0621)
=====> [Pb2+] = (3.2*10-28)/(0.0621) = 5.15*10-27
The solubility of Pb2+ in solution when [S2-] = 0.0621 M is 5.15*10-27 M.
We started with 0.334 M Pb(II) nitrate; therefore, the percentage of Pb2+ remaining in solution = (5.15*10-27 M)/(0.334 M)*100 = 1.54*10-24 % (ans).
2) Consider the dissociation of zinc phosphate Zn3(PO4)2 as
Zn3(PO4)2 (s) <=====> 3 Zn2+ (aq) + 2 PO43- (aq)
The solubility product constant of Zn3(PO4)2 is given as
Ksp = [Zn2+]3[PO43-]2
Given [Zn2+] = 0.0491 M and Ksp = 9.1*10-33, plug in values and obtain
9.3*10-33 = (0.0491)3*[PO43-]2
=====> [PO43-]2 = (9.3*10-33)/(0.0491)3 = 7.8567*10-29
=====> [PO43-] = 8.8638*10-15
The solubility of PO43- in solution when [Zn2+] = 0.0491 M is 8.8638*10-15 M.
We started with 0.255 M potassium phosphate; therefore, the percentage of PO43- remaining in solution = (8.8638*10-15 M)/(0.255 M)*100 = 3.476*10-12 % (ans).
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