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Part B: (19 points) Completion: for each of the following give a brief but complete answer. (8
points) The following table relates to solubility of given ionic compounds, for each open space
state how solubility is changed: increase, decease, remain the same, or cannot be determined. 1.
lonic Increasing the Decreasing the Adding EDTA -1 Adding Cro2 compoundpH to 13 Fe(OH)3
PbCrO4
Solution
Solubility studies for ionic compounds
Fe(OH)3
Solubility decreases by increasing pH at 13 (more basic). As solution becomes more basic, more
of Fe(OH)3 is precipitated out of solution by the reaction of Fe3+ and OH- in solution, thereby
reducing the solubility of salt. Reaction shifts to reactant (left handside) end.
Fe(OH)3(s) <===> Fe3+(aq) + 3OH-(aq)
Solubility increases by decreasing pH to 2 (more acidic). As solution becomes more acidic, OH-
in solution would react with added H+ (acidic) and concentration of OH- on right handisde of
equation reduces. Therefore, more of Fe(OH)3(s) would dissolve to produce OH- and reestablish
equilibrium again.Thereby increasing the solubility of salt at low pH. Reaction shifts to product
(right handside) end.
Fe(OH)3(s) <===> Fe3+(aq) + 3OH-(aq)
OH-(aq) + H+(added) <==> H2O(l)
---
PbCrO4
PbCrO4(s) <==> Pb2+(aq) + CrO4^2-(aq) ---- (1)
CrO4^2-(aq) + 2H+(aq) ---> H2CrO4(aq) [acidic pH to 2]
Pb2+(aq) + 2OH-(aq) <==> Pb(OH)2(s) [basic pH to 13]
As can be seen from above, raising pH to 13 (basic) forms insoluble Pb(OH)2, therefore
insoluble PbCrO4 forms insoluble Pb(OH)2, no change in solubility. Whereas, in acidic pH to 2,
soluble H2CrO4 forms, which increases solubility of PbCrO4.
Addition of EDTA increases solubility as more of Pb2+ in solution forms PbEDTA^2- complex
and reduces concentration of Pb2+ in solution. Therefore more of PbCrO4 dissolves to give
Pb2+ and reestablish equilibrium again
Pb2+ + EDTA <==> PbEDTA^2-
Adding CrO4^2- would increase concentration of CrO4^2- on right handside of equation (1),
thus Pb2+(aq) would react with added excess CrO4^2- to form insoluble PbCrO4. Solubility of
PbCrO4 reduces by adding common ion CrO4^2-.

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Part B- (19 points) Completion- for each of the following give a brief.docx

  • 1. Part B: (19 points) Completion: for each of the following give a brief but complete answer. (8 points) The following table relates to solubility of given ionic compounds, for each open space state how solubility is changed: increase, decease, remain the same, or cannot be determined. 1. lonic Increasing the Decreasing the Adding EDTA -1 Adding Cro2 compoundpH to 13 Fe(OH)3 PbCrO4 Solution Solubility studies for ionic compounds Fe(OH)3 Solubility decreases by increasing pH at 13 (more basic). As solution becomes more basic, more of Fe(OH)3 is precipitated out of solution by the reaction of Fe3+ and OH- in solution, thereby reducing the solubility of salt. Reaction shifts to reactant (left handside) end. Fe(OH)3(s) <===> Fe3+(aq) + 3OH-(aq) Solubility increases by decreasing pH to 2 (more acidic). As solution becomes more acidic, OH- in solution would react with added H+ (acidic) and concentration of OH- on right handisde of equation reduces. Therefore, more of Fe(OH)3(s) would dissolve to produce OH- and reestablish equilibrium again.Thereby increasing the solubility of salt at low pH. Reaction shifts to product (right handside) end. Fe(OH)3(s) <===> Fe3+(aq) + 3OH-(aq) OH-(aq) + H+(added) <==> H2O(l) --- PbCrO4 PbCrO4(s) <==> Pb2+(aq) + CrO4^2-(aq) ---- (1) CrO4^2-(aq) + 2H+(aq) ---> H2CrO4(aq) [acidic pH to 2]
  • 2. Pb2+(aq) + 2OH-(aq) <==> Pb(OH)2(s) [basic pH to 13] As can be seen from above, raising pH to 13 (basic) forms insoluble Pb(OH)2, therefore insoluble PbCrO4 forms insoluble Pb(OH)2, no change in solubility. Whereas, in acidic pH to 2, soluble H2CrO4 forms, which increases solubility of PbCrO4. Addition of EDTA increases solubility as more of Pb2+ in solution forms PbEDTA^2- complex and reduces concentration of Pb2+ in solution. Therefore more of PbCrO4 dissolves to give Pb2+ and reestablish equilibrium again Pb2+ + EDTA <==> PbEDTA^2- Adding CrO4^2- would increase concentration of CrO4^2- on right handside of equation (1), thus Pb2+(aq) would react with added excess CrO4^2- to form insoluble PbCrO4. Solubility of PbCrO4 reduces by adding common ion CrO4^2-.