Solutions # 5
Chapter 5

Dr. Robert Belford



All the questions on this quiz deal with the following solutions.

Solution A: 40.0 mL of 0.500 M AlCl3
Solution B: 60.0 mL of 1.000 M Ag2SO4

Treat the solubilities of these solutions as would be determined from the solubility rules in the text.   In reality, things are more complicated, as silver forms a AgCl2- ion in high chloride concentrations, which causes the precipitate to redissolve, giving lower yields than predicted.   Also, in reality, Silver sulfate is a weak electrolyte and will not form a 1M solution, but we will treat it as a strong electrolyte, in accordance with our solubility rules.  This is an important concept to understand, that is, these solubility rules are a simplification of things.   During the second semester of this course we will introduce the more advanced concept of Solubility Constants, which will allow us to deal rigorously with the actual concentrations of ionic compounds in aqueous solutions

Question 1
What are the respective aluminum and chloride concentrations in solution A?
(A)  1.0 M, 1.5M
(B)  0.5M, 0.5M
(C)  0.5M, 1.5M
(D)  Not enough information
 
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Question 2

What are the respective silver and sulfate ion concentrations in solution B?



(A)  1.000M, 1.000M
(B)  2.000M, 4.000M
(C)  2.000M, 1.000M
(D)  not enough information
 
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Question 3
Identify any precipitates which would form if the two solutions were mixed.
(A)  aluminum sulfate
(B)  silver chloride
(C)  aluminum sulfate and silver chloride
(D)  not enough information
 
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Question 4
Identify any spectator ions which would result if the two solutions were mixed.
(A)  aluminum and chloride
(B)  silver and sulfate
(C)  aluminum and sulfate
(D)  silver and chloride
 

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Question 5
Write the balanced general (molecular equation) for the reaction which would occur if the two solutions were mixed.
(A)  AlCl3 + Ag2SO4 --> Al2(SO4 )3 + AgCl
(B)  2AlCl3 + 3Ag2SO4 --> Al2(SO4 )3 +6 AgCl
(C)  AlCl3 + 2Ag2SO4 --> Al2(SO4 )3 + 3AgCl
(D)  none of the above
 

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Question 6
Write the total ionic equation for the reaction which would occur if the two solutions were mixed.
(A)  Al+3 + Cl- + Ag+ + SO4-2 --> Al+3 + SO4-2 + AgCl(s)
(B)  2Al+3 + 2Cl- + 3Ag+ + 3SO4-2 --> Al+3 + SO4-2 + 6AgCl(s)
(C)  2Al+3 + 6Cl- + 6Ag+ + 3SO4-2 --> 2 Al+3 + 3SO4-2 + 6AgCl(s)
(D)  none of these
 

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Question 7
Write the net ionic equation for the reaction which would occur if the two solutions were mixed.
(A)  6Ag+ + 6Cl- --> 6AgCl(s)
(B)  Ag+ + Cl- --> AgCl(s)
(C)  2Al+3 + 3SO4-2 --> 2 Al23SO4)3(s)
(D)  none of the above
 

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Question 8
Calculate the number of moles of all ions in solution A.
(A)  Al+3 = 0.5 mol, Cl- = 1.5
(B)  Al+3 = 0.2 mol, Cl- = 0.6
(C)  Al+3 = 0.02mol, Cl- = 0.06
(D)  none of the above
 
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Question 9
Calculate the number of moles of all ions in solution B.
(A)  1 mol sulfate, 2 mol silver
(B)  0.12 mol sulfate, 0.060 mol silver
(C)  0.06 mol sulfate, 0.12 mole silver
(D)  none of the above
 
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Question 10
What mass of precipitate is formed after the two solutions are mixed. Assume complete reaction.
(A)  17g
(B)  0.06g
(C)  8.6g
(D)  none of the above
 
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Question 11
What is the final concentration of the excess reagent after the two solutions are mixed?
(A)  0.06M
(B)  0.60M
(C)  0.12M
(D)  none of the above
 
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Question 12
What is the final concentration of the spectator ions after the two solutions are mixed?
(A)  aluminum is 0.20M, sulfate is 0.60 M
(B)  aluminum is 0.60M, sulfate is 0.60 M
(C)  aluminum is 0.20M, sulfate is 0.20 M
(D)  none of the above
 
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