Question about Ksp and ICE boxes

[angrybirds3]

What is the minimum concentration of chloride ion

necessary to induce precipitation upon addition of

the silver nitrate solution in the procedure outlined in

the passage?

A. 4.68 x 10–9 M

B. 3.12 x 10–11 M

C. 5.20 x 10–12 M

D. 6.24 x 10–12 M


AgCl (s ) --> Ag+ (aq ) + Cl– (aq ) Ksp = 1.56 10–10


This is from a Kaplan FL (#4, Q32). The answer key wants you to use the definition of KSp and calculate [Cl-] from KSp=[Cl-][Ag+]. It's pretty straight forward after you figure out that the concentration of Ag is 1/30M from the passage. The answer is A, which means that [Cl-] and [Ag+] are different. But according to ICE box, shouldn't both ion concentrations be x so that x^2= KSp????

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[amy_k]

It is not always the case that [Cl-] = [Ag+], just that the PRODUCT of the two ion concentrations equals Ksp. If we had a hypothetical Ksp of '12,' for two ions A+ and B- then the ion concentrations of ion A+ and B- could be

1M and 12M
2M and 6M
3M and 4M...etc, so long as [A+]*[B-] = 12.

Make sense?

 

[sazerac]

What is the minimum concentration of chloride ion

necessary to induce precipitation upon addition of

the silver nitrate solution in the procedure outlined in

the passage?

A. 4.68 x 10–9 M

B. 3.12 x 10–11 M

C. 5.20 x 10–12 M

D. 6.24 x 10–12 M


AgCl (s ) --> Ag+ (aq ) + Cl– (aq ) Ksp = 1.56 10–10


This is from a Kaplan FL (#4, Q32). The answer key wants you to use the definition of KSp and calculate [Cl-] from KSp=[Cl-][Ag+]. It's pretty straight forward after you figure out that the concentration of Ag is 1/30M from the passage. The answer is A, which means that [Cl-] and [Ag+] are different. But according to ICE box, shouldn't both ion concentrations be x so that x^2= KSp????

ICE box is not the appropriate tool to solve this problem. ICE box is for a solution that is Initially in equilibrium, some Changes occur, and then the solution Ends up in equilibrium again. This solution was never initially in silver chloride equilibrium in the first place, since there never was any chloride ion or silver chloride precipitate initially.

The 5 second solution to this problem is to note that Ksp=[Ag+][Cl-], note that [Ag+] is less than one, therefore [Cl-] > Ksp. Only one answer, "A", has an answer larger than Ksp. Boom, done. That's how the MCAT wants you to approach these problems.

 

[chemtopper]

What is the minimum concentration of chloride ion

necessary to induce precipitation upon addition of

the silver nitrate solution in the procedure outlined in

the passage?

A. 4.68 x 10–9 M

B. 3.12 x 10–11 M

C. 5.20 x 10–12 M

D. 6.24 x 10–12 M


AgCl (s ) --> Ag+ (aq ) + Cl– (aq ) Ksp = 1.56 10–10


This is from a Kaplan FL (#4, Q32). The answer key wants you to use the definition of KSp and calculate [Cl-] from KSp=[Cl-][Ag+]. It's pretty straight forward after you figure out that the concentration of Ag is 1/30M from the passage. The answer is A, which means that [Cl-] and [Ag+] are different. But according to ICE box, shouldn't both ion concentrations be x so that x^2= KSp????

Ksp does not define that the ions concentration should be equal.It defines that product of concentration (raised to power of their no of moles in the formula) of ions in the solution.If this product is more than ksp then you can see precipitates formed .Now these concentrations can be same or different but the product should exceed ksp.Now when concentrations are same as in case of AgCl ,then it is called as molar solubility of the ions.However in your case [Ag+] is already in the solution and now when you add [Cl-] then you can see start of the precipitation once the product of [Ag+][Cl-] exceeds Ksp .