Chad's GC quiz 5.04

[diene]

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What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8)?

This is the answer:
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2
3.2x10-8 = (0.032)[F-]2

I was doing (x)(2x)^2...can anyone explain this to me? Also, when does one use ICE charts?

 

[circa86]

someone else correct me if i'm wrong b/c its been awhile...you use (x)*(2x^2) when you're trying to figure out the Ksp, not the concentration of one ion in solution. Example.

Molar solubility of BiI3 = 1.32 x 10^-5 M. What is Ksp?
1. BiI3 <---> Bi + 3I
2. Ksp = [Bi]^3
3. Ksp= [x][3x^3]
4. Ksp= 27x^4 = 27(1.32x10^-5M)^4 = 5.4 x 10^-19

In the OP's problem above, you would stop at 2 because you're only trying to observe what one molecule is doing in solution---- [F]^2-----. Remember, when solving for Ksp it is important to incorporate the coefficient in the front of the x; in my problem, the 3 from [3x^3]. Solving for concentration however, you don't need to incorporate the 3 in front of the x.

 

[SN1]

What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8)?

This is the answer:
BaF2(s) &#8594; Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2
3.2x10-8 = (0.032)[F-]2

I was doing (x)(2x)^2...can anyone explain this to me? Also, when does one use ICE charts?

For this problem, you are asked to find the CONCENTRATION of fluoride ion. You are GIVEN the CONCENTRATION of Ba+2 in a form of Ba(NO3)2 so therefore you do NOT need to find the molar solubility (x) as you did above.

so
Ksp = [ Ba2+] [F-]^2

then just plug in everything and solve for [F]

In my opinion, you don't need to use ice charts for ksp problems. Even if it's a common ion effect problem you can just add whatever the common ion is and ignore the variable since it is going to be very small and it will also be a quadratic equation which won't show up on the DAT.

 

[SN1]

someone else correct me if i'm wrong b/c its been awhile...you use (x)*(2x^2) when you're trying to figure out the Ksp, not the concentration of one ion in solution. Example.

Molar solubility of BiI3 = 1.32 x 10^-5 M. What is Ksp?
1. BiI3 <---> Bi + 3I
2. Ksp = [Bi]^3
3. Ksp= [x][3x^3]
4. Ksp= 27x^4 = 27(1.32x10^-5M)^4 = 5.4 x 10^-19

In the OP's problem above, you would stop at 2 because you're only trying to observe what one molecule is doing in solution---- [F]^2-----. Remember, when solving for Ksp it is important to incorporate the coefficient in the front of the x; in my problem, the 3 from [3x^3]. Solving for concentration however, you don't need to incorporate the 3 in front of the x.

When you are solving for concentration. You DO need to incorporate the 3 in front of the x.

the reaction is
BiI3 <---> Bi + 3I

this means for every 1 mole of BiI3 you get 1 mole of Bi ions and 3 moles of Iodide ions.

you do NOT have to CUBE it because that is an expression for finding molar solubility.

So the concentration of iodide would be 3x
NOT 3x^3

 

[Candypaint]

For this problem, you are asked to find the CONCENTRATION of fluoride ion. You are GIVEN the CONCENTRATION of Ba+2 in a form of Ba(NO3)2 so therefore you do NOT need to find the molar solubility (x) as you did above.

so
Ksp = [ Ba2+] [F-]^2

then just plug in everything and solve for [F]

In my opinion, you don't need to use ice charts for ksp problems. Even if it's a common ion effect problem you can just add whatever the common ion is and ignore the variable since it is going to be very small and it will also be a quadratic equation which won't show up on the DAT.

Totally get it now. Gotta watch closly how they ask the questions... molar solubility vs concentration (precipitation)