Ksp question.

[Mistro4321]

Advertisement - Members don't see this ad

The Ksp for the CaCl2 is 7.1×10^−4. What is the concentration of the Ca2+ ion in a saturated solution?

Could someone help me out with this question. Since the question is asking for the concentration of an ion we wouldn't be solving for x (molar solubtility) right?

So that means that to solve this we would have the equation Ksp=[Ca+][Cl-]^2 and not Ksp=(x)(2x)^2?

So is the answer (7.1×10^−4)^(1/3) and not sqrt[(7.1×10^−4)/4]?

 

[theleatherwalle]

ksp=4x^3

 

[ktran17]

[Ca2+] = 2x

as stated by chad, he said that you can use the molar solubility to solve for concentrations as long as the respective ions only come from the original solid.

 

[GoBlue24]

The Ksp for the CaCl2 is 7.1×10^−4. What is the concentration of the Ca2+ ion in a saturated solution?

Could someone help me out with this question. Since the question is asking for the concentration of an ion we wouldn't be solving for x (molar solubtility) right?

So that means that to solve this we would have the equation Ksp=[Ca+][Cl-]^2 and not Ksp=(x)(2x)^2?

So is the answer (7.1×10^−4)^(1/3) and not sqrt[(7.1×10^−4)/4]?

Could you post the answer? I'm pretty sure you're right. Ksp is just an equilibrium expression so it would should be 7.1x10^4 = [Ca+][Cl-2]^2 = [X][X]^2 = [X]^3 and solve for X.

IF given the molar solubility (which you're not), then you could do 7.1X10^4 = [X][2X]^2 and figure out what 2X is to get the conc. of Cl- (but that's not the case here).

 

[Mistro4321]

The qvault answer is [(7.1×10^−4)/4]^(1/3). I guess im confused because of a question on one of chad's quizzes:

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

Solution:
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]^2
3.2x10-8 = (0.032)[F-]^2
(1.0x10-6)^1/2=[F-]
[F-]=1.0x10^-3M

If you look at the solution the fluoride ion concentration was not multiplied by its coefficient.

 

[Mistro4321]

[Ca2+] = 2x

as stated by chad, he said that you can use the molar solubility to solve for concentrations as long as the respective ions only come from the original solid.

I guess in Chad's questions that the fluoride ion that we are to solve for is not part of the original solid so you can't use it as "x". The multiplier can't therefore be a factor. Am I right?

 

[GoBlue24]

I agree with you. Someone else posted a question similar to this and I was confused as well.

Based on the answer you gave, that X shouldn't be the conc. of Cl- but rather the Ca+ (and how much of the ionic solid dissolves, 1 to 1 mol ratio)..... 😕

The qvault answer is [(7.1×10^−4)/4]^(1/3). I guess im confused because of a question on one of chad's quizzes:

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

Solution:
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]^2
3.2x10-8 = (0.032)[F-]^2
(1.0x10-6)^1/2=[F-]
[F-]=1.0x10^-3M

If you look at the solution the fluoride ion concentration was not multiplied by its coefficient.

 

[ktran17]

Sorry guys I made a mistake. I made Ca2+ = 2x its supposed to be only x. therefore, my method does agree with datqvault's answer. as stated before, the reason why you're able to use the molar solubility because the ions are only coming from the original solid and is saturated.
ksp= [ca2+][Cl-]
ksp = [x] [2x]^2
x= [(7.1x 10^-4)/4] ^ 1/3
where x = [Ca2+]

 

[ktran17]

I guess in Chad's questions that the fluoride ion that we are to solve for is not part of the original solid so you can't use it as "x". The multiplier can't therefore be a factor. Am I right?

exactly! Also when you're not using the multiplier, they have to give you the concentration of the other ion or else you will have too many variables =)

I hope this helps everyone!