help! easy solubility product question

[msu08]

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this is an easy one... these ones always confuse me for some reason though.
CaF2 is added to a .1M Ca(NO3) solution. At what concentration of F- will CaF2 begin to precipitate? Ksp of CaF2 is 4x 10^-11.

the answer is 2x10^-5 can anyone lay out the steps? Thanks!!!

 

[joonkimdds]

I am keep getting 1x10^-5 instead of 2x10^-5.

 

[yankees27th]

That question is from the Kaplan mid-term. I'm looking at it right now and the answer is 1x10^-5. You either looked at it wrong or you have an old version of something that has a typo. If you still need the question explained let me know.

 

[joonkimdds]

yay~ i got it right~! 🙂

errrrrrr i can't believe I spent last 40 min trying to figure out how it's 2x~ instead of 1x~ 🙁

 

[msu08]

heyy sorry guys i have an old test so it's a typo. my first guess was 1x 10^-5 as well. would anyone actually like to go thru the problem and explain it? i tend to look at the answers and guess for these...

 

[joonkimdds]

Use Joon's law. (brb)

 

[yankees27th]

First get the equation:
CaF2-->Ca + 2F
If CaF2 is x, then Ca is x and F is 2x
Ksp=(x)(2x)^2=4x10^-11 BUT you must take into account the Ca from the Ca(NO3)2 (which is .1M)
So the new equation is:
4x10^-11 = (x+.1)(4x^2)
Now, x going to be a small number, so in the term "x+.1", just get rid of x.
The equation is now:
4x10^-11=(.1)(4x^2)
Solve for x and you should have your answer.

 

[joonkimdds]

^ yeah~ that's the Joon's law 😛

 

[yankees27th]

Haha I guess I beat you to it Joon!

 

[TheBoondocks]

this is an easy one... these ones always confuse me for some reason though.
CaF2 is added to a .1M Ca(NO3) solution. At what concentration of F- will CaF2 begin to precipitate? Ksp of CaF2 is 4x 10^-11.

the answer is 2x10^-5 can anyone lay out the steps? Thanks!!!

I haven't had chem in 4 years. I just began review for mcat.

4*10^-11=(.1+x)(2x)^2 .1 is a lot bigger and so x can be ignored. In your head, this step is skipped.

4*10^-11=.1*4x^2 Divide by .1= 4*10^-10=4x^2

Divide by 4= 1*10^-10=x^2

Square root= 1*10^-5

this would by my rational. In ksp, all that matters is the product. You don't double x, because in this case x is F-.

 

[TheBoondocks]

I haven't had chem in 4 years. I just began review for mcat.

4*10^-11=(.1+x)(2x)^2 .1 is a lot bigger and so x can be ignored. In your head, this step is skipped.

4*10^-11=.1*4x^2 Divide by .1= 4*10^-10=4x^2

Divide by 4= 1*10^-10=x^2

Square root= 1*10^-5

this would by my rational. In ksp, all that matters is the product. You don't double x, because in this case x is F-.

Ha, I never submitted my answer and came back from study break and was too slow. I have a bunch of pre dent friends. Good luck in your studies.

 

[dleodyddlek]

actually 1*10^-5 will be the moles of CaF2 that will dissolve. But the question is asking for the concentration of F-, not CaF2. Since you have 2 F- for every mole of CaF2, your F- concentration will be 2*10^-5.

 

[yankees27th]

actually 1*10^-5 will be the moles of CaF2 that will dissolve. But the question is asking for the concentration of F-, not CaF2. Since you have 2 F- for every mole of CaF2, your F- concentration will be 2*10^-5.

Woops. You're right. Thanks.

 

[dg7903]

I mean, all the steps in calculation above except for getting the square root value, I can do it without a calculator, but how do you guys get x value from 4.0*10^-11= X^2 in your head?
I can't use a calculator on the DAT right?
Help please!!!!!