# qchem question again

Discussion in 'DAT Discussions' started by dontbam, Jul 22, 2006.

1. ### dontbam Member

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hey guys can you help me with this question from kaplan:

Q: The solubility of CaF2 (Ksp = 4 x 10^-11) in a 0.1M solution of Ca(No3)2 is approximately:

A: 4 x 10^-11 M
B: 2 x 10^-11 M
C: 1 x 10^-5 M
D: 5 x 10^-6 M
E: None of these bad boys

answer is C, can anyone explain?

3. ### dat_student Junior Member

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CaF2 <--> Ca2+ + 2F-
......(approx. 0.1) + 2S

(A) Ksp = (0.1)(2S)^2 = 4 * 10^-1 * S^2
(B) Ksp = 4 * 10 ^-11
(A) & (B) >> 4 * 10^-1 * S^2 = 4 * 10 ^-11

S^2 = 10^-10
S = 10^-5 (Answer choice C)

4. ### dontbam Member

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wait i dont really get how you come up with these numbers, can you explain it ? thx

5. ### Envision Envisioning...

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CaF2 -> Ca + 2F

from this, set up your solubility equation. what you are given is the solubility constant (ksp), and the solubility of Ca (.1 M). this is testing your knowledge of the common ion effect, because there is already .1M of Ca from Ca(NO3)2, it will be less soluble than if it was in water. anyway, the equation is as follow:

ksp= [Ca][2x]^2 the 2x is because of the stoichmetric coeficient, notice there is 2F for every 1 Ca.

from here, you already know ksp, and Ca is ~.1M. although there is a small amount due to CaF2, it plays a very small role so most will be from 0.1 M from Ca(no3)2.

4x10^-11=[.1][2x]^2....x=1x10^-5 (c)

hope that helps.