# Solubility problemm

#### JohnDoeDDS

##### Senior Member
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Can you guys explain this please. I'm having trouble following the Kaplan explanation and I understand ther eis a common ion effect ehre. Thanks

The Solubility of CaF2 (Ksp=4X10^-11) in a .1M solution of Ca(NO3)2 is approximately:

A) 4X10^-11 M
B) 2X10^-11 M
C) 1X10^-5 M
D) 5X10^-6 M

#### dat_student

##### Junior Member
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5+ Year Member
JohnDoeDDS said:
Can you guys explain this please. I'm having trouble following the Kaplan explanation and I understand ther eis a common ion effect ehre. Thanks

The Solubility of CaF2 (Ksp=4X10^-11) in a .1M solution of Ca(NO3)2 is approximately:

A) 4X10^-11 M
B) 2X10^-11 M
C) 1X10^-5 M
D) 5X10^-6 M

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

Ksp = 4 * 10^-11 = (0.1)(2S)^2
4 * 10^-10 = 4 S^2

S = 10^-5
C is the correct answer choice

#### Dr. Konfetka

##### Sveta
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Ok..I suck at chem..but here is how I did it.

CaF2-----> Ca+2 + 2F-, make the concentration of ca+2=x, this means that the [2F-]=2x. plugging into the ksp eqation: (x)*(2x)^2=ksp=4*10^-11, solve for x, you get x=10^-11/3 which is the concentraion of the ca+2.

For the common ion affect, Ca(NO3)2-----> ca+2 + 2NO3. so the ksp, or solubility, is equal to x(from above)*(.1, the concentration of the NO3 soln)^2. so, (10^-11/3)*(10^-2)^2=10^-15/3=10^-5.

hope this helps

JohnDoeDDS said:
Can you guys explain this please. I'm having trouble following the Kaplan explanation and I understand ther eis a common ion effect ehre. Thanks

The Solubility of CaF2 (Ksp=4X10^-11) in a .1M solution of Ca(NO3)2 is approximately:

A) 4X10^-11 M
B) 2X10^-11 M
C) 1X10^-5 M
D) 5X10^-6 M

#### keibee82

##### Blue_tooth...
10+ Year Member
JohnDoeDDS said:
Can you guys explain this please. I'm having trouble following the Kaplan explanation and I understand ther eis a common ion effect ehre. Thanks

The Solubility of CaF2 (Ksp=4X10^-11) in a .1M solution of Ca(NO3)2 is approximately:

A) 4X10^-11 M
B) 2X10^-11 M
C) 1X10^-5 M
D) 5X10^-6 M

CaF2 ----> Ca2+ + 2F-

Ksp = [Ca2+][F]^2

4E-11 = (0.1)(2x)^2

x = 1E-5M

#### DonExodus

##### Dentist in Virgin Islands
10+ Year Member
Unsure as to why its 2x² and not x². Isnt the 2 redundant, since its accounted for in the coefficient?

#### dat_student

##### Junior Member
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DonExodus said:
Unsure as to why its 2x² and not x². Isnt the 2 redundant, since its accounted for in the coefficient?
4X^2 (i.e. (2X)^2).

2X because for every mole that is dissociated you get 2 moles of F

#### DonExodus

##### Dentist in Virgin Islands
10+ Year Member
dat_student said:
4X^2 (i.e. (2X)^2).

2X because for every mole that is dissociated you get 2 moles of F
What about the Ksp for something like:
Ag2SO4 <===> 2Ag + SO4- ?

The Ksp there is [Ag+]²[SO4-]. Why is it [Ag]² and not 2[Ag]² like the preceding problem / your example? Im sorry, thats whats confusing me.

Also what about the [Ca]? Why is it not .1+x? Saying its .1 doesnt account for the Ca released from the CF2 to my understanding.

#### slayerdeus

##### Member
10+ Year Member
DonExodus said:
What about the Ksp for something like:
Ag2SO4 <===> 2Ag + SO4- ?

The Ksp there is [Ag+]²[SO4-]. Why is it [Ag]² and not 2[Ag]² like the preceding problem / your example? Im sorry, thats whats confusing me.

Also what about the [Ca]? Why is it not .1+x? Saying its .1 doesnt account for the Ca released from the CF2 to my understanding.

I can't answer your first question. But with the second, you will need to know the principle of common ion effect. If you add ion X to a solution already with that ion, then the concentration of that ion is not felt nearly as much as it would if there was no solution to begin with. That might sound confusing, hopefully someone else can explain it better, but essentially the concentration of what you refer to as x is negligable.

#### dat_student

##### Junior Member
10+ Year Member
5+ Year Member
DonExodus said:
What about the Ksp for something like:
Ag2SO4 <===> 2Ag + SO4- ?

The Ksp there is [Ag+]²[SO4-]. Why is it [Ag]² and not 2[Ag]² like the preceding problem / your example? Im sorry, thats whats confusing me.

....
#1
it should be:

Ag2SO4 <===> 2Ag+ + SO4 2-

#2
[Ag+]²[SO4 2-] simply means (concentration of Ag+)^2(Conc. of SO4 2-)

when you know conc. of Ag+ is 0.1 M you don't need to multiply it by anything (i.e. (0.1)^2). If you start with 0 Ag+ and X M Ag2SO4 because you know for every mole of dissociated Ag2SO4 you get 2 moles of Ag+ you need to multiply by 2 (i.e. (2 * dissociated amnt S)^2))....

OP
J

#### JohnDoeDDS

##### Senior Member
10+ Year Member
7+ Year Member
dat_student said:
#1
it should be:

Ag2SO4 <===> 2Ag+ + SO4 2-

#2
[Ag+]²[SO4 2-] simply means (concentration of Ag+)^2(Conc. of SO4 2-)

when you know conc. of Ag+ is 0.1 M you don't need to multiply it by anything (i.e. (0.1)^2). If you start with 0 Ag+ and X M Ag2SO4 because you know for every mole of dissociated Ag2SO4 you get 2 moles of Ag+ you need to multiply by 2 (i.e. (2 * dissociated amnt S)^2))....
You guys the f*cking best! I love you! I used to crack these problems easily in my gen chem class and thats exactly how I did it. The kaplan explanation just confused me.

But here is a follow up question:
If you know the M of Ag2SO4. I will get Ksp= (2x)^2 * X... And then what? I dont know the KSP I just know the concentration of the Ag2SO4 where will I go from there? Thanks again eveyone you are great!