Solubility

[EgyptianDoc]

So my math is not that great and I was wondering if I could get some help solving for the solubility of BaF2

Equation: BaF2 (s) <--> Ba2+ (aq) + 2F- (aq)
Ksp=2.4x10^-5
Ksp=[products]/[reactants]=[Ba2+][F-]^2

2.4x10^-5=(x)(2x)^2
2.4x10^-5=4x^3

Now my question is how do you quickly figure out the answer for x?

A: x=1.8x10^-2

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[sleepy425]

So my math is not that great and I was wondering if I could get some help solving for the solubility of BaF2

Equation: BaF2 (s) <--> Ba2+ (aq) + 2F- (aq)
Ksp=2.4x10^-5
Ksp=[products]/[reactants]=[Ba2+][F-]^2

2.4x10^-5=(x)(2x)^2
2.4x10^-5=4x^3

Now my question is how do you quickly figure out the answer for x?

A: x=1.8x10^-2

So you set it up right, so now it's pretty easy to divide 2.4x10^-5 by 4 in your head, so you get 6x10^-6. Now, you can't take the cube root of it in your head, but you know the answer has to be bigger than the cube root of 10^-6, which is 10^-2. You also know it has to be less than the cube root of 8x10^-6, which is 2x10^-2 (I just picked 8 because it is the smallest perfect cube after 1). So you know it's gotta be between 1 and 2x10^-2, and it is very unlikely that there would be multiple answer choices on the MCAT that would be that close together.

 

[EgyptianDoc]

So you set it up right, so now it's pretty easy to divide 2.4x10^-5 by 4 in your head, so you get 6x10^-6. Now, you can't take the cube root of it in your head, but you know the answer has to be bigger than the cube root of 10^-6, which is 10^-2. You also know it has to be less than the cube root of 8x10^-6, which is 2x10^-2 (I just picked 8 because it is the smallest perfect cube after 1). So you know it's gotta be between 1 and 2x10^-2, and it is very unlikely that there would be multiple answer choices on the MCAT that would be that close together.

Thanks Sleepy! Appreciate the help