GS-1 Physical Sciences Question 22

[student]

it asked: Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater solubility in terms of mol L-1?

Why is the answer Ca(OH)2? I said CaSO4 because there are less particles dissolved, so therefore each "s" has more concentration that is dissolved (since both have the same ksp).

Doesn't having a higher concentration of "s" imply a higher solubility?

[student]

sorry I think i explained #22 quite bad

for instance if ksp was 2

for Ca(OH)^2

(x)(2x)^2 = 2
x = 0.79

for CaSO4

x^2 = 1.41

so wouldnt CaSO4 be more soluble?

[admin]

You are right and wrong. Your logic is exactly right, but the math is the problem.

The key is that Ksp is always a very, very low number. The solubility product usually has an exponent which is negative alot! For example, 10 to the minus 8 to minus 60 would be quite normal.

Now try to work out your example but instead of a Ksp of 2, try 10 to the power of minus 2. You will see that because it is a fraction, you will get the opposite answer.

Keep up the good work!

[admin]

OK, let's work it out:

Given that Ksp = 10e-1

(i) solubility s = sq rt Ksp = 0.32

(ii) s = cube root of (1/4 x Ksp) = cube root of (25 x 10e-3) = 0.29

You would seem to have a point but Ksp's are very very low numbers. And the example above is not representative.

Let's say Ksp = 10e-28

(i) s = 10e-14

(ii) s = cube root of (25 x 10e-30) = 2.9 x 10e-10

Thus the solubility is higher for (ii).

[mcat_premed3832]

If you do not wish to accept the argument that Ksp's tend to be numbers that are incredibly small (in fact, common examples of Ksp range from 10 to the MINUS 25 to 60) then I'm out of explanations. But there is no doubt that the question and answer as written are correct.

[blechz2803]

Hi there...I just did this test, and got the questions wrong. Can you explain this to me?
Why in the world would it matter how small the ksp is or not? and isn't the ksp usually about 10-7 area, not 10-60??
Thanks

[mcat_premed3832]

It's OK, let's take a number from the neighborhood you mentioned: Ksp = 10^-8 (easier to work with than 10^-7)

(i) solubility s = sq rt Ksp = 10^-4

(ii) s = cube root of (1/4 x Ksp) = cube root of (0.25 x 10^-8 ) = cube root of (2.5 x 10^-9) = 1.4 x 10^-3

Thus the solubility is higher for (ii).


{NB: the step creating the exponent "10^-9" was done so a cube root can be taken without a calculator; cube root of 2.5 can simply be estimated as a number above 1 and then there is only one answer; the math is such that it has to be a small number, similar to real Ksp's; otherwise for large numbers, the opposite would be true}

[ozairmh8911]

Why is Ksp not dependent on temperature? Doesn't the solubility change with temperature?

[jellywing_2058]

Temperature is not relevant to this question. You can determine the answer by the fact that Ca(OH)2 gives 3 ions while CaSO4 gives only 2 ions.

[hua8986059]

I got this right thinking along the range of Jellywing_2058, 3 ions = more soluble. Don't need to do the cubic root etc. But now I know how to do that too based on this thread lol..

[shanabona7368]