Chem Aq. Equilibria...Please Help!

Question

Consider a salt of composition, A2B3. For this salt, pKsp = 19.94. What is the logarithm of the molar concentration of this salt?

I've seen others explain it...but not to proper understanding of where the calculations are coming from...please explain! Thank you in advance !

Answer 1

If the salt has composition A2B3 then in solution, what we care about since we are given the "solubility product":

A2B3 --> 2A3+  +  3B2-  (plus 3 ox-state for cation and minus 2 for anion)

What really matters is the moles of each, we have 2 moles of A in solution per mole of A2B3 and 3 moles of B in solution per mole of A2B3.

Based on this information, we can set up the solubility product, which is the concentration of product raised to their respective powers (based on # of moles) over the concentration of reactants (solid):

Ksp = [A3+]^2*[B2-]^3 ; Now!

Ksp = [2x]^2*[3x]^3 ; where x is the solubility of A2B3, so we get two moles of A and 3 of B from A2B3--Hence the integers before X.

Ksp = 4x^2 * 27x^3 = 108x^5

Now, you were given pKsp.  You will see "p" before many terms in chemistry and usually it means take the -log of the value (ie. pH = -log[H+] or pKa = -log(Ka)).

so the Ksp given is 10^-19.94 or 1.15 x 10^-20

Finally, we have 1.15 x 10^-20 = 108x^5:

x = 4 x 10^-5 M

Since it is not clear which logarithm is needed, I'll use base 10.

log (4 x 10^-5) = -4.39

Answer 2

If your salt is not so soluble the solubility product is

Ksp = [A]^2.[B]^3

and

pKsp = -logKsp = -2log[A] - 3log[B] = 19.94    equation 1

why take log anyway? what happens next depends on what you want? I don't really understand what you are asking.

My guess would be that since one mol of the salt gives 2 mol A and 3 mol B in solution

[A] = 2[A2B3]   and

[B] = 3[A2B3]

substituting for [A] and [B] in equation 1 gives you an expression with one variable that can be solved for log[A2B3].

19.96 = -2log 2[A2B3] - 3log 3[A2B3]