### Some additional molar solubility problem

Request: if you came here first without studying any of the other problems (or knowing what a Ksp expression is), I'd like to recommend you go study those problems first.

Problem #1: The Ksp of Sn3(BO3)4 is 6.128 x 10¯12. Determine the molar solubility.

Solution:

1) Write the dissociation equation:

Sn3(BO3)4(s) ⇌ 3Sn4+(aq) + 4BO33¯(aq)

2) Write the Ksp expression:

Ksp = [Sn4+]3 [BO33¯]4

3) Solve for the molar solubility:

6.128 x 10¯12 = (3s)3 (4s)4

6.128 x 10¯12 = (27s3) (256s4)

6.128 x 10¯12 = 6912s7

s = 0.007074 M

Note: three tin(IV) ions are produced in solution for every one Sn3(BO3)4 that dissolves. As well, 4 borate ions are produced for every one tin(IV) borate formula unit that dissolves.

Problem #2: Given that the Ksp value for MX, M2X and MX3 are the same, determine the solubility order.

(a) MX > M2X > MX3
(b) MX3 > M2X > MX
(c) M2X > MX3 > MX
(d) MX > MX3 > M2X

Solution:

1) Since the Ksp values for all three are the same, I will assume a Ksp value to use. Since Ksp values are quite small, I should pick a reasonable value. In other words:

Ksp = 1 x 10¯6 is reasonable

Ksp = 1 is not reasonable, because it is too large compared to what actual Ksp values are.

2) What I need to do is calculate each substance's molar solubility:

MX ---> s12 = 1.00 x 10¯6

M2X ---> 4s23 = 1.00 x 10¯6

MX3 ---> 27s34 = 1.00 x 10¯6

If you need to review why the solutions just above are set up the way they are, please go here. The link take you to the explanation for MX. Contained within that file are links to explanations concerning M2X and MX3.

s1 = 0.00100 M <--- molar solubility for MX

s2 = 0.00630 M <--- molar solubility for M2X

s3 = 0.0139 M <--- molar solubility for MX3

4) Sometimes, the answer choices use a 'less than' sign rather than a 'greater than' sign. If the were the case, here are the answer choices:

(a) MX < M2X < MX3
(b) MX3 < M2X < MX
(c) M2X < MX3 < MX
(d) MX < MX3 < M2X

If these were the answers, then choice (a) is the correct answer.

Problem #3: How much water is required to dissolve 25.2 mg of CaF2?

Solution:

1) We need to look up the Ksp of CaF2:

3.9 x 10¯11

Found here.

2) We need to know the solubility of CaF2 in mg/L. To do that, first calculate the molar solubility of CaF2, then convert to g/L and on to mg/L:

CaF2(s) ⇌ Ca2+(aq) + 2F¯(aq)

Ksp = [Ca2+] [F¯]2

3.9 x 10¯11 = (s) (2s)2

4s3 = 3.9 x 10¯11

s = 0.000213633 M

(0.000213633 mol/L) (78.074 g/mol) = 0.01668 g/L

(0.01668 g/L) (1000 mg / 1 g) = 16.68 mg/L

The solution to the g/L value is also given in the link given above.