### 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.

 25.2 mg 16.68 mg ––––––– = ––––––– x 1 L

x = 1.5 L (answer to two sig figs, based on the Ksp)

Problem #4: Two forms of barium oxalate (called A and B just below) have different Ksp values at 18 °C:

 (A)  BaC2O4 ⋅ 2H2O      1.2 x 10¯7 (B)  BaC2O4 ⋅ 1⁄2H2O     2.2 x 10¯7

Describe what will happen if 1.0 g of A and 1.0 g of B are added to 100. mL of water under 1.00 atm of pressure.

Solution:

1) Write the chemical equation and the Ksp expression:

BaC2O4(s) ⇌ Ba2+(aq) + C2O42¯(aq)

Ksp = [Ba2+] [C2O42¯]

2) The 1.0 g of A and the 1.0 g of B are added to the water and the entire system is stirred. What happens?

The instant before anything dissolves and ionizes, both A and B "see" pure water, so each starts dissolving.

A tiny amount of A and a tiny amount of B dissolve, then ionize. This results in a value for [Ba2+] and a value for [C2O42¯].

However, let us stipulate that Q (the reaction quotient) that results from these tiny amounts of A and B that dissolve is less than the Ksp.

Both A and B "see" a solution that can accept more dissolved ions. This means a tiny bit more BaC2O4 (some from A and some from B) dissolves, then ionizes, thus pushing the Q to a larger value, getting it even closer to the Ksp.

More of A and B dissolve and ionize, until Q equals the Ksp. At this point (where Q = Ksp), an equilibrium is established. This equilibrium is between the undissolved A, the undissolved B, and the ions in solution. The values of [Ba2+] and [C2O42¯] will stop increasing and will remain constant.

3) However, which Ksp value has been reached at equilibrium? The answer is the Ksp value for A, the less soluble form of barium oxalate.

Let us take the view of B, the more soluble form. B "sees" a solution that can accept more ions since its Ksp has not been reached. So, some of B dissolves and ionizes.

However, A now "sees" a solution where its Ksp has been exceeded. As a result, some ions reform into the solid state and the Ksp value for A is restored.

4) Let us assume that equal amounts of A and B dissolve to produce the Ksp value of A. One can perform a calculation to determine the tiny amount of A and of B that do dissolve. This exercise is left to the reader.

5) Calcium sulfate also shows this difference in solubilities between two hydrates. Barium iodate shows the difference between the anhydrate and one hydrate (as does calcium iodate). Magnesium carbonate (and strontium iodate) show differences in solubility between two hydrates and the anhydrate.