### Solving Ksp Problems: Part Five - 256s5

The generic problem is:

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

I'm going to assume you've gone through the other files first, so the presentations here will be a bit abbreviated.

Example #1: The Ksp of Sn(OH)4 has been determined to be 1 x 10¯57. Calculate its molar solubility.

Solution:

1) Write the dissociation equation for Sn(OH)4 and its Ksp expression:

Sn(OH)4 ⇌ Sn4+ + 4OH¯
Ksp = [Sn4+] [OH¯]4

2) Substitute into the Ksp expression and solve:

1 x 10¯57 = (s) (4s)4

256s5 = 1 x 10¯57

s = 1 x 10¯12 M (to one significant figure)

The value for s is the molar solubility of the tin(IV) ion. Since every Sn(IV) ion comes from a dissolved Sn(OH)4 formula unit, the value for s is the molar solubility of Sn(OH)4.

Example #2: Given that the Ksp of Ti(OH)4 is 7 x 10¯53, determine its solubility in grams per liter of saturated solution.

Solution:

1) Write the dissociation equation for Ti(OH)4 and its Ksp expression:

Ti(OH)4 ⇌ Ti4+ + 4OH¯
Ksp = [Ti4+] [OH¯]4

2) Substitute into the Ksp expression and solve:

7 x 10¯53 = (s) (4s)4

256s5 = 7 x 10¯53

s = 1.222845571 x 10¯11 M

The value for s is the molar solubility of the titanium(IV) ion. Since every Ti(IV) ion comes from a dissolved Ti(OH)4 formula unit, the value for s is the molar solubility of Ti(OH)4.

3) Determine concentration in grams per liter:

(1.222845571 x 10¯11 mol/L) (115.8946 g/mol) = 1 x 10¯9 g/L (to one sig fig)

Example #3: The log Ksp for thorium(IV) hydroxide has been determined to be -50.52 ± 0.08. What is the molar solubility for Th(OH)4?

Solution:

1) Write the dissociation equation for Th(OH)4 and its Ksp expression:

Th(OH)4 ⇌ Th4+ + 4OH¯
Ksp = [Th4+] [OH¯]4

2) Substitute into the Ksp expression and solve:

Ksp = 10¯50.52 = 3.0 x 10¯51

3.0 x 10¯51 = (s) (4s)4

256s5 = 3.0 x 10¯51

s = 2.6 x 10¯11 M