Two unusual molar solubility problems that didn't fit anywhere else

Solving Ksp Problems:
Part One - s2
Solving Ksp Problems:
Part Two - 4s3
Solving Ksp Problems:
Part Three - 27s4
Solving Ksp Problems:
Part Four - 108s5
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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 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


Problem #2: 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.
Solving Ksp Problems:
Part One - s2
Solving Ksp Problems:
Part Two - 4s3
Solving Ksp Problems:
Part Three - 27s4
Solving Ksp Problems:
Part Four - 108s5
Back to Equilibrium Menu