Solving K _{sp}Problems:

Part One - s^{2}Solving K _{sp}Problems:

Part Two - 4s^{3}Solving K _{sp}Problems:

Part Three - 27s^{4}Solving K _{sp}Problems:

Part Four - 108s^{5}Back to Equilibrium Menu

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

**Problem #1:** The log K_{sp} 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 K_{sp} expression:

Th(OH)_{4}⇌ Th^{4+}+ 4OH¯

K_{sp}= [Th^{4+}] [OH¯]^{4}

2) Substitute into the K_{sp} expression and solve:

K_{sp}= 10¯^{50.52}= 3.0 x 10¯^{51}3.0 x 10¯

^{51}= (s) (4s)^{4}256s

^{5}= 3.0 x 10¯^{51}s = 2.6 x 10¯

^{11}M

**Problem #2:** The K_{sp} of Sn_{3}(BO_{3})_{4} is 6.128 x 10¯^{12}. Determine the molar solubility.

**Solution:**

1) Write the dissociation equation:

Sn_{3}(BO_{3})_{4}(s) ⇌ 3Sn^{4+}(aq) + 4BO_{3}^{3}¯(aq)

2) Write the K_{sp} expression:

K_{sp}= [Sn^{4+}]^{3}[BO_{3}^{3}¯]^{4}

3) Solve for the molar solubility:

6.128 x 10¯Note: three tin(IV) ions are produced in solution for every one Sn^{12}= (3s)^{3}(4s)^{4}6.128 x 10¯

^{12}= (27s^{3}) (256s^{4})6.128 x 10¯

^{12}= 6912s^{7}s = 0.007074 M

Solving K _{sp}Problems:

Part One - s^{2}Solving K _{sp}Problems:

Part Two - 4s^{3}Solving K _{sp}Problems:

Part Three - 27s^{4}Solving K _{sp}Problems:

Part Four - 108s^{5}Back to Equilibrium Menu