Solving K _{sp}Problems:

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

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

Part Four - 108s^{5}Solving K _{sp}Problems:

Part Five - 256s^{5}Back to Equilibrium Menu

The generic problem is this:

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 (Part One and Part Two) first, so the presentations here will be a bit abbreviated.

Keep in mind that the coefficient in front of each ion does two things:

1) it puts a power on the concentration which represents that particular ion

2) it puts a coefficient in front of the 's'

**Example #1:** Iron(III) hydroxide, K_{sp} = 2.64 x 10¯^{39}

**Solution:**

Fe(OH)_{3}(s) ⇌ Fe^{3+}(aq) + 3OH¯(aq)K

_{sp}= [Fe^{3+}] [OH¯]^{3}2.64 x 10¯

^{39}= (s) (3s)^{3}27s

^{4}= 2.64 x 10¯^{39}s = 9.94 x 10¯

^{11}M

How's that for abbreviated? Remember, the 3s comes from the fact that there are three hydroxides produced for every Fe(OH)_{3} that dissolves.

**Example #2:** Silver phosphate, K_{sp} = 8.88 x 10¯^{17}

**Solution:**

Ag_{3}PO_{4}(s) ⇌ 3Ag^{+}(aq) + PO_{4}^{3}¯(aq)K

_{sp}= [Ag^{+}]^{3}[PO_{4}^{3}¯]8.88 x 10¯

^{17}= (3s)^{3}(s)27s

^{4}= 8.88 x 10¯^{17}s = 4.26 x 10¯

^{5}M

**Example #3:** Silver arsenate, K_{sp} = 1.03 x 10¯^{22}

**Solution:**

Ag_{3}AsO_{4}(s) ⇌ 3Ag^{+}(aq) + AsO_{4}¯(aq)K

_{sp}= [Ag^{+}]^{3}[AsO_{4}¯]1.03 x 10¯

^{22}= (3s)^{3}(s)27s

^{4}= 1.03 x 10¯^{22}s = 1.40 x 10¯

^{6}M

Comment: at the high school level, arsenate may be one of those unusual polyatomic ions, one that you didn't learn in the nomenclature section. Be aware that teachers are sometimes known to teach material using well-known formulas and then ask an unusual one on the test.

**Example #4:** Aluminum hydroxide, K_{sp} = 3.5 x 10¯^{24}

**Solution:**

Al(OH)_{3}(s) ⇌ Al^{3+}(aq) + 3OH¯(aq)K

_{sp}= [Al^{3+}] [OH¯]^{3}3.5 x 10¯

^{24}= (s) (3s)^{3}27s

^{4}= 3.5 x 10¯^{24}s = 6.0 x 10¯

^{7}M

**Example #5:** Yttrium iodate, Y(IO_{3})_{3}, K_{sp} = 1.12 x 10¯^{10}

**Solution:**

Y(IO_{3})_{3}(s) ⇌ Y^{3+}(aq) + 3IO_{3}¯(aq)K

_{sp}= [Y^{3+}] [IO_{3}¯]^{3}1.12 x 10¯

^{10}= (s) (3s)^{3}1.12 x 10¯

^{10}= 27s^{4}s

^{4}= 4.148 x 10¯^{12}s = 1.43 x 10¯

^{3}M (to three sig figs)

**Example #6:** CeF_{3}, K_{sp} = 8 x 10¯^{16}

The solution is left to the reader.

**Example #7:** Calculate the molar solubility of Co(OH)_{3} (K_{sp} = 2.51 x 10¯^{43}) in moles per liter

**Solution:**

1) Write the chemical equation for the dissolving of cobalt(III) hydroxide. Then, write the K_{sp} expression for Co(OH)_{3}.

Co(OH)_{3}(s) ⇌ Co^{3+}(aq) + 3OH¯(aq)K

_{sp}= [Co^{3+}] [OH¯]^{3}

2) Solve the K_{sp} expression for the molar solubility of Co(OH)_{3}:

2.51 x 10¯^{43}= (s) (3s)^{3}27s

^{4}= 2.51 x 10¯^{43}s = 1.58 x 10¯

^{11}M

Solving K _{sp}Problems:

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

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

Part Four - 108s^{5}Solving K _{sp}Problems:

Part Five - 256s^{5}Back to Equilibrium Menu