Solving K _{sp}Problems:

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

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

Part Three - 27s^{4}Back to Equilibrium Menu

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:** Bismuth sulfide, K_{sp} = 1.82 x 10¯^{99}

**Solution:**

Bi_{2}S_{3}(s) ⇌ 2Bi^{3+}(aq) + 3S^{2}¯(aq)K

_{sp}= [Bi^{3+}]^{2}[S^{2}¯]^{3}1.82 x 10¯

^{99}= (2s)^{2}(3s)^{3}108s

^{5}= 1.82 x 10¯^{99}s = 7.00 x 10¯

^{21}M

Note how 's' is the moles of the Bi_{2}S_{3} that dissolved.

There may also be a bit of a calculator problem. When you divide the K_{sp} by 108, the answer is 1.685 x 10¯^{101}. Some older calculators (as well as some low-end newer calculators) are not equipped to handle three digit exponents. If you have one of those, you can (1) use a spreadsheet on your computer, (2) access a calculator on-line, or (3) go shopping.

**Example #2:** Copper(II) phosphate, K_{sp} = 1.93 x 10¯^{37}

**Solution:**

Cu_{3}(PO_{4})_{2}(s) ⇌ 3Cu^{2+}(aq) + 2PO_{4}^{3}¯(aq)K

_{sp}= [Cu^{2+}]^{3}[PO_{4}^{3}¯]^{2}1.93 x 10¯

^{37}= (3s)^{3}(2s)^{2}108s

^{5}= 1.93 x 10¯^{37}s = 1.78 x 10¯

^{8}M

**Example #3:** Iron(III) sulfide, K_{sp} = 1 x 10¯^{88}

**Solution:**

1) Here's the dissociation equation:

Fe_{2}S_{3}(s) ⇌ 2Fe^{3+}(aq) + 3S^{2}¯(aq)

2) The K_{sp} expression:

K_{sp}= [Fe^{3+}]^{2}[S^{2}¯]^{3}

3) Solve the K_{sp} expression

1 x 10¯^{88}= (2s)^{2}(3s)^{3}108s

^{5}= 1 x 10¯^{88}s = 9.847 x 10¯

^{19}M

4) When you round the answer off to one sig figs, you get:

s = 1 x 10¯^{18}M

**Example #4:** Magnesium phosphate, K_{sp} = 9.86 x 10¯^{25}

**Solution:**

Mg_{3}(PO_{4})_{2}(s) ⇌ 3Mg^{2+}(aq) + 2PO_{4}^{3}¯(aq)K

_{sp}= [Mg^{2+}]^{3}[PO_{4}^{3}¯]^{2}9.86 x 10¯

^{25}= (3s)^{3}(2s)^{2}9.86 x 10¯

^{25}= 108s^{5}s = 6.20 x 10¯

^{6}M

**Example #5:** Magnesium arsenate, K_{sp} = 2.1 x 10¯^{20}

**Solution:**

Mg_{3}(AsO_{4})_{2}(s) ⇌ 3Mg^{2+}(aq) + 2AsO_{4}^{3}¯(aq)K

_{sp}= [Mg^{2+}]^{3}[AsO_{4}^{3}¯]^{2}2.1 x 10¯

^{20}= (3s)^{3}(2s)^{2}2.1 x 10¯

^{20}= 108^{5}s = 0.000045 M

Here are two more substances, the dissociation equations, and their K_{sp} values.Determine the molar solubilities, following the pattern shown in the above five examples.

Substance Dissociation Equation K _{sp}Ni _{3}(PO_{4})_{2}Ni _{3}(PO_{4})_{2}⇌ 3Ni^{2+}+ 2PO_{4}^{3}¯4.74 x 10¯ ^{32}Pb _{3}(PO_{4})_{2}Ni _{3}(PO_{4})_{2}⇌ 3Ni^{2+}+ 2PO_{4}^{3}¯1.0 x 10¯ ^{54}

Solving K _{sp}Problems:

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

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

Part Three - 27s^{4}Back to Equilibrium Menu