Solubility Lesson 3 Calculating Ksp

The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s) ⇌ Ba2+ + CO32- Ba2+ CO32- BaCO3(s) s s s Ksp = [Ba2+][CO32-] Ksp = [s][s] Ksp = s2 Ksp = (5.1 x 10-5)2 Ksp = 2.6 x 10-9

Ksp Solubility Product Saturated solutions- at equilibrium No Units Increasing Temperature increases the Ksp

2. The solubility of PbBr2 is 0.012 M @ 25 0C. Calculate the Ksp. dissociation equation PbBr2(s) ⇌ Pb2+ + 2Br- solubility s s 2s equilibrium expression Ksp = [Pb2+][Br-]2 substitute & solve Ksp = [s][2s]2 Ksp = 4s3 Ksp = 4(0.012)3 Ksp = 6.9 x 10-6 Note that the Br- is doubled and then squared!

3. If 0. 00243 g of Fe2(CO3)3 is required to saturate 100 3. If 0.00243 g of Fe2(CO3)3 is required to saturate 100.0 mL of solution. What is the solubility product? Fe2(CO3)3 ⇌ 2Fe3+ + 3CO32- s 2s 3s Ksp = [Fe3+]2[CO32-]3 s = 0.00243 g x 1 mole 291.6 g Ksp = [2s]2[3s]3 0.100 L Ksp = 108s5 = 8.333 x 10-5 M Ksp = 108(8.333 x 10-5)5 Ksp = 4.34 x 10-19

4. A 200. 0 mL sample of a saturated solution of Mg(OH)2 weighs 222 4. A 200.0 mL sample of a saturated solution of Mg(OH)2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp. Mass of Beaker + Mg(OH) 22.1213 g note sig figs- 4th decimal -Mass of Beaker -22.1200 g Mass of Mg(OH)2 0.0013 g Mg(OH)2⇌ Mg2+ + 2OH- s s 2s 0.0013 g x 1 mole Ksp = [Mg2+][OH-]2 s = 58.3 g = [s][2s]2 = 4s3 0.2000 L = 4(1.1149 x 10-4)3 = 1.1149 x 10-4 M = 5.5 x 10-12

5. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. Titration 2HCl + 1Ba(OH)2 0.02910 L 0.0400 L 0.300 M ? M 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl [Ba(OH)2] = 0.0400 L s = 0.1091 M

Ksp Ba(OH)2 ⇌ Ba2+ + 2OH- s s 2s Ksp = [Ba2+][OH-]2 = [s][2s]2 = 4s3 = 4(0.1091)3 = 5.20 x 10-3