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Solubility Lesson 3 Calculating Ksp

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 BaCO 3 is 5.1 x 10 -5 M @ 25 0 C. Calculate the solubility product or Ksp.

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Solubility Lesson 3 Calculating Ksp

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  1. Solubility Lesson 3 Calculating Ksp

  2. The Molar Solubilityis 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.

  3. The Molar Solubilityis 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. Ba2+ CO32- BaCO3(s) BaCO3(s)⇌ Ba2+ + CO32- s s s Ksp = [Ba2+][CO32-] Ksp = [s][s] Ksp = s2 Ksp = (5.1 x 10-5)2 Ksp = 2.6 x 10-9

  4. Ksp Solubility Product Saturated solutions- at equilibrium No Units Increasing Temperatureincreasesthe Ksp

  5. 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 doubledand then squared!

  6. 3.If0.00243 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

  7. 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)20.0013 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

  8. 5. 40.00 mL of a saturatedBa(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 x 1mole Ba(OH)2 0.02910 L HCl x 0.300 moles 2 moles HCl 1 L [Ba(OH)2] = 0.0400 L s= 0.1091 M

  9. 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

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