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Solubility Lesson 3 Calculating Ksp. The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature.

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Presentation on theme: "Solubility Lesson 3 Calculating Ksp. The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature."— Presentation transcript:

1 Solubility Lesson 3 Calculating Ksp

2 The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature.

3 The Molar Solubility is the molarity required to saturate of 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.

4 The Molar Solubility is the molarity required to saturate of 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. Ba 2+ CO 3 2- BaCO 3(s)

5 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- Ba 2+ CO 3 2- BaCO 3(s)

6 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ba 2+ CO 3 2- BaCO 3(s)

7 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ba 2+ CO 3 2- BaCO 3(s)

8 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ksp = [s][s] Ba 2+ CO 3 2- BaCO 3(s)

9 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ksp = [s][s] Ksp = s 2 Ba 2+ CO 3 2- BaCO 3(s)

10 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ksp = [s][s] Ksp = s 2 Ksp = (5.1 x 10 -5 ) 2 Ba 2+ CO 3 2- BaCO 3(s)

11 The Molar Solubility is the molarity required to saturate of 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. BaCO 3(s) ⇌ Ba 2+ + CO 3 2- s s s Ksp = [Ba 2+ ][CO 3 2- ] Ksp = [s][s] Ksp = s 2 Ksp = (5.1 x 10 -5 ) 2 Ksp = 2.6 x 10 -9 Ba 2+ CO 3 2- BaCO 3(s)

12 Ksp Solubility Product Special Keq Saturated solutions No Units Only Changes with Temperature

13 2.The solubility of PbBr 2 is 0.012 M @ 25 0 C. Calculate the Ksp. dissociation equation PbBr 2(s) ⇌ Pb 2+ + 2Br - solubility ss2s equilibrium expressionKsp = [Pb 2+ ][Br - ] 2 substitute & solveKsp = [s][2s] 2 Ksp = 4s 3 Ksp = 4(0.012) 3 Ksp = 6.9 x 10 -6 Note that the Br - is doubled and then squared!

14 3.If 0.00243 g of Fe 2 (CO 3 ) 3 is required to saturate 100.0 mL of solution. What is the solubility product? Fe 2 (CO 3 ) 3 ⇌ 2Fe 3+ +3CO 3 2- s2s3s s=0.00243 g x 1 mole 291.6 g 0.100 L =8.333 x 10 -5 M Ksp = [Fe 3+ ] 2 [CO 3 2- ] 3 Ksp = [2s] 2 [3s] 3 Ksp = 108s 5 Ksp = 108(8.333 x 10 -5 ) 5 Ksp = 4.34 x 10 -19

15 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) 2 22.1213 g -Mass of Beaker-22.1200 g Mass of Mg(OH) 2 0.0013 gnote sig figs s=0.0013 gx1 mole 58.3 g 0.2000 L =1.1149 x 10 -4 M Mg(OH) 2 ⇌ Mg 2+ + 2OH - ss2s Ksp=[Mg 2+ ][OH - ] 2 = [s][2s] 2 = 4s 3 =4(1.1149 x 10 -4 ) 3 =5.5 x 10 -12

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