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Monday, April 11 th : “A” Day Agenda Homework questions/collect Finish section 14.2: “Systems At Equilibrium” Homework: Section 14.2 review, pg. 511: #1-9 Concept Review: Systems at Equilibrium, #15-20

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The Solubility Product Constant, K sp The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water. Solubilities can be expressed in moles of solute per liter of solution (mol/L or M). –For example, the solubility of calcium fluoride in water is 3.4 × 10 −4 mol/L. –So, 0.00034 mol of CaF 2 will dissolve in 1 L of water to give a saturated solution. –If you try to dissolve 0.00100 mol of CaF 2 in 1 L of water, 0.00066 mol of CaF 2 will remain undissolved. (0.00100 – 0.00034 = 0.00066)

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The Solubility Product Constant, K sp Like most salts, calcium fluoride is an ionic compound that dissociates into ions when it dissolves in water Calcium fluoride is one of a large class of salts that are said to be slightly soluble in water. The ions in solution and any solid salt are at equilibrium. Since solids are not part of the equilibrium constant expression, K eq = [Ca 2+ ] [F − ] 2, which is equal to a constant.

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The Solubility Product Constant, K sp Solubility product constants, K sp : the equilibrium constant for a solid that is in equilibrium with the solid’s dissolved ions. K sp = [Ca 2+ ][F − ] 2 = 1.6 10 −10 K sp values have NO units, just like K eq values. This relationship is true whenever calcium ions and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves.

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The Solubility Product Constant, K sp For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates. The net ionic equation for this precipitation is the reverse of the dissolution. This equation is the same equilibrium. So, the K sp for the dissolution of CaF 2 in this system is the same and is 1.6 × 10 −10

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Solubility Product Constants at 25°C

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Rules for Determining K sp 1. Write a balanced chemical equation. The solubility product is only for salts that have low solubility. Soluble salts, like NaCl, do not have K sp values. Make sure that the reaction is at equilibrium. Equations are always written so that the solid salt is the reactant and the ions are the products.

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Rules for Determining K sp 2. Write a solubility product expression. Write the product of the ion concentrations. Concentrations of any solid or pure liquid are omitted. 3. Complete the solubility product expression. Raise each concentration to a power equal to the substance’s coefficient in the balanced chemical equation. (Remember: K sp values depend on temperature)

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Sample Problem C, pg. 509 Calculating K sp from solubility Most parts of the oceans are nearly saturated with CaF 2. The mineral fluorite, CaF 2, may precipitate when ocean water evaporates. A saturated solution of CaF 2 at 25°C has a solubility of 3.4 X 10 −4 M. Calculate the solubility product constant for CaF 2. CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ] [F - ] 2

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Sample Problem C, pg. 509 Calculating K sp from solubility CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ] [F - ] 2 The solubility of CaF 2 is 3.4 X 10 -4 M That means [Ca 2+ ] = 3.4 X 10 -4 From the balanced equation [F - ] = 2[Ca 2+ ] So [F - ] = 2 (3.4 X 10 -4 ) = 6.8 X 10 -4 K sp = (3.4 X 10 -4 ) (6.8 X 10 -4 ) 2 = 1.6 X 10 -10

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Additional Practice Calculate the solubility product constant, K sp, of HgI 2 if the Hg 2+ concentration in a saturated solution is 1.9 X 10 -10 M. HgI 2 (s) Hg 2+ (aq) + 2 I - (aq) K sp = [Hg 2+ ] [ I - ] 2 [Hg 2+ ] = 1.9 X 10 -10 [ I - ] = 2 [Hg 2+ ] = 2 (1.9 X 10 -10 ) = 3.8 X 10 -10 K sp = (1.9 X 10 -10 ) (3.8 X 10 -10 ) = 2.7 X 10 -29

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Sample Problem D, pg. 510 Calculating Ionic Concentrations Using K sp Copper (I) chloride has a solubility product constant of 1.2 × 10 −6 and dissolves according to the equation below. Calculate the solubility of this salt in ocean water in which the [Cl − ] = 0.55 K sp = 1.2 X 10 -6 = [Cu + ] [Cl - ] 1.2 X 10 -6 = [Cu + ] (0.55) [Cu + ] = 2.2 X 10 -6 Solubility of CuCl = 2.2 X 10 -6 M

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Additional Practice A chemist wishes to reduce the silver ion concentration in saturated AgCl solution to 2.0 X 10 -6 M. What concentration of Cl – would achieve this goal? AgCl (s) Ag + (aq) + Cl – (aq) K sp = [Ag + ] [Cl – ] [Ag + ] = 2.0 X 10 -6 From table 3 in book: K sp of AgCl = 1.8 X 10 -8 1.8 X 10 -8 = 2.0 X 10 -6 [Cl – ] [Cl – ] = 9.0 X 10 -5

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Using K sp to Make Magnesium Though slightly soluble hydroxides are not salts, they have solubility product constants. Magnesium hydroxide is an example: K sp =[Mg 2+ ][OH − ] 2 = 1.8 × 10 −11 This equilibrium is the basis for obtaining magnesium from seawater.

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Using K sp to Make Magnesium To get magnesium, calcium hydroxide is added to sea water, which raises the hydroxide ion concentration to a large value so that [Mg 2+ ][OH − ] 2 would be greater than 1.8 × 10 −11 As a result, magnesium hydroxide precipitates and can be collected.

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Homework Section 14.2 review, pg. 511: #1-9 Concept Review: Systems at Equilibrium, #15-20 Please use your class time wisely… Be ready for a quiz over this section next time..

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Solubility Equilibria

Solubility Equilibria

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