Monday, April 11th: “A” DayAgenda

Homework questions/collectFinish section 14.2: “Systems At Equilibrium”Homework:

Section 14.2 review, pg. 511: #1-9Concept Review: Systems at Equilibrium, #15-20

The Solubility Product Constant, Ksp

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 CaF2 will dissolve in 1 L of

water to give a saturated solution.– If you try to dissolve 0.00100 mol of CaF2 in

1 L of water, 0.00066 mol of CaF2 will remain undissolved. (0.00100 – 0.00034 = 0.00066)

The Solubility Product Constant, KspLike 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.

2 –2CaF ( ) Ca ( ) 2F ( ) s aq aq

Since solids are not part of the equilibrium constant expression, Keq = [Ca2+] [F−]2 , which is equal to a constant.

The Solubility Product Constant, Ksp

Solubility product constants, Ksp: the equilibrium constant for a solid that is in equilibrium with the solid’s dissolved ions.

Ksp = [Ca2+][F−]2 = 1.6 10−10

Ksp values have NO units, just like Keq values.This relationship is true whenever calcium ions

and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves.

2 –2CaF ( ) Ca ( ) 2F ( ) s aq aq

The Solubility Product Constant, KspFor 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.

2 –2Ca ( ) 2F ( ) CaF ( ) aq aq s

This equation is the same equilibrium. So, the Ksp for the dissolution of CaF2 in this system is the same and is 1.6 × 10−10

Solubility Product Constants at 25°C

Rules for Determining Ksp

1. Write a balanced chemical equation. The solubility product is only for salts that

have low solubility. Soluble salts, like NaCl, do not have Ksp 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.

Rules for Determining Ksp2. 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: Ksp values depend on temperature)

Sample Problem C, pg. 509Calculating Ksp from solubility

Most parts of the oceans are nearly saturated with CaF2. The mineral fluorite, CaF2, may precipitate when ocean water evaporates. A saturated solution of CaF2 at 25°C has a solubility of

3.4 X 10−4 M. Calculate the solubility product constant for CaF2.

CaF2(s) Ca2+(aq) + 2F-(aq)

Ksp = [Ca2+] [F-]2

Sample Problem C, pg. 509Calculating Ksp from solubility

CaF2(s) Ca2+(aq) + 2F-(aq)

Ksp = [Ca2+] [F-]2

The solubility of CaF2 is 3.4 X 10 -4 M That means [Ca 2+] = 3.4 X 10 -4 From the balanced equation [F-] = 2[Ca2+] So [F-] = 2 (3.4 X 10 -4 ) = 6.8 X 10 -4

Ksp = (3.4 X 10 -4 ) (6.8 X 10 -4)2 = 1.6 X 10 -10

Additional PracticeCalculate the solubility product constant, Ksp, of

HgI2 if the Hg2+ concentration in a saturated solution is 1.9 X 10 -10 M.

HgI2 (s) Hg2+(aq) + 2 I-(aq)

Ksp = [Hg2+] [ I-]2

[Hg2+] = 1.9 X 10 -10

[ I-] = 2 [Hg2+] = 2 (1.9 X 10 -10) = 3.8 X 10 -10

Ksp= (1.9 X 10 -10) (3.8 X 10 -10) = 2.7 X 10 -29

Sample Problem D, pg. 510Calculating Ionic Concentrations Using Ksp

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

–CuCl( ) Cu ( ) Cl ( ) s aq aq

Ksp = 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

Additional PracticeA 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)

Ksp = [Ag+ ] [Cl– ][Ag+] = 2.0 X 10 -6 From table 3 in book: Ksp of AgCl = 1.8 X 10 -8

1.8 X 10 -8 = 2.0 X 10 -6 [Cl– ][Cl– ] = 9.0 X 10 -5

Using Ksp to Make MagnesiumThough slightly soluble hydroxides are not salts,

they have solubility product constants.Magnesium hydroxide is an example:

2 –2Mg(OH) ( ) Mg ( ) 2OH ( ) s aq aq

Ksp =[Mg2+][OH−]2 = 1.8 × 10−11

This equilibrium is the basis for obtaining magnesium from seawater.

Using Ksp to Make MagnesiumTo get magnesium, calcium hydroxide is added

to sea water, which raises the hydroxide ion concentration to a large value so that [Mg2+][OH−]2 would be greater than 1.8 × 10−11

As a result, magnesium hydroxide precipitates and can be collected.

2 –2Mg(OH) ( ) Mg ( ) 2OH ( ) s aq aq

HomeworkSection 14.2 review, pg. 511: #1-9Concept Review: Systems at Equilibrium, #15-20

Please use your class time wisely…

Be ready for a quiz over this section next time..