Transcript Monday, April 11th: “A” Day Agenda Homework questions/collect Finish section 14.2: “Systems At Equilibrium” Homework: Section 14.2 review, pg.

Monday, April 11th: “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
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, Ksp
 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.
2
–


CaF2 (s ) 
Ca
(
aq
)

2F
(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.
2
–


CaF2 (s ) 
Ca
(
aq
)

2F
(aq )

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.
The Solubility Product Constant, Ksp
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.

 CaF2 (s )
Ca (aq )  2F (aq ) 

2
–
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 Ksp
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: Ksp values depend on temperature)
Sample Problem C, pg. 509
Calculating 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. 509
Calculating 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 Practice
Calculate 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. 510
Calculating 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(s )  Cu (aq )  Cl (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 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)
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 Magnesium
Though slightly soluble hydroxides are not salts,
they have solubility product constants.
Magnesium hydroxide is an example:
2
–


Mg(OH)2 (s )  Mg (aq )  2OH (aq )
Ksp =[Mg2+][OH−]2 = 1.8 × 10−11
This equilibrium is the basis for obtaining
magnesium from seawater.
Using Ksp 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 [Mg2+][OH−]2 would be greater than
1.8 × 10−11
2
–


Mg(OH)2 (s ) 
Mg
(
aq
)

2OH
(aq )

As a result, magnesium hydroxide precipitates
and can be collected.
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..