Updates

1
Updates

• Assignment 06 is due today (in class)
• Midterm 2 is THIS Thurs., March 15 and will cover
  Chapters 16 17
• Huggins 10, 7-8pm
• For conflicts ELL 221, 6-7pm (must arrange at
  least one week in advance)

2
Acid-Base Equilibria andSolubility Equilibria

• Chapter 17

3
Precipitation and separation of ions

• Predicting what precipitate might form from a
  mixture of ions (solubility rules, pg. 97)
• Using quantitative means to decide whether a
  precipitate will form
• Predicting selective precipitation of an ion from
  a mixture of ions

4
Will a Precipitate Form?
Remember Q is the reaction quotient, which is
obtained by substituting the initial
concentrations into the equilibrium expression.

• In a solution,
• If Q Ksp, the system is at equilibrium and the
  solution is saturated.
• If Q lt Ksp, more solid will dissolve until Q
  Ksp.
• If Q gt Ksp, the salt will precipitate until Q
  Ksp.

5
If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
0.100 M CaCl2, will a precipitate form?

1. Note what ions present in solution Na, OH-,
   Ca2, Cl-.

1. Note that the only possible precipitate is
   Ca(OH)2 (solubility rules).

1. Is Q gt Ksp for Ca(OH)2?

Ca20 0.100 M
OH-0 4.0 x 10-4 M
0.10 x (4.0 x 10-4)2 1.6 x 10-8
Ksp Ca2OH-2 8.0 x 10-6
Q lt Ksp
No precipitate will form
17.6
6
What concentration of Ag is required to
precipitate ONLY AgBr in a solution that contains
both Br- and Cl- at a concentration of 0.02 M?

1. Use Ksp for AgBr to figure out the solubility of
   Ag when Br- is 0.02 M

1. Use Ksp for AgCl to figure out the solubility of
   Ag when Cl- is 0.02 M

1. Is it possible to choose a concentration between
   the two solubility values where only AgBr
   precipitates? This will only work if the Ksp for
   AgCl is greater than that for AgBr since the
   concentrations of the counterions are identical
   in the problem.

17.7
7
What concentration of Ag is required to
precipitate ONLY AgBr in a solution that contains
both Br- and Cl- at a concentration of 0.02 M?
AgBr is less soluble than AgCl so a
selective precipitation should be possible
When Ag is greater than 3.9 x 10-11 M,
AgBr will precipitate when Ag is greater than
8.0 x 10-9, AgCl will precipitate.
3.9 x 10-11 M lt Ag lt 8.0 x 10-9 M
17.7
8
Factors that affect solubility

• We have considered the solubility of ionic
  compounds in pure water we noted that
  temperature and ionic strength has an effect, but
  we did not discuss these influences further
• We will now examine three factors that affect the
  solubility of ionic compounds in water
• Presence of common ions
• pH of solution
• Presence of complexing agents

9
The common ion effect is the shift in equilibrium
caused by the addition of a compound having an
ion in common with the dissolved substance.
Consider mixture of CH3COONa (strong electrolyte)
and CH3COOH (weak acid).
17.2
10
What is the pH of a solution containing 0.30 M
HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
0.30
0.00
0.52
-x
x
x
0.30 - x
x
0.52 x
Common ion effect
0.30 x ? 0.30
4.01
0.52 x ? 0.52
HCOOH pKa 3.77
17.2
11
Common ions affect solubility

• CaF2 Ca2 F-
• The presence of either Ca2 or F- (from another
  source) reduces the solubility of CaF2, shifting
  the solubility equilibrium of CaF2 to the left

12
The Common Ion Effect and Solubility
Ksp 7.7 x 10-13
The solubility of Ag and Br ions equals
(Ksp)1/2 only when AgBr is the only source of Ag
and Br ions (a)!
s2 Ksp
s 8.8 x 10-7
17.8
13
The Common Ion Effect and Solubility
Br- 0.0010 M
Ag s
Br- 0.0010 s ? 0.0010
Ksp 0.0010 x s
s 7.7 x 10-10
17.8
14
(No Transcript)
15
pH and Solubility
At pH less than 10.45
Ksp Mg2OH-2 1.2 x 10-11
Lower OH-
Ksp (s)(2s)2 4s3
4s3 1.2 x 10-11
Increase solubility of Mg(OH)2
s 1.4 x 10-4 M
At pH greater than 10.45
OH- 2s 2.8 x 10-4 M
pOH 3.55 pH 10.45
Raise OH-
Decrease solubility of Mg(OH)2
17.9
16
Complex Ions Affect Solubility

• Complex Ions
• The formation of these complex ions increases the
  solubility of these salts.

17
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central
metal cation bonded to one or more molecules or
ions.
The formation constant or stability constant (Kf)
is the equilibrium constant for the complex ion
formation.
17.10
18
17.10
19
17.11
20
Selective Precipitation of Ions

• Common cations can be divided into five groups
• Insoluble chlorides
• Acid-insoluble sulfides
• Base-insoluble sulfides and hydroxides
• Insoluble phosphates
• Alkali metal ions and NH4 remain in solution
  each ion can be tested for individually using a
  flame test

21
Qualitative Analysis of Cations
17.11
22
lithium
sodium
potassium
copper
17.11
23
Chemistry In Action How an Eggshell is Formed
24

• explain why as the polarity of H-X bonding
  increases, the acid strength increases?
  especially the section involving oxoacids

25
Binary acids (HX, H2X, H3X, H4X)

• Bond strength determines acidity within the same
  group (column), size
• Bond polarity determines acidity within the same
  period (row), electronegativity

26
Oxyacids Central atoms derived from same group
(same oxidation state)

• More electronegative central atom polarizes the
  OH bond more, facilitating ionization (effect is
  weakening the O-H bond)
• More electronegative central atom better able to
  stablize resulting negative charge following
  ionization, making a happier (more stable)
  conjugate base

27

• 16.41) Calculate the concentrations of all the
  species (HCN, H, CN- and OH-) in a 0.15 M HCN
  solution.

If H is 8.6 x 10-6, then OH-
If H CN- 8.6 x 10-6, then we have lost
this amount of HCN, so HCN 0.15 (8.6 x
10-6) 0.15 M
28

• 16.97) Henrys law constant for CO2 at 38oC is
  2.28 x 10-3 mol/L.atm. Calculate the pH of a
  solution of CO2 at 38oC in equilibrium with the
  gas at a partial pressure of 3.20 atm.

Remember that Henrys law describes the effect of
pressure on the solubility of gases. The
solubility of CO2 can be calculated from Henrys
law 2.28 x 10-3 mol/L.atm x 3.20 atm 7.30 x
10-3 mol/L.
Remember that CO2 dissolves in water to form
H2CO3. Therefore, the pH will depend on the
extent of ionization of H2CO3, which can be found
from Ka (4.2 x 10-7) 4.2 x 10-7 x2/(7.30 x
10-3 M) 5.54 x 10-5 M pH -log x, pH 4.26.
29

• 17.47) insert