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Applications of Aqueous Equilibria. Buffered Solutions. A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt. Acid/Salt Buffering Pairs.

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slide2

Buffered Solutions

  • A solution that resists a change in pH when either hydroxide ions or protons are added.
  • Buffered solutions contain either:
    • A weak acid and its salt
    • A weak base and its salt
slide3

Acid/Salt Buffering Pairs

The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

slide4

Base/Salt Buffering Pairs

The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

slide5

Weak Acid/Strong Base Titration

A solution that is

0.10 M CH3COOH

is titrated with

0.10 M NaOH

Endpoint is above pH 7

slide6

Strong Acid/Strong Base Titration

Endpoint is at

pH 7

A solution that is

0.10 M HCl is titrated with

0.10 M NaOH

slide7

Strong Acid/Strong Base Titration

A solution that is

0.10 M NaOH is titrated with

0.10 M HCl

Endpoint is at

pH 7

It is important to recognize that titration curves are not always increasing from left to right.

slide8

Strong Acid/Weak Base Titration

A solution that is

0.10 M HCl is titrated with

0.10 M NH3

Endpoint is below

pH 7

slide9

Titration of an Unbuffered Solution

A solution that is

0.10 M HC2H3O2

is titrated with

0.10 M NaOH

slide10

Titration of a Buffered Solution

A solution that is

0.10 M HC2H3O2 and

0.10 M NaC2H3O2 is titrated with 0.10 M NaOH

slide11

Comparing Results

Unbuffered

Buffered

  • In what ways are the graphs different?
  • In what ways are the graphs similar?
slide12

Comparing Results

Buffered

Unbuffered

buffer capacity
Buffer capacity
  • The best buffers have a ratio [A-]/[HA] = 1
  • This is most resistant to change
  • True when [A-] = [HA]
  • Make pH = pKa (since log1=0)
general equation
General equation
  • Ka = [H+] [A-] [HA]
  • so [H+] = Ka [HA] [A-]
  • The [H+] depends on the ratio [HA]/[A-]
  • taking the negative log of both sides
  • pH = -log(Ka [HA]/[A-])
  • pH = -log(Ka)-log([HA]/[A-])
  • pH = pKa + log([A-]/[HA])
using the henderson hasselbalch equation
Using the Henderson-Hasselbalch Equation
  • pH = pKa + log([A-]/[HA])
  • pH = pKa + log(base/acid)
  • Calculate the pH of the following mixtures
  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
  • 0.25 M NH3 and 0.40 M NH4Cl
  • (Kb = 1.8 x 10-5)
prove they re buffers
Prove they’re buffers
  • What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the following solutions.
    • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
    • 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
  • What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

To do this I must introduce the BAAM table!

buffer capacity1
Buffer capacity
  • The pH of a buffered solution is determined by the ratio [A-]/[HA].
  • As long as this doesn’t change much the pH won’t change much.
  • The more concentrated these two are the more H+ and OH- the solution will be able to absorb.
  • Larger concentrations bigger buffer capacity.
buffer capacity2
Buffer Capacity
  • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following:
  • 5.00 M HAc and 5.00 M NaAc
  • 0.050 M HAc and 0.050 M NaAc
  • Ka= 1.8x10-5
summary
Summary
  • Strong acid and base just stoichiometry.
  • Determine Ka, use for 0 mL base
  • Weak acid before equivalence point
      • Stoichiometry first
      • Then Henderson-Hasselbach
  • Weak acid at equivalence point Kb
  • Weak base after equivalence - leftover strong base.
summary1
Summary
  • Determine Ka, use for 0 mL acid.
  • Weak base before equivalence point.
      • Stoichiometry first
      • Then Henderson-Hasselbach
  • Weak base at equivalence point Ka.
  • Weak base after equivalence - leftover strong acid.
solubility equilibria

Solubility Equilibria

Will it all dissolve, and if not, how much?

slide25
All dissolving is an equilibrium.
  • If there is not much solid it will all dissolve.
  • As more solid is added the solution will become saturated.
  • Solid dissolved
  • The solid will precipitate as fast as it dissolves .
  • Equilibrium
watch out
Watch out
  • Solubility is not the same as solubility product.
  • Solubility product is an equilibrium constant.
  • it doesn’t change except with temperature.
  • Solubility is an equilibrium position for how much can dissolve.
  • A common ion can change this.
slide28

Solving Solubility Problems

For the salt AgI at 25C, Ksp = 1.5 x 10-16

AgI(s)  Ag+(aq) + I-(aq)

O

O

+x

+x

x

x

1.5 x 10-16 = x2

x = solubility of AgI in mol/L = 1.2 x 10-8 M

slide29

Solving Solubility Problems

For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5

PbCl2(s)  Pb2+(aq) + 2Cl-(aq)

O

O

+2x

+x

2x

x

1.6 x 10-5 = (x)(2x)2 = 4x3

x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

relative solubilities
Relative solubilities
  • Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions.
  • The bigger the Ksp of those the more soluble.
  • If they fall apart into different number of pieces you have to do the math.
the common ion effect
The Common Ion Effect
  • When the salt with the anion of a weak acid is added to that acid,
  • It reverses the dissociation of the acid.
  • Lowers the percent dissociation of the acid.
  • The same principle applies to salts with the cation of a weak base..
  • The calculations are the same as last chapter.
slide32

Solving Solubility with a Common Ion

For the salt AgI at 25C, Ksp = 1.5 x 10-16

What is its solubility in 0.05 M NaI?

AgI(s)  Ag+(aq) + I-(aq)

0.05

O

0.05+x

+x

0.05+x

x

1.5 x 10-16 = (x)(0.05+x)  (x)(0.05)

x = solubility of AgI in mol/L = 3.0 x 10-15 M

ph and solubility
pH and solubility
  • OH- can be a common ion.
  • More soluble in acid.
  • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.
  • CaC2O4 Ca+2 + C2O4-2
  • H+ + C2O4-2 HC2O4-
  • Reduces C2O4-2 in acidic solution.
precipitation
Precipitation
  • Ion Product, Q =[M+]a[Nm-]b
  • If Q>Ksp a precipitate forms.
  • If Q<Ksp No precipitate.
  • If Q = Ksp equilibrium.
  • A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?
selective precipitations
Selective Precipitations
  • Used to separate mixtures of metal ions in solutions.
  • Add anions that will only precipitate certain metals at a time.
  • Used to purify mixtures.
  • Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.
selective precipitation
Selective Precipitation
  • In Basic adding OH-solution S-2 will increase so more soluble sulfides will precipitate.
  • Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3
selective precipitation1
Selective precipitation
  • Follow the steps first with insoluble chlorides (Ag, Pb, Ba)
  • Then sulfides in Acid.
  • Then sulfides in base.
  • Then insoluble carbonate (Ca, Ba, Mg)
  • Alkali metals and NH4+ remain in solution.
slide39

Complex Ions

A Complex ion is a charged species composed of:

1. A metallic cation

2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

the addition of each ligand has its own equilibrium
The addition of each ligand has its own equilibrium
  • Usually the ligand is in large excess.
  • And the individual K’s will be large so we can treat them as if they go to equilibrium.
  • The complex ion will be the biggest ion in solution.
slide42

Coordination Number

  • Coordination number refers to the number of ligands attached to the cation
  • 2, 4, and 6 are the most common coordination numbers
slide43

Complex Ions and Solubility

AgCl(s)  Ag+ + Cl- Ksp = 1.6 x 10-10

Ag+ + NH3 Ag(NH3)+ K1 = 1.6 x 10-10

Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 1.6 x 10-10

K = KspK1K2

AgCl + 2NH3 Ag(NH3)2+ + Cl-