The CommonIon Effect in AcidBase Equilibria

1
The Common-Ion Effect in Acid-Base Equilibria

• The Common-Ion Effect describes the effect on an
  equilibrium by a second substance that furnishes
  ions that can participate in that
  equilibrium.The added ions are said to be common
  to the equilibrium.

The equilibrium will shift according to
LeChâteliers Principle. AgCl(s) Ag(aq)
Cl-(aq)
2
The Common Ion Effect

• QUESTION What is the effect on the pH of adding
  0.1moles NH4Cl to 0.25 M NH3(aq)?
• NH3(aq) H2O NH4(aq) OH-(aq)
• Here we are adding an ion COMMON to the
  equilibrium. Le Chatelier predicts that the
  equilibrium will shift.
• The pH will change
• (Recall NH4 is an acid!)

3
The Common Ion Effect

• QUESTION What is the effect on the pH of adding
  0.1 moles NH4Cl to 0.25 M NH3(aq)?
• NH3(aq) H2O NH4(aq) OH-(aq)

First calculate the pH of a 0.25 M NH3
solution. NH3 NH4
OH- Initial 0.25 0 0 change -x x
x equilib 0.25 - x x x
4
The Common Ion Effect

• QUESTION What is the effect on the pH of adding
  0.1 moles NH4Cl to 0.25 M NH3(aq)?
• NH3(aq) H2O NH4(aq)
  OH-(aq)

Assuming x is ltlt 0.25, we have OH- x
Kb(0.25)1/2 0.0021 M This gives pOH
2.67 and so pH 14.00 - 2.67 11.33 for 0.25
M NH3
5
The Common Ion Effect

• QUESTION What is the effect on the pH of adding
  0.1 moles NH4Cl to 0.25 M NH3(aq)?
• NH3(aq) H2O NH4(aq) OH-(aq)

Now take the added NH4Cl into account. We expect
that the pH will decline on adding NH4Cl. Lets
test that! NH3 NH4
OH- initial 0.25 0.10 0 change
-x x x equilib 0.25 - x
0.10 x x
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The Common Ion Effect

• QUESTION What is the effect on the pH of adding
  0.1 moles NH4Cl to 0.25 M NH3(aq)?
• NH3(aq) H2O NH4(aq) OH-(aq)

OH- x (0.25 / 0.10)Kb 4.5 x 10-5 M This
gives pOH 4.35 and pH 9.65 pH drops from
11.33 to 9.65 on adding a common ion.
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Common Ion Effect

• How will addition of sodium acetate to an acetic
  acid solution affect the pH?
• a. It will lower the pH.
• b. The pH will not change.
• c. The solution becomes hotter.
• d. The pH cannot be measured.
• It will raise the pH.
• I truly do not care, please leave me alone

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Common Ion Effect with an ACID
100 mL of 0.500 M aqueous acetic acid has 4.10 g
of sodium acetate added. What is the change in
pH? (acetate OAc- CH3CO2-)
First calculate pH of acetic acid
Now calculate pH of acetic acid acetate
4.10 g / 82.03 (g/mol) 0.0500 mol / 0.100 L
0.500 M CH3CO2 Na
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100 mL of 0.500 M aqueous acetic acid has 4.10 g
of sodium acetate added. What is the change in
pH? (acetate OAc- CH3CO2-)
Now calculate pH of acetic acid acetate
HOAc(aq) H2O(liq) OAc-(aq) H3O(aq)
Initial 0.500 0.500 Change -x x x Equil
0.500-x 0.500 x x
x 1.8x10-5 pH 4.74 DpH 4.74 2.52 2.22
10
Buffered Solution Characteristics

• Buffers contain relatively large amounts of weak
  acid and corresponding base.
• Added H reacts to completion with the weak base.
• Added OH- reacts to completion with the weak
  acid.
• The pH is determined by the ratio of the
  concentrations of the weak acid and weak base.

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A Buffered Solution

• The function of a buffer is to resist changes in
  the pH of a solution when either H or OH- are
  added.

Buffer Composition Weak Acid Conj.
Base HOAc OAc- H2PO4- HPO42- Weak
Base Conj. Acid NH3 NH4
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Buffer Solutions

• Consider CH3CO2H CH3CO2- in a solution.

H3O CH3CO2-
Ka
1.8x10-5

CH3CO2H
pH -logH3O -logKa -log(1.8x10-5) 4.74
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Buffer Solutions
Problem What is the pH of a buffer that has
HOAc 0.700 M and OAc- 0.600 M? HOAc
H2O OAc- H3O Ka 1.8 x 10-5

• HOAc OAc- H3O
• initial 0.700 0.600 -
• change -x x x
• equilib 0.700-x 0.600x x

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Buffer Solutions
Problem What is the pH of a buffer that has
HOAc 0.700 M and OAc- 0.600 M? HOAc
H2O OAc- H3O Ka 1.8 x 10-5

• HOAc OAc- H3O
• equilib 0.700 - x 0.600 x x
• Assuming that x ltlt 0.700 and 0.600, we have
• H3O 2.1 x 10-5 and pH 4.68

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Buffer Solutions

• Notice that the expression for calculating the H
  conc. of the buffer is
• and this leads to a general equation for finding
  the H or OH- of a buffer.
• Notice that the H or OH- depends on K and
  the ratio of acid and base concentrations.

Orig. conc. of HOAc


H
O





K
3
a
-
Orig. conc. of OAc
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How Far Buffering Works?
When there is a SMALL amount of A-, adding acid
or base changes the ratio drastically. Therefore
H (i.e. pH) changes drastically
When there is a LARGE amount of A-, addition
doesnt change the ratio drastically. Therefore
H (i.e. pH) doesnt change drastically
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A Buffered Solution
To the buffer 100 mL solution of 0.500 M HOAc/
0.500 M NaOAc, 1.00x10-3 mol of NaOH was added.
What is the change in pH?

• This is a two-step problem
• stoichiometry of acid-base reaction
• Work out the neutralization of the added acid or
  base
• equilibrium calculation
• Calculate the new equilibrium for your weak acid
  or base

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To the buffer solution of HOAc/NaOAc, 1.00x10-3
mol of NaOH was added. What is the new pH?
Recall that the pH of this solution is 4.74 (see
previous slide)
We added 1.00x10-3 mol / 0.100 L 0.01 M OH-
Before 0.500 0.01 0.500 After 0.490 0.510
NEUTRALIZATION
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To the buffer solution of HOAc/NaOAc, 1.00x10-3
mol of NaOH was added. What is the new pH?
Now turn on the equilibrium
Initial 0.490 0.510 Change -x x x Equil
0.490-x 0.510 x x
x 1.73x10-5 pH 4.76
Since the initial pH was 4.74 from a previous
example, 1.00x10-3 mol of NaOH raised the pH 0.02
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Adding Base to Acetic Acid without buffering
To 100 mL of a 0.500 M solution of HOAc,
1.00x10-3 mol of NaOH was added. What is the
change in pH?
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To the 0.500 M solution of HOAc, 1.00x10-3 mol of
NaOH was added. What is the new pH?

• The initial pH is calculated using the following
• HOAc(aq) H2O(liq) OAc-(aq) H3O(aq)
• Ka 1.8x10-5
• 1.8x10-5 x2/(0.500-x)
• x 0.003
• pH 2.52

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To the 0.500 M solution of HOAc, 1.00x10-3 mol of
NaOH was added. What is the new pH?
1.00x10-3 mol / 0.100 L 0.01 M OH-
(volume)
HOAc(aq) OH-(aq) OAc-(aq) H2O
Before 0.500 0.010 After 0.490 0.010
NEUTRALIZATION
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Solution of Strong/Weak AcidSuppression of
Ionization of a Weak Acid
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Solution of Strong/Weak BaseSuppression of
Ionization of a Weak Base
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Solutions of Weak Acids and Their Salts
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Solutions of Weak Bases and Their Salts
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Buffer Solutions ?

• HCl is added to pure water.

• HCl is added to a solution of a weak acid H2PO4-
  and its conjugate base HPO42-.

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To the 0.500 M solution of HOAc, 1.00x10-3 mol of
NaOH was added. What is the new pH?
Initial 0.490 0.010 Change -x x x Equil
0.490-x 0.010 x x
x 8.15x10-4 pH 3.05
Since the initial pH was 2.52, 1.00x10-3 mol of
NaOH raised the pH by 0.53
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Key Points on Buffered Solutions

• 1. They are weak acids or bases containing a
  common ion.
• 2. After addition of strong acid or base, deal
  with stoichiometry first, then equilibrium.

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Henderson-Hasselbalch Equation

• useful for calculating pH when the A-/HA
  ratios are known.

pH pKa log A-/HA
pKa -log Ka
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Henderson-Hasselbalch Equation
conjugate base
pKa log
pH
acid

• Only useful when you can use initial
  concentrations of acid and salt.
• This limits the validity of the equation.
• Limits can be met by

A-
0.1 lt
lt 10
HA
A- gt 100 x Ka and HA gt 100 x Ka
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Henderson-Hasselbalch in ACTION!
25.0 g of propionic acid and 36.2 g of sodium
propionate are mixed in a 1.00 L of solution.
What is the pH?
Ka 1.30x10-5
25.0 g / 74.1 (g/mol) 0.337 mol / 1.00 L
0.337 M propionic acid 36.2 g / 96.1 (g/mol)
0.377 mol/ 1.00 L 0.377 M sodium propionate
pH pKa log A-/HA
pH -log(1.30x10-5) log (0.377 / 0.337)
4.89 0.0487 4.94
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Six Methods of Preparing Buffer Solutions
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Buffer Capacity and Range

• Buffer capacity is the amount of acid or base
  that a buffer can neutralize before its pH
  changes appreciably.
• Maximum buffer capacity exists when HA and A-
  are large and approximately equal to each other.
• Buffer range is the pH range over which a buffer
  effectively neutralizes added acids and bases.
• Practically, range is 2 pH units around pKa