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Acid Base Equilibria

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In a 0.01 M solution of butyric acid the acid is 4 % ionized at 20 0C. Calculate Ka and pKa for butyric acid at this temperature. 9/7/09 ... – PowerPoint PPT presentation

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Title: Acid Base Equilibria


1
Acid Base Equilibria
  • Chapter 17

2
Topics of Discussion
  • Solutions of a weak acid and base
  • Acid-Ionization Equilibria
  • Polyprotic Acids
  • Base-Ionization Equilibria
  • Acid-base Properties of Salt Solutions
  • Solutions of Weak Acid/Base with Another Solute
  • Common-Ion Effect
  • Buffers
  • Acid-Base Titration Curves

3
Weak acid
  • HA H2O ? H3O A-
  • Ka is acid dissociation constant
  • HA ? H A-

4
Some weak acids
5
Weak Base
  • B H20 ? BH OH-
  • NH3 H2O ? NH4 OH-

6
Some weak bases
7
Equilibrium calculations
  • Ka and Kb from percentage ionization
  • ionization (amount ionized)/(amount
    available)x100
  • Example
  • In a 0.01 M solution of butyric acid the acid is
    4 ionized at 20 0C. Calculate Ka and pKa for
    butyric acid at this temperature.

8
Solution
  • HBu H2O ? H3O Bu-
  • Ka (0.0004)2/(0.00996)
  • 1.6x10-5
  • pKa 4.8

9
Equilibrium calculations
  • Ka and Kb from initial concentrations and pH
  • Example
  • In a 0.1 M solution of formic acid, the pH is
    2.38 at 25 0C. Calculate the Ka and pKa for
    formic acid at this temperature.
  • HCOOH H2O ? HCOO- H3O

10
Solution
  • H3O 10-2.38 0.0042 M

Ka 1.84x10-4
11
Simplifying assumption
  • HA ? H A-
  • Accuracy of calculations may be set within 5
  • H/HAinitial lt 0.05
  • H2/HAinitial Ka
  • HAinitial gt Ka 400
  • HAequilibrium HAinitial - x (x is
    negligible)
  • HAequilibrium HAinitial

12
Example
  • What is the pH of 0.010 M solution of dimethyl
    amine, (CH3)2NH? Kb 9.6x10-4
  • (CH3)2NH H2O ? (CH3)2NH2 OH-
  • Is simplification possible?
  • (CH3)2NHinitial gt 400 Kb
  • 0.01 gt (400)(9.6x10-4) 0.384
  • No simplification. Quadratic solution.

13
Solution
  • x2 9.6x10-4x - 9.6x10-6 0
  • x 2.65x10-3
  • pOH -log(2.65x10-3) 2.58
  • pH 11.42

14
Polyprotic Acids
  • Acids with two or more acidic protons
  • H2SO3(aq) ? HSO3-(aq) H(aq) Ka1 1.3x10-2
  • HSO3-(aq) ? SO32-(aq) H(aq) Ka2 6.3x10-8
  • For polyprotic acids Ka1gt Ka2gt Ka3
  • Number of protons to be dissociated are larger.
  • First proton separates from singly negative
    charged ion.
  • For 0.25 M sulfuroz acid pH?, ?SO32-??

15
Polyprotic Acids-II
  • xapp 0.06, quadratic equation is needed to
    obtain x.
  • ?SO32-??

16
Acid-base Properties of Salt Solutions
  • Ionization of formic acid
  • HCOOH H2O ? H3O HCOO-
  • Hydrolysis of formate ion
  • HCOO- H2O ? HCOOH OH-

17
For any acid-base conjugate pairs
  • Ka x Kb KW
  • pKa pKb pKW 14
  • The value of Ka 1.8x10-5 for CH3COOH. What is
    the value of Kb for CH3COO-?
  • Solution

18
Solutions of salts of weak acids and bases
  • HCOOH H2O ? HCOO- H3O
  • Strong acid ? weak conjugate base
  • HCl ? H Cl-
  • H2O ? H OH-

acid
conjugate base
Very strong acid
Very weak base
Very weak acid
Very strong base
19
Example
  • What is the pH of 0.1 M solution of NaOCl?
  • Ka 3.0x10-8
  • Solution
  • OCl- H2O ? HOCl OH-

(x2/0.1) 3.3x10-7 HOCl OH- x
1.8x10-4 pOH -log(1.8x10-4) 3.74 pH 14.00 -
3.74 10.26
20
Ammonia and its conjugate acid ammonium ion
  • NH3 H2O ? NH4 OH-
  • NH4 H20 ? NH3 H3O

21
Examples of some weak acids and bases
  • Metal ions with high charge densities are weak
    acids
  • Al(H2O)63 H2O ? Al(H2O)5OH2 H3O
  • Metal ions with small charges are nonacids
  • Aquous ions of Gr IA Li, Na, K, Rb, or Cs
    (except Be2) and Gr IIA Mg2, Ca2, Sr2, and
    Ba2 do not affect pH of solution.

22
Acid-base properties of a salt
  • If only the cation of salt is acidic, the
    solution will be acidic.
  • If only the anion of the salt is basic, the
    solution will be basic.
  • If a salt has a cation that is acidic and an
    anion that is basic, the pH of the solution is
    determined by the relative strength of the acid
    or base.

23
Example
  • Predict whether the following salts are acidic,
    basic or neutral.
  • NaOCl
  • NaNO2
  • KCl
  • NH4Br
  • NaOCl ? Na OCl-
  • OCl- H2O ? HOCl OH- (basic)
  • NaNO2 (basic)
  • KCl (neutral)
  • NH4Br (acidic)

24
Common-Ion Effect
  • The common-ion effect is the shift in an ionic
    equilibrium cause by the addition of a solute
    that provides an ion that takes part in the
    equilibrium.
  • What is the degree of ionization of 0.10 M formic
    acid, HCOOH, solution? Ka 1.84x10-4
  • a. When it is together with 0.20 M HCl. b.When
    it is pure.

25
Buffers
  • Buffer solutions are used to keep the pH
    constant. Dilution of solution, addition of acid
    or base to buffer solution affects the pH very
    little.
  • Buffer consists of a weak Bronsted acid and its
    conjugate base or vice versa.
  • Examples
  • NH3 and NH4Cl (NH3 / NH4)
  • CH3COOH and CH3COONa (CH3COOH / CH3COO-)

26
How a buffer work?
  • Buffer HA ? H A-
  • If extra acid is added
  • H A- ? HA (extra acid is neutralized by
    conjugate base A-)
  • If extra base is added
  • OH- HA ? H2O A- (extra base is neutralized
    by acid HA)

27
Acetic acid / acetate buffer
  • What is the pH of a buffered solution made up
    0.015 M sodium acetate and 0.10 M acetic acid? Ka
    1.8x10-5
  • H x 1.2x10-4
  • pH 3.92

28
Ammonia / ammonium ion buffer
  • Calculate the pH of a buffer solution prepared by
    dissolving 0.10 mol NH3 and 0.2 mol NH4Cl in 1 L
    water. Kb 1.8x10-5

29
Henderson-Hasselbalch Equation
  • HA ? H A-
  • -logKa logHA - logH - logA-

30
Preparation of a buffer solution with a given pH
  • How do you prepare an acetate/acetic acid buffer
    solution with a pH 5.0? (Ka 1.8x10-5)
  • Solution
  • pH pKa log(A-/HA)
  • 5.0 -log(1.8x10-5) log(A-/HA)
  • 10(5.00 - 4.74) A-/HA
  • A-/HA 1.82

31
Selecting a weak acid for the preparation of a
buffer solution
  • A buffer solution is effective if
  • 0.1 lt HA / A- lt 10
  • pH pKa log(A-/HA)
  • pH pKa 1

32
Example
  • How a buffer solution with pH 8.00 be prepared?
  • Solution
  • pH pKa log(A-/HA)
  • Hypochlorous acid, HOCl Ka 3.0x10-8
  • pKa 7.52
  • 8.00 7.52 log(OCl-/HOCl)
  • OCl-/HOCl 3.02

33
Buffer capacity
  • The ratio A-/HA defines the pH of buffer
    solution.
  • The magnitude of concentrations A- and HA
    defines the buffer capacity.
  • It is the ability of a buffer solution to
    compensate extra added acid or base.

34
Example (Buffer capacity)
  • A 1 L buffer solution is prepared using 1.00 M
    acetic acid, HA, and 1.00 M NaAc. In an
    experiment 0.11 mol of hydroxide ion is generated
    without any volume change. Can the buffer
    solution handle this without a pH change of 0.1
    unit?
  • Ka 1.8x10-5

35
Solution
HA H2O ? H3O A-
  • pH 4.84 (buffer solution just tolerates excess
    base)

36
Acid-base titrations
  • The pH of an unknown acid may be found by
    titrating it with a standard base solution until
    the end point is reached.
  • The equivalence point occurs when
    stoichiometrically equivalent acid and base
    reacts.
  • Titration curves are drawn, pH vs added reagent,
    and equivalence point is determined.

37
Titration of 25 mL of 0.2 M HCl with 0.2 M NaOH
38
Titration curve
39
Titration of weak acid by strong base
  • CH3COOH OH- ? CH3COO- H2O
  • Before the titration begins
  • Dissociation of weak acid
  • During the titration but before the equivalence
    point
  • Formation of a buffer solution
  • At the equivalence point
  • Hydrolysis of conjugate base
  • After the equivalence point
  • Concentration of OH-

40
Before the titration begins
  • What is the pH of 0.2 M CH3COOH?
  • Ka 1.8x10-5
  • CH3COOH H2O ? CH3COO- H3O
  • Solution

H3O 1.9x10-3 M pH 2.72
41
Before the equivalence point...
  • Adding 10 mL 0.2 M NaOH to 25 mL 0.2 M CH3COOH.

CH3COOH (3 mmol)/(35 mL) 0.0857 M CH3COO-
(2 mmol)/(35 mL) 0.0571 M
H 2.7x10-5 M pH 4.57
42
At the equivalence point
  • 25 mL 0.2 M NaOH is added to 25 mL 0.2 M CH3COOH.
    What is the pH of solution?
  • Acetic acid is converted to acetate ion
    completely.
  • CH3COO- H2O ? CH3COOH OH-
  • OH- 7.5x10-6 M
  • pOH 5.12 pH 14.00 - 5.12 8.88

43
After the equivalence point
  • 35 mL 0.2 M NaOH is added to 25 mL 0.2 M CH3COOH.
    What is the pH of solution?
  • OH- neutralizes CH3COOH. Excess OH- defines the
    pH.
  • OH- (35x0.2-25x0.2 mmol)/(60 mL) 0.033M
  • pOH 1.48 pH 12.5

44
pH 8.88 at equivalence pt
45
Titration of a weak base by strong acid
  • NH3 H2O ? NH4 OH-
  • Before the titration begins
  • Dissociation of weak base
  • During the titration but before the equivalence
    point
  • Formation of a buffer solution
  • At the equivalence point
  • Hydrolysis of conjugate acid
  • After the equivalence point
  • Concentration of H

46
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48
Acid-base indicators
  • HIn(aq) ? H(aq) In-(aq)
  • pH pKa log(In-/HIn)
  • pH gt pKa 1 (In-/HIn 10, base color)
  • pH lt pKa - 1 (In-/HIn 0.1, acid color)

Acid form color1
Base form color2
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