Title: Acid Base Equilibria
1Acid Base Equilibria
2Topics 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
3Weak acid
- HA H2O ? H3O A-
- Ka is acid dissociation constant
- HA ? H A-
4Some weak acids
5Weak Base
- B H20 ? BH OH-
- NH3 H2O ? NH4 OH-
6Some weak bases
7Equilibrium 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.
8Solution
- Ka (0.0004)2/(0.00996)
- 1.6x10-5
- pKa 4.8
9Equilibrium 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
10Solution
Ka 1.84x10-4
11Simplifying 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
12Example
- 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.
13Solution
- x2 9.6x10-4x - 9.6x10-6 0
- x 2.65x10-3
- pOH -log(2.65x10-3) 2.58
- pH 11.42
14Polyprotic 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-??
15Polyprotic Acids-II
- xapp 0.06, quadratic equation is needed to
obtain x. - ?SO32-??
16Acid-base Properties of Salt Solutions
- Ionization of formic acid
- HCOOH H2O ? H3O HCOO-
- Hydrolysis of formate ion
- HCOO- H2O ? HCOOH OH-
17For 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
18Solutions 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
19Example
- 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
20Ammonia and its conjugate acid ammonium ion
- NH3 H2O ? NH4 OH-
- NH4 H20 ? NH3 H3O
21Examples 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.
22Acid-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.
23Example
- 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)
24Common-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.
25Buffers
- 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-)
26How 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)
27Acetic 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
28Ammonia / 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
29Henderson-Hasselbalch Equation
- HA ? H A-
- -logKa logHA - logH - logA-
30Preparation 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
31Selecting 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
32Example
- 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
33Buffer 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.
34Example (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
35Solution
HA H2O ? H3O A-
- pH 4.84 (buffer solution just tolerates excess
base)
36Acid-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.
37Titration of 25 mL of 0.2 M HCl with 0.2 M NaOH
38Titration curve
39Titration 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-
40Before 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
41Before 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
42At 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
43After 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
44pH 8.88 at equivalence pt
45Titration 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
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48Acid-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