1. Dr. Fred Omega Garces Chemistry 201 Miramar College
17.3 Weak Acids Weak Bases Titration Titration of Weak Acid with Strong Base Titration of Base Acid with Strong Acid
Dr. Fred Omega Garces
Chemistry 201
Miramar College

2. Weak Acid (or Weak Base) with Strong Base (or strong Acid)
Experimental technique and the concept is similar to that of the titration of a strong acid with a strong base (or vice versa) with equilibrium concept applied.
pH calculation involves 4 different type of calculations.
i) The analyte alone
(equilibrium calculation)
ii) Buffer region
(Henderson-Hasselbach eqn)
iii) Equivalence point
(Hydrolysis)
iv) Excess titrant
(Stoichiometric calculation)
Equilb
Buffer
Hydrolysis
Stoic Excess
Click for simulation

3. Titration (WA-SB): Weak Acid (or Weak Base) with Strong Base (or strong Acid)
Consider the titration problem:
Titration curve for a weak acid (HOCl) and an strong base (KOH).
Generate a titration curve upon addition of 0%-, 50%-, 95%-, 100%- and 105% of equivalent point.
Analyte : ml M HOCl: Titrant: M KOH
HOCl + KOH  H2O + OCl K+
HOCl = 0.400M • 10.00ml = 4.0 mmol HOCl
Volume KOH corresponding to 0%-, 50%-, 95%-, 100%- and 105%
0.400 M KOH = ml, 5.0ml, 9.5ml, ml, ml
Misc. Info.: HOCl: Ka = 3• pKa = 7.5

4. Type i: Weak Acid with Strong Base
0 % Type 1: Calculation EQUILBRIUM

5. Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)
50 % Type 2: Calculation BUFFER, Henderson- Hasselbalch Eqn

6. Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)
95% Type II Calculation BUFFER, Henderson- Hasselbalch Eqn

7. Titration (4iii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)
100% Type III Calculation HYDROLYSIS, conjugate back to original

8. Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid)
105% Type IV Calculation Stoichiometry

9. i [H+] = ([HOCl]o * Ka)½ iii [OH-] = ([OCl-]o * Kb)½
Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid) i
iv. [OH-] = ([OH-]o – [HOCl])/VT ii
iii [OH-] = ([OCl-]o * Kb)½
iii
ii pH = pKa + log (Cb/Ca)
i [H+] = ([HOCl]o * Ka)½ iv

10. [OH-] = ([OCl-]o * Kb)½ pH = pKa + log (Cb/Ca) [H+] = ([HOCl]o * Ka)½
Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid)
[OH-] = ([OH-]o – [HOCl])/VT
[OH-] = ([OCl-]o * Kb)½
pH = pKa + log (Cb/Ca)
[H+] = ([HOCl]o * Ka)½

11. Titration Curve: Strong Acid / Strong Base
Titration of 20.0 mL 0.500M HCl with 0.500M NaOH
A strong acid-strong base titration curve, showing how the pH increases as M NaOH is added to 20.00mL of M HCl. The equivalence point pH is The steep portion of the curve includes the transition intervals of bromothymol blue, phenolphthalein and bromophenol blue .
Titration of M HCl with M NaOH.

12. Titration Curve Features: WA - SB
Titration of mL M HCH3CH2OOO with M NaOH Weak acid - Strong base
Titration curve for a weak acid by a strong base: mL of M CH3CH2OOOH by M NaOH. When exactly one-half the acid is neutralized, [CH3CH2COOH] = [CH3CH2COO-]
and pH = pKa = 8.80
The equivalent point is above 7.00 because the solution contains the weak base CH3CH2COO-. Phenolphthalein is a suitable indicator for this titration but Methyl red is not because its color changes over a large volume range. .

13. Titration Curve Features: Monoprotic AcidB C D E
Titration curve for a series of acids (A - F) being titrated with a strong base
Acid F is the strongest acid, Acid E is the next strongest acid followed by acid D, acid C, Acid B and acid A. Acid A is the weakest among the weak acid. The Ka’s of each acid is determined by reading the pH half way to the equivalent volume for each acid.

14. Titration Curve Features: WB -SA
Titration of M NH3 with M HCl Weak base - Strong acid
A weak base-strong acid titration curve, showing how the pH decreases as M HCl is added to 40.00mL of M NH3. When exactly one-half the base is neutralized, [NH3] = [NH4+] and pOH = pKb of NH3 (4.76) or the pH = 14 - pKb of NH3 (9.24).
Note that this pH value (9.24) is actually the pka of NH4+, the conjugate of NH3.
Note that the equivalent point is below 7.00 because the solution contains the weak acid NH4+ . Methyl red is a suitable indicator for this titration but phenolphthalein is not because its color changes over a large volume range.

15. Titration Curve: Polyprotic
Titration Curve of 0.100M H2SO3 with M NaOH
Curve for the titration of a weak polyprotic acid. Titrating 40.00mL of M H2SO3 with M NaOH leads to a curve with two buffer regions and two equivalence points. Because the Ka values are separated by several orders of magnitude, in effect the titration curve looks like two weak acid-strong base curves attached. The pH of the first equivalence point is below 7 because the solution contains HSO3-, which is a stronger acid than it is a base. Ka of HSO3- = 6.5•10-8; Kb of HSO3- = 7.1•10-13

16. Titration Curve: Poly-Basic
Titration Curve of 0.10 M Na2CO3 with 0.10 M HCl.
1. CO H+ g HCO3- + H2O
2. HCO3- + H+ g H2CO3 + H2O

17. Various Titration Curve
Example of Various: Titration Curves
Analytes
Acids
Base
Strong Weak Poly

18. Summary The strategy to solve them are:
There are two main type of titration problems.
The strategy to solve them are:
1) SA-SB: Strong acid being titrated with a strong base (or vice versa). This is a stoichiometry type of problem. The pH at the equivalent point = 7.0.
2a) Weak acid being titrated with a strong base. In this type of problem, the there are four sub-problems that must be solved.
a) At 0% titrant, the problem is an equilibrium type. The Ka of the weak acid will determine the extent of ionization and therefore the pH.
b) At 1-99% titrant, the problem is a buffer type. Use the Henderson-Hasselbach Equation to solve for the pH
c) At equivalence point, this is now a hydrolysis problem. Recall that the conjugate base of the weak acid now reacts with water to produce OH-. The solution will be basic at the equivalent point.
d) Pass the equivalence point, this is now a stoichiometry problem. The excess titrant base will dictate the pH of the solution.
2b) Weak base being titrated with a strong acid. This is the same type of problem as (2a) above except in this case a weak base is being neutralized by the strong acid titrant.