17.3 Weak Acids Weak Bases Titrationfaculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch201/aMy_FileLec/...HOCl + KOH a! H2O + OCl- + K+ ... NaOH A strong acid-strong base titration curve,

March 181 Weak Acids Weak Bases Titration

17.3 Weak Acids Weak BasesTitration

Titration of Weak Acid with Strong BaseTitration of Base Acid with Strong Acid

Dr. Fred Omega GarcesChemistry 201Miramar College


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

Simulation_WAWB.xls


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 KOH @ 0%-, 50%-, 95%-, 100%- and 105% of

equivalent point. Analyte : 10.00 ml 0.400M HOCl: Titrant: 0.400 M KOH

HOCl + KOH g 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 = 0ml, 5.0ml, 9.5ml, 10.0ml, 10.5 ml

Misc. Info.: HOCl: Ka = 3•10-8 pKa = 7.5


Type i: Weak Acid with Strong Base0 % Type 1: Calculation EQUILBRIUM0% KOH added (TYPE 1 Weak acid calc.) 0%, VT = 10.0 mlWeak acid pKa - (Type 1 Calculation)

pH of solution is determined by the dissociation of the weak acid.

HOCl + H2O !Ka

H3O+ + OCl-

i 0.4M Ex 0 0C -x -x +x +xe 0.4M-x Ex +x +x

€

Ka = 3 ⋅10−8 =x2

.4M −x , 0.4M−x ≈0.4M

x = .4 3•10−8( )H30+[ ]= x = 1.10• 10-4 M

pH = 3.96


Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

50 % Type 2: Calculation BUFFER, Henderson- Hasselbalch Eqn50% KOH added (TYPE 2 Buffer) 50%, VT = 15.0 ml

HOCl + KOH ! H2O + OCl- + K+ s 4mmol 2mmol Ex 0 - R -2 -2 +2 +2 - f 2mmol 0 Ex+2 2mmol - c .133M - Ex .133M

Notice that the concentration of the acid and its conjugate are equal. In the mass action expression these two terms cancel. pH = pKa. Buffer situation-(Type-2) A. Long Approach Note that the excess 2mmol of HOCl will dissociate in water to

HOCl + H2O " Ka

H3O+ + OCl-

i .133M Ex 0 0.133M C -x -x +x +x e .133-x Ex x .133+x

The calculation is a simple equilibrium analysis-

€

Ka = 3⋅10−8 =x 0.133+x( )

0.133-x( ) , 0.133M ±x ≈ 0.133M

x = 3⋅10−8 ,Assumption checks!

H30+[ ]= 3⋅10−8M

pH = 7.5

B Simple ApproachBuffer solution using Henderson Hasselbalchequation and the sRfc table above

€

pH = pKa + logCbCa

pH = 7.5 + log.133M.133M

pH = 7.5


Titration (4ii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

95% Type II Calculation BUFFER, Henderson- Hasselbalch Eqn95% KOH added (TYPE 2 Buffer): 95%, VT = 19.5 ml

HOCl + KOH ! H2O + OCl- + K+ s 4 mmol 3.8 mmol Ex 0 - R -3.8 -3.8 +3.8 +3.8 - f 0.2 mmol 0 Ex 3.8 mmol - c 1.03•10-2M - Ex 0.195M

Noticed that the acid and conjugate are present in the same solution. This is a common ion effect or buffer type problem. Use the HH equation. Buffer situation-(Type-2) A. Long Approach Note that the excess 0.2mmol (or 1.03•10-2M) of HOCl will dissociate

in water HOCl + H2O "

Ka H3O+ + OCl-

i 1.03•10-2M Ex 0 0.195M

C -x -x +x +x e 1.03•10-2M - x Ex x 0.195M+x

€

Ka = 3⋅10−8 =x .195 +x( )

1.03⋅10-2 -x( )

0.195M +x ≈ 0.195M1.03⋅10-2M − x ≈ 1.03⋅10-2M

1.03⋅10-2 3⋅10−8% & ' (

) *

0.195= x

1.58 ⋅10-9 = x ,Assumption checks!

1.58 ⋅10-9M = H30++ , -

. / 0

8.80 = pH

B Simple ApproachBuffer solution using HendersonHasselbach equation and sRfc table

€

pH = pKa + logCbCa

pH = 7.5 + log 0.195M1.03⋅10-2 M

pH = 7.5+ 1.28pH = 8.78


Titration (4iii): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

100% Type III Calculation HYDROLYSIS, conjugate back to original

100% KOH added (TYPE 3 Hydrolysis): 100%, VT = 20.0 ml

HOCl + KOH ! H2O + OCl- + K+ s 4mmol 4mmol Ex 0 - R -4 -4 +4 +4 - f 0 0 Ex 4mmol - c - - Ex 0.2M -

How is the pH or pOH calculated since there are no H30+ or OH-?

Actually the excess HOCl does react with water to form H3O+. Equivalence point calculation, Hydrolysis (Type-3) Note that all the HOCl acid is neutralize by the base. HOCl

cannot dissociate in water since there is no excess. But the conjugate base OCl- can react with water in a hydrolysis reaction according to-

OCl- + H2O " Kb

OH- + HOCl

i 0.2M Ex 0 0 C -x -x +x +x e 0.2-x Ex x x

€

Kb =KwKa

= 1 ⋅10−14

3⋅10−8 =x2

0.2-x( ) 0.2 -x ≈ 0.2M

x2

0.2= 3.33⋅10−7 x = 2.58 ⋅10−4M

OH−% & '

( ) * = 2.58 ⋅10−4M, pOH = 3.59

pH = 10.41


Titration (4iv): Weak Acid (or Weak Base) with Strong Base (or strong Acid)

105% Type IV Calculation Stoichiometry105% KOH added (TYPE 4 Strong Base)

HOCl + KOH ! H2O + OCl- + K+ s 4mmol 4.2mmol Ex 0 - R -4 -4 +4 +4 - f 0 0.2mmol Ex 4mmol - c - 9.76•10-3M Ex .195M -

Strong Base calculation (Type-4) Since the excess is KOH, a strong base, then the pH (or pOH in

this case) is determine by the following dissociation of KOH:

KOH ! OH- + K

i 9.76•10-3M 0 0 C -9.76•10-3M +9.76•10-3M +9.76•10-3M e - +9.76•10-3M +9.76•10-3M

€

[OH-] 1 = 9.76 ⋅ 10−3MpOH = 2.01pH = 11.99

Note that the 0.195 M of OCl- will contribute negligible

amounts of OH- as it back reacts with water in a hydrolysis type reaction.

A simple check shows that [OH-]2 is negligible. OCl- + H2O !

Kb OH- + HOCl

i 0.195 M Ex 9.76e-3 0

C -x -x +x +x e 0.195 M-x Ex 9.76e-3 +x x

Kb=

Kw

Ka

= 1 ⋅10−14

3 ⋅10−8=

(9.76 ⋅10-3 + x) x0.195-x( )

, 0.195-x ≈ 0.195 M

(9.76 ⋅10-3x + x2)= 3.33 ⋅10−7(0.195)

Solve Quadratic a=1, b=9.76 ⋅10-3, c=−6.49 ⋅10-8 x = [OH-]2 = 6.64 ⋅10−6M

Therefore you see that -

[OH−]Total

= [OH-]1 + [OH-]

2 = 2.54 ⋅10−4M + 9.76 ⋅10−3

[OH−]Total

= 1.00 ⋅10−2 ≈ 9.76 ⋅10−3M

pOH = 2.010 pH = 11.990


Titration Curve: Strong Acid / Strong Base

Titration of 0.100 M HCl with 0.100 M NaOH.

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 0.5000M NaOH is added to 20.00mL of 0.5000M HCl. The equivalence point pH is 7.00. The steep portion of the curve includes the transition intervals of bromothymol blue, phenolphthalein and bromophenol blue .


Titration Curve Features: WA - SB

. Titration of 40.00 mL 0.1000MHCH3CH2OOO with 0.1000 M NaOHWeak acid - Strong base

Titration curve for a weak acid by a strong base: 40.00mL of 0.1000M CH3CH2OOOH by 0.1000M NaOH. When exactly one-half the acid is neutralized, [CH3CH2COOH] = [CH3CH2COO-] and pH = pKa = 8.80The 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.


Titration Curve Features: Monoprotic Acid

Titration curve for a series of acids (A - F) being titrated with a strong base

F

A

B

C

D

E

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.


Titration Curve Features: WB -SA

Titration of 0.1000 M NH3

with 0.1000 M HClWeak base - Strong acid

A weak base-strong acid titration curve, showing how the pH decreases as 0.1000M HCl is added to 40.00mL of 0.1000M 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.


Titration Curve: Polyprotic

Titration Curve of 0.100M H2SO3

with 0.100 M NaOHCurve for the titration of a weak polyproticacid. Titrating 40.00mL of 0.1000M H2SO3

with 0.1000M 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


Titration Curve: Poly-Basic

Titration Curve of 0.10 M Na2CO3with 0.10 M HCl.

1. CO32- + H+ g HCO3

- + H2O

2. HCO3- + H+ g H2CO3 + H2O


Various Titration Curve

Example of Various: Titration Curves

Strong Weak PolyAnalytes

Acids

Base


SummaryThere 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.

17.3 Weak Acids Weak Bases Titrationfaculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch201/aMy_FileLec/...HOCl + KOH a! H2O + OCl- + K+ ... NaOH A strong acid-strong base titration curve,

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