ANALYTICAL CHEMISTRY CHEM 3811

CHAPTER 10

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 10

ACID-BASE TITRATIONS

STRONG ACID – STRONG BASE

- Write balanced chemical equation between titrant and analyte

- Calculate composition and pH after each addition of titrant

- Construct a graph of pH versus titrant added

- Consider titration of base with acid

H+ + OH- → H2O

K = 1/Kw = 1/10-14 = 1014

- Equilibrium constant = 1014

- Reaction goes to completion

STRONG ACID – STRONG BASE

H+ + OH- → H2O

At equivalence point

moles of titrant = moles of analyte

(V titrant)(M titrant) = (V analyte)(M analyte)

STRONG ACID – STRONG BASE

Consider 50.00 mL of 0.100 M NaOH with 0.100 M HCl

- Three regions of titration curve exists

- Before the equivalence point where pH is determined by excess OH- in the solution

- At the equivalence point where pH is determined by dissociation of water (H+ ≈ OH-)

- After the equivalence point where pH is determined byexcess H+ in the solution

STRONG ACID – STRONG BASE

First calculate the volume of HCl needed to reach the equivalence point

(V HCl)(0.100 M) = (50.00 mL)(0.100 M)

Volume HCl = 50.00 mL

STRONG ACID – STRONG BASE

Before the equivalence point

Initial amount of analyte (NaOH) = 50.00 mL x 0.100 M = 5.00 mmol

After adding 1.00 mL of HClmmol H+ added = mmol OH- consumed

mmol H+ = (1.00 mL)(0.100 M) = 0.100 mmolmmol OH- remaining = 5.00 – 0.100 = 4.90 mmol

STRONG ACID – STRONG BASE

Before the equivalence point

Total volume = 50.00 mL + 1.00 mL = 51.00 mL[OH-] = 4.90 mmol/51.00 mL = 0.0961 M

pOH = - log(0.0961) = 1.017

pH = 14.000 - 1.017 = 12.983

STRONG ACID – STRONG BASE

- Repeat calculations for all volumes added

- Increments can be large initally but must be reduced just beforeand just after the equivalence point (around 50.00 mL in this case)

- Sudden change in pH occurs near the equivalence point

- Greatest slope at the equivalence point

STRONG ACID – STRONG BASE

At the equivalence point

- pH is determined by the dissociation of water

H2O ↔ H+ + OH-

Kw = x2 = 1.0 x 10-14

x = 1.0 x 10-7

pH = 7.00 (at 25 oC)

STRONG ACID – STRONG BASE

After the equivalence point

- Excess H+ is present

After adding 51.00 mL of HClExcess HCl present = 51.00 – 50.00 = 1.00 mL

Excess H+ = (1.00 mL)(0.100 M) = 0.100 mmol

Total volume of solution = 50.00 + 51.00 = 101.00 mL[H+] = 0.100 mmol/101.00 mL = 9.90 x 10-4 M

pH = -log(9.90 x 10-4) = 3.004

STRONG ACID – STRONG BASE

STRONG ACID – STRONG BASEp

H

Volume of HCl added (mL)

7

50.00

Equivalence point(maximum slope or point of inflection)

pH

Volume of NaOH added (mL)

7

50.00

Equivalence point(maximum slope or point

of inflection)

TITRATION CURVE

Compare titration of HCl with NaOH and H2SO4 with NaOH(same volume and same concentration of acid)

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

Net ionic equation in both cases:H+(aq) + OH-(aq) → H2O(l)

orH3O+(aq) + OH-(aq) → 2H2O(l)

1 mol HCl conctributes 1 mol H3O+ 1 mol H2SO4 contributes 2 mol H3O+

STOICHIOMETRY AND TITRATION CURVE

pH

Volume of NaOH added (mL)

STOICHIOMETRY AND TITRATION CURVE

HCl H2SO4

WEAK ACID – STRONG BASE

Consider 50.00 mL 0f 0.0100 M acetic acid with 0.100 M NaOH

pKa of acetic acid = 4.76

HC2H3O2 + OH- → C2H3O2- + H2O

For the reverse reactionKb = Kw/Ka = 5.8 x 10-10

Equilibrium constant = 1/Kb = 1.7 x 109

So large that we can assume the reaction goes to completion

WEAK ACID – STRONG BASE

Determine volume of base at equivalence point

mmol HC2H3O2 ≈ mmol OH-

(V NaOH)(0.100 M) = (50.00 mL)(0.0100 M)

Volume NaOH = 5.00 mL

WEAK ACID – STRONG BASE

Four types of calculations to be considered

Before OH- is added

- pH is determined by equilibrium of weak acid

HA ↔ H+ + A-

x]-[F

][xK

2

a

x = 4.1 x 10-4

pH = 3.39

x]-[0.0100

][x2

WEAK ACID – STRONG BASE

Before equivalence point

By adding OH- a buffer solution of HA and A- is formed

After adding 0.100 mL OH-

HA + OH- → A- + H2OInitial mmol 0.500 0.0100 0Final mmol 0.490 0 0.0100

WEAK ACID – STRONG BASE

Before equivalence point

[HA]

][AlogpKpH

-

a

07.3[0.490]

[0.0100]log4.76pH

WEAK ACID – STRONG BASE

Before equivalence point

If volume of OH- added is half the volume at equivalence pointHA = A- = 0.250 mmol

[0.250]

[0.250]log4.76pH

pH = pKa = 4.76

WEAK ACID – STRONG BASE

At equivalence point

Volume of OH- = 5.00 mL

mmol OH- = (5.00 mL)(0.100 M) = 0.500 mmol

HA is used up and [HA] = 0

mmol A- = 0.500 mmol

WEAK ACID – STRONG BASE

At equivalence point

Only A- is present in solution

A- + H2O ↔ HA + OH-

[A-] = (0.500 mmol)/(50.00 mL + 5.00 mL) = 0.00909 M

WEAK ACID – STRONG BASE

At equivalence point

x]-[F

][xK

2

b

Kb = Kw/Ka = 5.8 x 10-10

x = [OH-] = 2.3 x 10-6

pOH = 5.64pH = 14.00 – 5.64 = 8.36

pH is not 7.00 but greater than 7.00(pH at equivalence point increases with decreasing strength of acid)

WEAK ACID – STRONG BASE

After equivalence point

- Strong base (OH-) being added to weak base (A-)

- pH is determined by the excess [OH-] (approximation)

- After adding 5.10 mL OH-

[OH-] = (0.10 mL)(0.100 M)/(50.00 mL + 5.10 mL)= 1.81 x 10-4

pH = 14.00 – pOH = 14.00 – 3.74 = 10.26

WEAK ACID – STRONG BASEp

H

Volume of NaOH added (mL)

8.36

5.00

Equivalence point(maximum slope or point of inflection)

pH = pKa

Minimum slope

pKa

2.50

STRONG ACID – WEAK BASE

- The reverse of weak acid and strong base

B + H+ → BH+

- Similarly assume reaction goes to completion

Consider 50.00 mL of 0.0100 M pyridine with 0.100 M HClKb of pyridine = 1.6 x 10-9

Determine volume of acid at equivalence point

mmol pyridine ≈ mmol H+

(V HCl)(0.100 M) = (50.00 mL)(0.010 M)

Volume HCl = 5.00 mL

STRONG ACID – WEAK BASE

Four types of calculations to be considered

Before H+ is added

- pH is determined by equilibrium of weak base(determined using Kb)

B + H2O ↔ BH+ + OH-

STRONG ACID – WEAK BASE

Before equivalence point

- By adding H+ a buffer solution of B and BH+ is formed

STRONG ACID – WEAK BASE

][BH

[B]logpKpH a

- When volume of H+ added = half the volume at equivalent point

pH = pKa (for BH+)

At equivalence point

- B has been converted into BH+

- B is used up and [B] = 0

pH is calculated by considering BH+

BH+ ↔ B + H+

pH is not 7.00 but less than 7.00

STRONG ACID – WEAK BASE

After equivalence point

- Strong acid (H+) is being added to weak acid (BH+)

pH is determined by the excess [H+] (approximation)

STRONG ACID – WEAK BASE

STRONG ACID – WEAK BASEp

H

Volume of HCl added (mL)

5.00

Equivalence point(maximum slope or point of inflection)

pH = pKa

Minimum slopepKa

2.50

END POINT

Use of Indicators

- Indicators are acids or bases so a few drops of dilute solutions are used to minimize indicator errors

- Acidic color if pH ≤ pKHIn - 1- Basic color if pH ≥ pKHIn + 1

- A mixture of both colors if pKHIn - 1 ≤ pH ≤ pKHIn + 1

- Use an indicator whose transition range overlaps the steepestpart of the titration curve

END POINT

Use of pH Electrodes

- The end point is where the slope of the curve is greatest

- The end point is the volume at which the first derivative of a titration curve is maximum

- The end point is the volume at which the second derivative of a titration curve is zero