ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

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


H+ + OH- → H2O

At equivalence point

moles of titrant = moles of analyte

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


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


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


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


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


- 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


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)


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 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


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


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


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


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



[HA]

][AlogpKpH

-

a

07.3[0.490]

[0.0100]log4.76pH



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



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



Only A- is present in solution

A- + H2O ↔ HA + OH-

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



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)


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


8.36

5.00


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


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-



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


][BH

[B]logpKpH a

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

pH = pKa (for BH+)


- 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


After equivalence point

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

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


STRONG ACID – WEAK BASEp

H

Volume of HCl added (mL)

5.00


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

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ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university