1st Acid-base Lecture (1)

PHARMACEUTICAL ANALYTICAL CHEMISTRY I

Code No: PHCH323

Lecturer: Dr.Zeinab A. Sheribah

Course Schedule:

Lecture, Wednesday All groups , 3rd & 4th Slots

Practical, Saturday, group 2, 1st & 2nd Slots

group 1, 3rd & 4th Slots

Course Assessment (Evaluation)‏

Practical course 30%

lab. reports & attendance 10%

Three practical exams. 20%

Theoretical course 70%

Quizzes (20%) (2 quizzes).

{Announced one week in advance}

Final term exam. (50%)

Text books

Analytical Chemistry by Gary D. Christian Publisher: Wiley.

Fundamental of Analytical chemistry by Skoog /West /Holler.

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Objectives

Scheme for Analytical Chemistry.

Define volumetric (titrimetric) reaction.

Standard solutions, Types & preparations.

Define neutralization reaction.

Undergo calculations involving acid-base mixtures.

Aqueous Acid – Base Theories.

Hydrogen exponent pH.

Buffer solutions.

Neutralization Indicators.

Construct titration curves of strong acids with strong bases.

Applications.

Analytical Chemistry Science dealing with the

qualitative and quantitative

analysis of substances

Gravimetry by

weighing Volumetry by

titration

Quantitative analysis deals with the

determination of

the quantity of substances

Qualitative analysis deals with the identification

of inorganic substances

Cations Anions Manual Instrumental

Redox

Titration

Acid – Base

Titration

Preciptimetric

Titration

Complexometric

Titration

II- Electron transfer reactions :

-In which electron transfer from one reactant to another. It is called

(oxidation -reduction reactions) eg:

i.e. Fe+2 Fe+3 + e oxidation (loss of electrons)

Ce+4 + Fe+2 Ce+3 + Fe+3

Ag+ + 2 CN- [Ag(CN)2]-

Ca+2 + H2Y

-2 [EDTA] 2H+ + CaY-2 [Ca-EDTA complex]

Volumetric analysis To understand volumetric analysis, we must understand the types of

reaction that happen in it.

Types of reactions used in volumetric analysis :

I- Ionic combination reactions:- -The reaction goes to completion due to formation of slightly ionized or

slightly insoluble products.

a- Neutralization reaction :

In which acid reacts with base to form slightly ionized water.

H+ + OH- (H2O) H2O (H3O+)

b- Formation of precipitate :

Ag+ + Cl- AgCl

c- Formation of slightly ionizable complex :

Ce+4 + e Ce+3 reduction (gain of electrons)

Requirements for titrimetric

reactions

• The reaction must be simple and can be expressed by chemical equation.

• A single reaction must occur between the desired sample and titrant.

• The reaction must be instantaneous. If slow it must be catalyzed.

• Suitable standard solution must be available as titrant

• The end point of the reaction should be easily detected either by the use of

indicator which changes its color at the end point.

• Molecular weight (M.wt) / liter (L) 1M

• e.g. 1 M NaCl (atomic wt. Na=23, Cl=35.5) = 58.5 gm NaCl in 1 litre

1-Molar solution

• Equivalent weigh/liter 1N

• Equivalent wt. of acid = mol.wt / no. of replaceable H+

• Equivalent wt of base = mol.wt./ no. of replaceable OH-

• Equivalent wt. of salt =mol.wt./no.of replaceable cations x its valance or no.of anions x its valance

2- Normal solution (N)

• 1mL of it reacts with a definite amount of the substance to be analyzed

3- Empirical solution

Standard solutions

Standard solution is a solution of exactly known concentration

• Types of Standard Solutions

Preparation of standard solutions

a. Direct method b- Indirect method

• Accurately weighed

amount of the solute is

transferred into

a volumetric flask,

dissolved in the solvent

then completed to the

required volume

•The solute must be a

chemical of primary

standard quality.

• Prepare solution of

approximate concentration

which is standardized against

1ry standard

•The solute is not primary

standard

Characteristics of primary standard

chemicals

1-They must be easily obtained in very high grade of

purity and of known composition

2-They must be easily tested for

impurity by simple test

3-They must react with other

substances quantitatively according to a

balanced chemical equation. i.e. react stoichiometrically

4-They must be stable, i.e. not

absorbing water or CO2

6- They should have high equivalent

weight to minimize error due to

incomplete transfer during weighing.

6- They must be readily soluble in

the solvent

By measuring volume

of standard solution

(titrant) used for

complete reaction

with the sample

Standard solution

Sample solution

Volume consumed

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Acid-base reactions An acid-base reaction is often called neutralization

reaction. When just enough base is added to react exactly with the acid in a solution, we say the acid has been neutralized.

Performing calculations for acid-base reactions

Write the balanced equation for the reaction resulting from the species present in solution.

Calculate the moles of reactants.

Determine the limiting reactant or product.

Calculate the moles of the required reactant or product.

Convert to grams or volume if required in the problem.

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Example What volume of a 0.100M HCl solution is

needed to neutralize 25.0 mLof 0.350M NaOH?

HCl + NaOH NaCl + H2O

The No. of moles of NaOH = MV= 25.0 x 0.350= 8.75

milli mole

The No. of moles of HCl should be the same as

that of NaOH and since they react in a 1:1ratio,

8.75 milli moles of HCl are required.

(MV)HCl = 8.75, substitute by the HCl concentration

(0.100 V) = 8.75, thus the volume of HCl = 87.5 mL

eg. NH3 + H2O NH4+ + OH-

base acid conj.acid conj.b

N.B. Water behave as acid or base because it is neutral (AMPHOTERIC).

ACID – BASE

Acid- Base Theories

1- Arrhenius theory :- Acid : Is the substance which ionize to give H+ eg: HCl

Base : Is the substance which ionize to give OH- eg: NaOH

2- Brönsted - Lowry theory :- Acid : Is the substance which loses or donates proton.

Base : Is the substance which gains or accepts proton.

Every acid has a conjugate base and the base has a conjugate acid.

eg. HCl + H2O Cl- + H3O+

Acid base conj.base conj.acid

Water as both base and acid. One H2O

acts as a base and gains H+ to become

H3O+; the other H2O acts as an acid and

loses H+ to become OH-.

Acid-Base Titration in Aqueous Medium Solutions are classified into :-

Electrolytes : Which dissociate (ionize) and conduct electricity.

Non electrolytes : Which doesn't ionize and doesn't conduct electricity.

3- Lewis theory :-

Acid : Is a substance which accepts lone pair of electrons eg. BF3, AlCl3.

Base : Is a substance which donates lone pair of electrons eg NH3, amines.

Dissociation of water

H2O H+ + OH-

Dissociation const. Kw = [ H+] [OH-] / [H2O]

- Since H2O is weakly dissociated , therefore; H2O is considered unity.

therefore; Kw = [H+] [OH-] = 10 -14 at 25oc

Kw : it is called ionic product of water.

At 25oc [H+] =[OH-] = 10-7

If [H+] = [OH-] , therefore; solution is neutral

If [H+] > 10-7 eg 10-6, 10-5 , therefore; solution is acidic

If [H+] < 10-7 eg 10-8, 10-9 , therefore; solution is alkaline.

Al

Cl N

H H H

Cl Cl

+ adduct

Lewis acid Lewis base

pH of Acids and Bases 1- pH of strong acids :- Since strong acids are strongly ionized.

Therefore; pH = pCa where; Ca ( conc. of acid)

i.e. 0.1 N HCl pH = - log 0.1= - log 10-1 = 1

2- pH of strong bases :- Since strong bases are completely ionized.

Therefore; pOH = pCb where; Cb (conc.of base)

pH = pKw – pOH i.e. pH = pKw – pCb.

i.e. 0.1 N NaOH pH = 14 _ 1 = 13

Hydrogen Exponent : pH pH = -log [H+]

i.e. If [H+] = 10-7 pH = - log 10-7 = 7

In acidic side i.e. If [H+] = 10-6 pH = - log 10-6 = 6

In basic side i.e. If [H+] = 10-8 pH = - log 10-8 = 8

i.e. as pH value increase [H+] conc. decrease.

Therefore; ► acid solution has pH less than 7

► alkaline solution has pH more than7

► neutral solution has pH = p OH = 7

3- pH of weak acids : pH = 1/2 pCa + 1/2 pKa

4- pH of weak bases :- pH =pKw - 1/2 pCb - 1/2 pKb

Buffer Solutions

-Definiton : They are solnution which resist changes in pH upon addition of

small amount of acid or base.

-They consist of weak acid and its salt or weak base and its salt

Type 1: weak acid and its salt eg. HAc and Na Ac

pH of this buffer is calculated from the eq.:

pH = pKa + log [salt] / [acid]

5- pH of salts :- a- Salt of strong acid and strong base eg. NaCl

Always neutral i.e. pH = 7

b- Salt of strong acid and weak base eg.NH4Cl

Always pH is in the acidic side , calculated from eq.

pH = 1/2 pKw - 1/2 pKb + 1/2pCs

where; Kb (dissociation constant of weak base) Cs (conc. of salt)

c- Salt of weak acid and strong base eg. Na Ac

Always pH is in the alkaline side, calculated from eq.:

pH = 1/2 pKw + 1/2 pKa - 1/2pCs

where; Ka (dissociation constant of weak acid) Cs (conc. of salt)

d- Salt of weak acid and weak base eg. NH4Ac

pH is calculated from eq. :

pH = 1/2 pKw + 1/2 pKa - 1/2pKb

Neutralization Indicators

Color indicators: - Substance which change their color with change in pH are used as

neutralization indicators.

eg. Phenolphthalein “ph.ph" (one color indicator),

methyl orange "M.O" (2 color indicator)

eg. ph.ph. = 8-10 M.O. = 3.3-4.4 M.R. = 4-6

N.B. the indicator is chosen according to pH of the product.

Type 2: weak base and its salt eg. NH4OH and NH4Cl

pH of this buffer is calculated from the eq.:

pH = pKw - pKb - log [salt]/[base]

log [salt]/[acid] or log [salt/base] is called buffer ratio

if [salt] = [acid] therefore; pH =pKa

Examples

1- Calculate the pH of a buffer solution containing 0.1 M acetic acid and

0.1 M sodium acetate pKa =4.76

solution.

pH = pKa + log [salt] / [acid]

pH = 4.76 + log 0.1 / 0.1 = 4.76

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Titrations and pH curves

A titration is commonly used to determine the amount of acid or base in a solution.

This process involves solution of known concentration (the titrant) delivered from a burette into the unknown solution until the substance being analyzed is just consumed.

The stoichiometric (equivalence point) is often signaled by the color change of the indicator.

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Titrations and pH curves (cont.)

In this section, we will discuss the changes that occur in pH during an acid base titration.

The progress of an acid-base titration is often monitored by plotting the pH of the solution being analyzed as a function of the amount of titrant added.

Such a plot is called a pH curve or a titration curve.

Neutralization Titration Curves

For neutralization reaction. The titration curve is plot of pH versus

the mL of titrant.

Types of neutralization curves: 1- Strong acid -strong base titration :-

eg. HCl and NaOH

we have sample of 100 mL HCl and titrate against NaOH.

HCl +NaOH NaCl + H2O

pH is due to NaCl i.e. pH= 7 (salt of strong acid and strong base)

c. After the equivalence point :

pH is due to excess titrant i.e. NaOH (strong base) pH = pKw-pCb

N.B. The pH rises slowly till 99.9 % of acid is titrated by adding

0.1 mL NaOH, pH rises from 4 to 7 then another 0.1 mL

after equivalence point pH rises from 7 to 10

i.e. at equivalence point pH rises from 4 to 10.

So we can use M.O.(3.3 - 4.4), M.R.(4 - 6), and ph.ph.(8 - 10) indicators

a.Before the titration pH is due to

the sample i.e. HCl (strong acid)

Therefore; pH = pCa

b. At the equivalence point :

HAc + NaOH NaAc + H2O

HAc will be present ( not completely neutralized) and NaAc i.e. weak

acid and its salt. (acidic buffer)

Therefore; pH is calculated from

pH = pKa + log [salt]/ [acid] = 4.74 + log 40/60 = 4.53

c. At the equivalence point :

all HAc is neutralized and only NaAc is present

i.e.(salt of weak acid and strong base)

Therefore; pH is calculated from:

pH = 1/2 pKw + 1/2 pKa -1/2 pCs

= 7 + 2.37 -1/2 (- log 100 x 0.1 /200) = 8.72

d- After equivalence point the pH is calculated from excess titrant

i.e. NaOH (strong base).

N.B. The end point is at the alkaline side and abrupt change in the curve

is from pH 7 to pH 11. i.e. M.O and M.R. indicator are not suitable.

Therefore; use ph.ph. indicator or any other of pH range on the

alkaline side.

2- Weak acid - Strong base titration :-

eg. CH3COOH (pKa = 4.74) and NaOH

We have sample of 100 mL 0.1 N HAc and titrate against 0.1 N NaOH

a.Before the titration pH is due to

the sample i.e. HAc (weak acid)

Therefore; pH = 1/2pKa + 1/2 pCa

= 2.37 + 0.5 = 2.87

b. During titration: eg. after adding 40 mL NaOH

NH4 OH + HCl NH4Cl + H2O

NH4OH will be present(not completely neutralized)

and NH4Cl i.e. weak base and its salt.(basic buffer)

Therefore; pH is calculated from:

pH = pKw -pKb - log [salt]/ [base]

= 14 - 4.74 - log 90/10 = 8.31

c. At the equivalence point:

all NH4OH is neutralized and NH4Cl is only present

i.e.( salt of strong acid and weak base)

Therefore; pH is calculated from :

pH = 1/2 pKw- 1/2 pKb + 1/2 pCs

= 7 - 2.37 + 1/2 (- log 100 x 0.1 /200) = 5.13

d. After the equi.point: the pH is calculated from excess titrant i.e. HCl

N.B. The end point is at the acidic side = 5.13 .The abrupt change is

From pH 6 to 4 , i.e. ph.ph. indicator cannot be used. Therefore; use M.O.

or M.R. or any indicator of pH range at acidic side.

3- Strong acid - weak base titration :-

eg. NH4OH (pKb = 4.74)and HCl

We have sample 100 mL 0.1 N NH4OH and titrate against 0.1 NHCl

a. Before titration: pH is due to the sample i.e. NH4OH (weak base)

Therefore; pH is calculated from eq. of.

pH = pKw - 1/2 pKb -1/2 pCb

= 14 – 2.37 - 0.5 =11.13

b. During titration: eg. After adding 90 mL HCl

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Strong Acid- Strong Base titrations

The net ionic reaction for a strong acid-strong base titration is

H+(aq) + OH-

(aq) H2O(l)

To compute [H+] at a given point in the titration, we must determine the amount of H+ that remains at that point and divide by the total volume of solution.

Since titrations usually involve small quantities (burettes are graduated in milliliters) we are going to use the mmoles = volume (in mL) X molarity.

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CASE STUDY: Consider the titration of 50.0 mL of 0.2M HNO3

with 0.1M NaOH. We will calculate the pH of the solution at

selected points during the course of the titration, where specific

volumes of 0.1M NaOH have been added.

A. No NaOH has been added.

Since HNO3 is a strong acid, the major species are H+, NO3

- and H2O

[H+] = 0.2 M and pH = 0.669

B. 10.0 mL of 0.1M NaOH has been added.

H+ + OH- H2O

Before reaction 50.0x0.2 10x0.1

= 10.0 mmol =1.0 mmol

After reaction 10.0-1.0 1.0-1.0

=9.0 mmol = 0

After reaction,

[H+] = mmol of H+ left/ volume of solution mL = 9.0/ (50.0+10.0) = 0.15 M

pH = - log (0.15) = 0.82

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(cont.)

C. 20 mL (total) of 0.1 M NaOH has been added.

H+ + OH- H2O

Before reaction 50.0x0.2 20x0.1

= 10.0 mmol =2.0 mmol

After reaction 10.0-2.0 2.0-2.0

=8.0 mmol = 0

After reaction,

[H+] = mmol of H+ left/volume of solution mL = 8.0/ (50.0+20.0) = 0.11 M

pH = - log (0.11) = 0.942

D. 50.0 mL (total) of 0.1M NaOH has been added.

Proceeding exactly as for point B and C, the pH is found to be 1.301

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(cont.)

E. 100.0 mL (total) of 0.1M NaOH has been added.

At this point the amount of NaOH that has been added

is 100mL x 0.1M = 10.0 mmol.

The original amount of nitric acid was

50.0 mL x 0.2M = 10.0 mmol

Enough OH- has been added to react exactly

with the H+ from the nitric acid.

This is the stoichiometric point or equivalence

point, of the titration.

At that point, the major species are Na+, NO3-

and H2O, since Na+ has no acid or base properties and

NO3- is the anion of strong acid and therefore is a very

weak base, neither NO3- nor Na+ affects the pH

and the solution is neutral. (The pH is 7.0)

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(cont.)

F. 150 mL (total) of 0.1m NaOH has been added

H+ + OH- H2O

Before reaction 50.0x0.2 150x0.1

= 10.0 mmol =15.0 mmol

After reaction 10.0-10.0 15.0-10.0

= 0 mmol = 5.0(excess OH- added)

After reaction,

[OH-]= mmol of OH- excess/volume of solution

mL = 5.0/ (50.0+150.0) = 0.025 M

[H+] = 1.0x10-14/2.5x10-2 = 4.0x10-13M pH = 12.40

G. 200.0 mL (total) of 0.1 M NaOH has been

added

Proceeding as for point F, the pH is found to be 12.60

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(cont.) The results of the previous steps are

summarized by the pH curve shown in fig.

Note that the pH changes very gradually

until the titration is close to the equivalence point, where a dramatic change occurs.

This behavior is due to the fact that early in the titration, there is relatively a large

amount of H+ in the solution, and the addition of a given amount of OH- thus produces a small change in pH.

However, near the equivalence point [H+] is relatively small and the addition of a small amount of OH- produces a large change.

1- Direct titration methods Direct titration is useful for :-

A- Strong acid B- Strong base

C- Weak acid or base if Ka and Kb not less than 10-7

Determination of acids 1. Strong acids can be titrated against strong alkali using ph.ph. or M.O.

On titrating weak acid only ph.ph. is suitable.

2. Acids like benzoic acid , salicylic acid which are not insoluble in water are

dissolved in neutral ethanol then add water and titrate against NaOH using

Ph.ph. as indicator.

3. Boric acid: weak acid is a monobasic acid i.e. release 1 H+,can be titrated

against NaOH only after potentiation by adding any poly hydroxy compound

eg. glycerol using ph.ph. as indicator.

Determination of bases 1- Strong base can be titrated against strong acids using M.O. or ph.ph .

For weak bases we use M.O. indicator or any indicator of pH range on the

acidic side.

Na2CO3 HCl NaHCO3

HCl NaCl + H2O +CO2

pH = 8.3 3.8

ph.ph M.O.

2- Displacement titration methods It is used for easily hydrolysable salts :-

A- Salt of strong base and weak acid eg. borax, Na2CO3

B- Salt of weak base and strong acid eg. FeCl3, Al2(SO4)3

N.B. Always titrate the strong part of the salt.

eg.1. KCN We titrate KOH by standard acid eg. HCl

eg.2. Borax Na2B4O7

Borax hydrolyze in water to give:

Na2B4O7 + 7 H2O 4H3BO3 + 2 NaOH

v.weak acid tit.

2 HCl using M.O

titrate HCl using M.O indicator.(let reading = x)

eg.3. Na2CO3 sodium carbonate.

2Na2CO3 +2HCl 2NaHCO3 + 2 NaCl

Na2CO3 + 2HCl 2NaCl + CO2 +H2O

Na2CO3 can be determined by titration against HCl by 2 methods :

a- Using M.O. as indicator it will give total CO3-2

b- Using ph.ph. as indicator it will give 1/2 CO3-2 and considered as half

neutralization step.

But care that the 1st step to NaHCO3 takes place on 2 separate steps;

Na2CO3 +CO2 + H2O 2 NaHCO3

Therefore; For 1/2 neutralization, we must prevent the escape of CO2 by :

1- cooling 2- dilution 3-stirring

4- dipping the nozzle of burette under the surface of the solution.

NH4Cl + NaOH NaCl + NH3 + H2O

Add known excess NaOH , then boil to remove

NH3 and titrate excess NaOH using HCl and M.R. as indicator.

eg.2: Determination of nitrogen in organic compounds :

(by Kjeldahl's method) :

Organic compound Conc.H2SO4 Δ NH3 (NH4)2SO4

-Nitrogen of organic compound is reduced to NH3 by digestion with conc.

H2SO4 in the presence of K2SO4 or Na2SO4 (to raise the boiling point of acid)

and CuSO4 or HgO as a catalyst.

3- Indirect or back (residual titration) - In it we add Known excess of standard to the sample and titrate the excess

non-reacted standard.

Conc. of sample A = Known excess standard - b

When do we use back titration ? 1. When sample is volatile .eg. NH4

+ salts, formic acid.

2. When sample is insoluble eg. ZnO, CaO, CaCO3, BaCO3

3. When reaction require heat of standard solution.

4. When reaction proceed only in presence of excess reagent

eg. lactic acid.

eg.1: Determination of inorganic ammonium salts

4(NH4)2SO4 + 6HCHO (CH2)6N4 + 2H2SO4 + 6H2O

N.B. HCHO must be neutralized from any formic acid which formed

due to aerial oxidation. .

-The organic compound is oxidized to CO2 and the acid is reduced to SO2

and nitrogen to NH3 which is fixed with excess acid as (NH4)2SO4 then add

known excess NaOH and titrate excess unreacted alkali against HCl as before.

4- Other indirect Titrations : Determination of esters:e.g Aspirin

Esters are hydrolyzed by reflux with known excess of NaOH, cool and the

excess unreacted NaOH is titrated against standard HCl using Ph.Ph. indicator.

Determination of ammonium salt and amino acids (Formol titration)

Another method for determination of ammonium salts is formol titration .

When formaldehyde is added to the sample, hexamethylene tetramine

(hexamine) which is neutral is formed and equivalent amount of acid which

can be titrated against NaOH.

4NH4Cl + 6HCHO (CH2)6N4 + 4HCl + 6H2O