acids, bases and salts chemistry 21a dr. dragan marinkovic for thousands of years people have known...

ACIDS, BASES and SALTS

Chemistry 21A Dr. Dragan Marinkovic

For thousands of years people have known that

vinegar, lemon juice and many other foods taste sour. However, it was not until a few hundred years ago that it was discovered why these things taste

sour - because they are all acids.

The term acid, in fact, comes from the Latin term acere, which means

"sour".

Acids taste sour, are corrosive to metals,

change litmus (a dye extracted from lichens) red,

and become less acidic when mixed with bases.

Bases feel slippery, change litmus blue,

and become less basic when mixed with acids.

Acids react with bases to form

salts.



Arrhenius Acids:Substances that when placed in water,

will dissociate to produce H+ ions:HCl(aq) → H+

(aq) + Cl-(aq)

Arrhenius Bases:Substances that when placed in water

will dissociate to yield OH- ions:NaOH(aq) → Na+

(aq) + OH-(aq)

Swedish chemist Svante Arrhenius, received the Nobel Prize in Chemistry in 1903 One of the founders of the science of Physical Chemistry

1884

Nitric Acid - HNO3

Chloric Acid - HClO3

Perchloric Acid - HClO4

Sulfuric Acid - H2SO4

Phosphoric Acid - H3PO4

Acetic Acid - HC2H3O2

Potassium Hydroxide – KOHCalcium Hydroxide - Ca(OH)2

Barium Hydroxide - Ba(OH)2

How these acids and bases dissociate?



Brønsted Acids:Any substance that can transfer a proton (H+) to another substance

Brønsted Bases:Any substance that can accept a proton (H+) from another substance

the Brønsted-Lowry theory is an acid-base theory, proposed independently by Danish Johannes Nicolaus Brønsted and English Thomas Martin Lowry in 1923. In this system, an acid is defined as any chemical species (molecule or ion) that is able to lose, or "donate" a hydrogen ion (proton), and a base is a species with the ability to gain or "accept" a hydrogen ion (proton). It follows that if a compound is to behave as an acid, donating a proton, there must be a base to accept the proton. So the Brønsted–Lowry concept can be defined by the reaction:

acid + base conjugate base + conjugate acid

CH3CO2H + H2O CH3CO2- + H3O

+

H2O + NH3 OH- + NH4+

There is strong evidence that the hydrogen ion is never found free as H+. The bare proton is so strongly attracted by the electrons of surrounding water molecules that H30

+

forms immediately.



Hydrochloric Acid - HClChloric Acid - HClO3





Potassium Hydroxide – KOHCalcium Hydroxide - Ca(OH)2


Show acid/base conjugate pairs.



AMPHIPROTIC a compound or ion that can

either donate or accept H+ ions,i.e. can act both as an acid and as a base

H2O, HSO4- , HPO4

2-, HSO3- etc.

HSO4- + H3O+ H2SO4 + H2O

HSO4- + OH- SO4

2- + H2O

CH3CO2H + H2O CH3CO2- + H3O

+

H2O + NH3 OH- + NH4+



Classical acid (and salt) naming system:

Anion (Salt) Prefix

Anion (Salt) Suffix

Acid Prefix Acid Suffix Example

per ate per ic acid perchloric acid (HClO4)

ate ic acid chloric acid (HClO3)

ite ous acid chlorous acid (HClO2)

hypo ite hypo ous acidhypochlorous acid (HClO)

ide hydro ic acid hydrochloric acid (HCl)

sulfnitrphosphcarbonbromiod

sulfurnitrphosphorcarbonbromiod



Hydrofluoric Acid - HFHydrochloric Acid - HClHydrobromic Acid - HBrHydroiodic Acid - HI Hydrosulfuric Acid - H2S

Nitric Acid - HNO3

Nitrous Acid - HNO2

Hypochlorous Acid - HClOChlorous Acid - HClO2

Chloric Acid - HClO3



Sulfurous Acid - H2SO3


Phosphorous Acid - H3PO3

Carbonic Acid - H2CO3


Oxalic Acid - H2C2O4

Boric Acid - H3BO3

Silicic Acid - H2SiO3

fertilizers, explosives

stomach acid

car batteries

soft drinks, seltzer water

bleach

vinegar, pickles

kidney and bladder stones

treatment of skin irritations; insecticide

glass etching

rotten eggs smell

involved metabolism such as adenosine diphosphate (ADP) and triphosphate (ATP); DNA, RNA

orthosilicic acid H4SiO4, is the form predominantly absorbed by humans

and is found in numerous tissues including bone, tendons, aorta, liver

and kidney.

Strong Acids Hydrohalic acids: HCl, HBr, HI

Nitric acid: HNO3 Sulfuric acid: H2SO4

Perchloric acid: HClO4



SULFURIC ACID

Hot concentrated sulfuric acid is an oxidizing agent Fe(s) + 2H2SO4(conc) FeSO4(aq) + 2H2O(l) + SO2(g)

Concentrated sulfuric acid is very good at removing the water from sugars.

C12H22O11(s) + nH2SO4 (l) 12C(s) + 11H2O (l) + nH2SO4(l)

Making hydrogen peroxide (H2O2) by reacting barium peroxide with sulfuric acid.

BaO2(s) + H2SO4(aq) BaSO4(s) + H2O2(aq)

World production in 2001 was 165 million tonnes, with an approximate value of US$8 billion.



Some important organic acids:

acetic acid

ascorbic acid, vitamin C

lactic acid

(milk acid)

citric acid

acetylsalicylic acid

non-steroidal anti-inflammatory drug (NSAID)

Scurvy is a disease resulting from a deficiency of vitamin C, which is required for the synthesis of collagen in humans

CH3COOH

CH3CHOHCOOH

OH CH2

CH2

OH

OOHO

O

OH

CH

CHCH

CH

OHO

O CH3

O

OCH O

OHOH

CH

OH

CH2

OH



MONOSODIUM GLUTAMATE (MSG) as a food ingredient has been the subject of health studies. A report from the Federation of American Societies for Experimental Biology (FASEB) compiled in 1995 on behalf of the FDA concluded that MSG was safe for most people when "eaten at customary levels.

an α-amino acid, with the amino group on the left and the carboxyl group on the right

L- and D-alanine

glycine



A polypeptide is a chain of amino acids.

glycine glycine

glycylglycine

When two amino acids react (“head-to-tail”) they form a peptide bond (in reaction between the acid of one molecule and amine of another molecule). Thus, a PEPTIDE (bond) is formed,



Bases Sodium Hydroxide - NaOHPotassium Hydroxide - KOHAmmonium Hydroxide - NH4OH

Calcium Hydroxide - Ca(OH)2

Magnesium Hydroxide - Mg(OH)2


Aluminum Hydroxide - Al(OH)3

Ferrous Hydroxide or Iron (II) Hydroxide - Fe(OH)2

Ferric Hydroxide or Iron (III) Hydroxide - Fe(OH)3

Zinc Hydroxide - Zn(OH)2

Lithium Hydroxide - LiOH

antacids, deodorants

glass cleaner

lye, oven and drain cleaner

laxatives, antacids

caustic lime, mortar, plaster

an absorbent in surgical dressings

Strong basesSodium Hydroxide - NaOHPotassium Hydroxide – KOHCalcium Hydroxide - Ca(OH)2




CH3CH2NH2

CH3NH2

methylamine

ethylamine

atropine

All organic bases (like inorganic ones) react with acids to form salts.

Injections of ATROPINE are used in the treatment of bradychardia (an extremely low heart rate)

the first effective treatment for malaria

Morphine, C17H19NO3, is the most abundant of opium’s 24 alkaloids, accounting for 9 to 14% of opium-extract by mass. Named after the Roman god of dreams, Morpheus.

VINCRISTINE, one of the most potent ANTILEUKEMIC DRUGS in use today, was isolated in a search for diabetes treatments from Vinca rosea (now Catharanthus roseus) in the 1950's

morphine

Atropine occurs in the deadly nightshade plant (Atropa belladonna)



2H2O H3O+ + OH-



2H2O H3O+ + OH-

at 25oC

KW = const.

[H2O] = const. = 55.5 M

Kw = [H3O+][OH-] = 10-14

Keqx[H2O]2 = [H3O+][OH-]

Keqx[H2O]2 = Kw

The product of water



2H2O H3O+ + OH-

at 25oCKw = [H3O

+][OH-] = 10-14

[H3O+] = [OH-] = 10-7


T (°C) Kw pKw neutral pH

0 0.114 × 10-14 14.94 7.47

5 0.186 × 10-14 14.73 7.37

20 0.681 × 10-14 14.17 7.08

70 15.85 × 10-14 12.80 6.40

100 51.3 × 10-14 12.29 6.14

The product of water

if

THE SOLUTION ISNEUTRAL



2H2O H3O+ + OH-

pH = -log[H+] or -log[H3O+]

pH = -log [1 x 10-7] = -(-7.00) = 7.00

pOH = -log[OH-]

[H3O+] = 10-pH

at 25oC

at 25oC

KW = const.

[H2O+] = const. = 55.5 M

Kw = [H3O+][OH-] = 10-14

[H3O+] = [OH-] = 10-7


Keqx[H2O]2 = Kw

THE pH CONCEPT

IN THE NEUTRALSOLUTION

For convenience instead of exponential numbers, negative

logarithms of these numbers are used.

ACIDS, BASES

and SALTS


[H+] pH Example

Acids(acidicSolutions)

1 x 100 0 HCl

1 x 10-1 1 Stomach acid

1 x 10-2 2 Lemon juice

1 x 10-3 3 Vinegar

1 x 10-4 4 Soda (Coca-Cola)

1 x 10-5 5 Rainwater

1 x 10-6 6 Milk

Neutral 1 x 10-7 7 Pure water

Bases(basicSolutions)

1 x 10-8 8 Egg whites

1 x 10-9 9 Baking soda

1 x 10-10 10 Tums® antacid

1 x 10-11 11 Ammonia

1 x 10-12 12 Mineral lime - Ca(OH)2

1 x 10-13 13 Drano®

1 x 10-14 14 NaOH

Note: concentration is commonly abbreviated by using square brackets, thus

[H+] = hydrogen ion concentration.

When measuring pH, [H+] is in units of moles of H+ per liter of solution.

pH = -log [H+]

The pH of blood is maintained within the narrow range of

7.35 to 7.45.

Normal urine pH averages about 6.0. Saliva has a pH between 6.0 and 7.4.

Tear pH was measured in 44 normal subjects. The normal pH range was

6.5 to 7.6; the mean value was 7.0.



pH in living systemsCompartment pH

Gastric acid 0.7

Lisosomes 4.5

Granules of chromaffin cells 5.5

Urine 6.0

Neutral H2O at 37 °C 6.81

Cytosol 7.2

Cerebrospinal fluid (CSF) 7.3

Blood 7.34 – 7.45

Mitochondrial matrix 7.5

Pancreas secretions 8.1



Calculate the molar concentration of H3O+ in water solutions with the following OH- molar concentrations:

Calculate the molar concentration of OH- in water solutions with the following H3O+ molar concentrations:

a) 6.9x10-5

b) 0.074

a) 0.0087b) 9.9x10-10




Calculate the molar concentration of OH- in water solutions with the following H3O+ molar concentrations:

a) 6.9x10-5

b) 0.074

a) 0.0087b) 9.9x10-10

Kw = [H3O+][OH-] = 10-14

[H3O+] = 10-14/[OH-] = 10-14/[6.9x10-5] = 1.449x10-10 M

[H3O+] = 10-14/[OH-] = 10-14/[7.4x10-2] = 1.35x10-13 M




Calculate the molar concentration of OH- in water solutions with the following OH- H3O+ molar concentrations:

a) 6.9x10-5

b) 0.074

a) 0.0087b) 9.9x10-10

Kw = [H3O+][OH-] = 10-14

[H3O+] = 10-14/[OH-] = 10-14/[6.9x10-5] = 1.449x10-10 M

[H3O+] = 10-14/[OH-] = 10-14/[7.4x10-2] = 1.35x10-13 M

[OH-] = 10-14 /[H3O+] = 10-14/[8.7x10-3] = 1.15x10-12 M

[OH-] = 10-14 /[H3O+] = 10-14/[9.9x10-10] = 1x10-5 M



Determine the pH of water solutions with the following characteristics. Classify each solution as acidic, basic or neutral.

a) [H+] = 7.5x10-6

b) [OH-] = 2.5x10-4

c) [OH-] = 8.6x10-10

Convert the following pH values in both [H+] and [OH-] values.

a) pH = 3.95b) pH = 4.00c) pH = 11.86




a) [H+] = 7.5x10-6

b) [OH-] = 2.5x10-4

c) [OH-] = 8.6x10-10

pH = -log [H+]

pOH = -log [OH-]


a) pH = 3.95b) pH = 4.00c) pH = 11.86

a) pH = -log [H+] = -log(7.5x10-6) = 5.12

pH + pOH = 14

b) pOH = -log [OH-] = -log(2.5x10-4) = 3.6 pH = 14 - pOH = 14 - 3.6 = 10.4




a) [H+] = 7.5x10-6

b) [OH-] = 2.5x10-4

c) [OH-] = 8.6x10-10

pH = -log [H+]

pOH = -log [OH-]


a) pH = 3.95b) pH = 4.00c) pH = 11.86

pH = -log [H+] = -log(7.5x10-6) = 5.12

pH + pOH = 14

pOH = -log [OH-] = -log(2.5x10-4) = 3.6 pH = 14 - pOH = 14 - 3.6 = 10.4

[H3O+] = 10-pH

[H3O+] = 10-pH = 10-3.95 = 1.12x10-4

[OH-] = 10-pOH

[OH-] = 10-pOH = 10-10.05 = 8.91x10-11



pH meter



Types of Acids

Monoprotic - a solution that produces one mole of H+ ions per mole of acid HCl , HNO3

Diprotic - a solution that produces two moles of H+ ions per mole of acid H2SO4

Triprotic - a solution that produces three moles of H+ ions per mole of acid H3PO4

Polyprotic - two ore more H+ per mole of acid

STRONG ACIDS Acids that are essentially 100% Ionized in aqueous solutions

Hydrohalic acids: HCl, HBr, HI Nitric acid: HNO3

Sulfuric acid: H2SO4 Perchloric acid: HClO4

WEAK ACIDSAcids that are partially ionized

(usually less than 5%) in equilibrium.

Properties of Acids

Nitrous Acid - HNO2





Boric Acid - H3BO3




Properties of AcidsAcids react with: Metals Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Metal oxides FeO(s) + 2HCl(aq) → FeCl2(aq) + H2O(l)

Hydroxides/bases Al(OH)3(s) + 3CH3COOH(aq) → Al(CH3COO)3(aq) + 3H2O(l)

Carbonates 3CaCO3(s) + 2H3PO4(aq) → Ca3(PO4)2(aq) + 3CO2(g) + 3H2O(l)

Bicarbonates KHCO3(aq) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l)

Strong acids react with salts of weak acids Na3BO3(aq) + 3HBr(aq) → H3BO3(aq) + 3NaBr(aq)

The major products of al these reactions (metal compounds with acids) are SALTS.



Properties of Acids

Making dilutions from stock solutions:

If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L?

M1 = 3 mol/L, V1 = 1 L, V2 = 6 L

M1V1 = M2V2 M1V1/V2 = M2

M2 = (3 mol/L x 1 L) / (6 L) = 0.5 MWhy does the formula work?

Because we are equating mol to mol: M1V1 = 3 mol M2V2 = 3 mol



Properties of Acids

What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl?

M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 LM1V1 = M2V2 M1V1/M2 = V2

V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L



Properties of Acids

1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting

solution is 1.5 M?4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl?5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl?6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out, what is the concentration of the remaining HF?



Properties of Bases

Strong BasesNaOH - Sodium Hydroxide KOH - Potassium HydroxideCa(OH)2 - Calcium HydroxideBa(OH)2 - Barium Hydroxide

Weak Bases Aluminum Hydroxide - Al(OH)3

Iron (II) Hydroxide - Fe(OH)2

Iron (III) Hydroxide - Fe(OH)3


Bases react with:

Acids Al(OH)3(s) + 3CH3COOH(aq) → Al(CH3COO)3(aq) + 3H2O(l)

“Acidic oxides” 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l)

Slimy or soapy feel on fingers, due to saponification of the lipids in human skin.

Concentrated or strong bases are caustic (corrosive) on organic matter

and react violently with acidic substances

NaOH(s) + H2O(l) → Na+(aq) + OH-

(aq)



Salts

Saline solution for intravenous infusion. The white port at the base of the bag is where additives can be injected with a hypodermic needle. The port with the blue cover is where the bag is spiked with an infusion set.

In medicine, saline (also saline solution) is a general term referring to a sterile solution of sodium chloride (table salt) in water. It is used for intravenous infusion, rinsing contact lenses, and nasal irrigation.

In medicine, normal saline (NS) is the commonly-used term for a solution of 0.91%

w/v of NaCl, about 300 mOsm/L. Less commonly, this solution is referred to as

physiological saline or isotonic saline,

NaCl(s) + H2O(l) → Na+(aq) + OH-

(aq)

NaOH HCl



SaltsAccording to chemistry, the term "salt" is used for ionic compounds that is composed of positively charged cations (usually metal or ammonium ions) and the negatively charged anions, so that the product remains neutral and without a net charge. The anions may be inorganic (Cl-) as well as organic (CH3COO-) and monoatomic (F-) as well as polyatomic ions (SO4

2-).

Salt's solution in water is called electrolytes. Both, the electrolytes and molten salts conduct electricity. Salts with OH- are basic salts (CaOHCl, BaOHNO3)and with H+ are acidic salts (NaHSO4).

Usually salts are solid crystals having high melting point.

Taste - It differes from salt to salt. It can elicit all the five basic tastes, like salty (sodium chloride), sweet (lead diacetate very toxic!),sour (potassium bitartrate), bitter (magnesium sulfate), and umami or savory (monosodium glutamate).



Salts

Anode: The positive terminal of an electrical current flow.

Cathode: The negative terminal of an electric current system.

Diagram of a Hoffman voltameter used for the electrolysis of water to produce hydrogen and oxygen gases

+ -

2H2O → 2H2 + O2

Cathode (reduction): 2H+(aq) + 2e− → H2(g)

Anode (oxidation): 2H2O(l) → O2(g) + 4H+(aq) + 4e

Silver electroplating. The anode is a silver bar and the cathode is an iron spoon.

anode

cathode



Salts

Anode: The positive terminal of an electrical current flow.

Cathode: The negative terminal of an electric current system.

Anions – negative ions – in aqueous solutions move towards ANODE, e.g. Cl-, NO3

-, SO42-

CATIONS – positive (usually metal) ions, in aqueous solutions move towards CATHODE, K+, Al3+, Ba2+

EQUIVALENT OF SALT is the amount that will produce 1 mol of positive electrical charge when dissolved and dissociated.

Number of salt equivalents in 1 L of 1 M of MgCl2 is 2(+) x 1 M = 2 eq

Number of salt equivalents in 3.12x10-2 mol of Fe(NO3)3 is 3(+) x 3.12x10-2 = 9.36x10-2 eq = 93.6 meq



Salts

When crysralluized from aqueous solutions

many salts crystallise as hydrates:CuSO4•5H2O - copper (II) sulfate pentahydrate

CoCI2•6H2O - cobalt (II) chloride hexahydrate

SnCl2•2H2O - stannous (tin II) chloride dihydrate

When heated, these salts lose their crystalline water and become

“anhydrous salts”.

HYDRATE is a salt containing specific numbers of water molcules as part of solid crystalline structure.

WATER OF HYDRATION is water retained as part of the solid crystalline structure of some salts.

CuSO4•5H2O

CoCI2CoCI2•6H2O

CuSO4



Salts

Metals Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Metal oxides FeO(s) + 2HCl(aq) → FeCl2(aq) + H2O(l)

Hydroxides/bases Al(OH)3(s) + 3CH3COOH(aq) → Al(CH3COO)3(aq) + 3H2O(l)

Carbonates 3CaCO3(s) + 2H3PO4(aq) → Ca3(PO4)2(aq) + 3CO2(g) + 3H2O(l)

Bicarbonates KHCO3(aq) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l)

Strong acids react with salts of weak acids Na3BO3(aq) + 3HBr(aq) → H3BO3(aq) + 3NaBr(aq)

Two soluble salts when mixed forming a new insoluble salt BaCl2(aq) + K2SO4(aq) → BaSO4(s) + 2KCl(aq)

“Acidic oxides” 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l)

Syntheses of salts – reactions yielding salts

Metal + nonmetal Fe(s) + S(s) → FeS(s)



Types of Acids

Monoprotic - a solution that produces one mole of H+ ions per mole of acid HCl , HNO3

Diprotic - a solution that produces two moles of H+ ions per mole of acid H2SO4

Triprotic - a solution that produces three moles of H+ ions per mole of acid H3PO4

Polyprotic - two ore more H+ per mole of acid

STRONG ACIDS Acids that are essentially 100% Ionized in aqueous solutions

Hydrohalic acids: HCl, HBr, HI Nitric acid: HNO3

Sulfuric acid: H2SO4 Perchloric acid: HClO4

The strength of Acids and Bases

% dissociation

6.734

0.21.30.01

Nitrous Acid - HNO2





Boric Acid - H3BO3


WEAK ACIDSAcids that are partially ionized

(usually less than 5%) in equilibrium.

moderately weak




ACID DISSOCIATION CONSTANTThe equilibrium constant for the dissociation of an acid.

HA + H2O A− + H3O+

n all but the most concentrated solutions it can be assumed that the concentration of water, [H2O], is constant, approximately 55 mol·dm−3. On dividing K by the constant terms and writing [H+] for the concentration of the hydronium ion the expression

CH3COOH + H2O CH3COO- + H3O+




Polyprotic acids are acids that can lose more than one proton. The constant for dissociation of the first proton may be denoted as Ka1 and the constants for dissociation of

successive protons as Ka2, etc.

Phosphoric acid, H3PO4, is an example of a

polyprotic acid as it can lose three protons.

equilibrium

H3PO4 H2PO4− + H+ Ka1

H2PO4− HPO4

2− + H+ Ka2

HPO42− PO4

3− + H+ Ka3




Strong BasesNaOH - Sodium Hydroxide KOH - Potassium HydroxideCa(OH)2 - Calcium HydroxideBa(OH)2 - Barium Hydroxide

Weak Bases Aluminum Hydroxide - Al(OH)3

Iron (II) Hydroxide - Fe(OH)2

Iron (III) Hydroxide - Fe(OH)3


Slimy or soapy feel on fingers, due to saponification of the lipids in human skin.

Concentrated or strong bases are caustic (corrosive) on organic matter

and react violently with acidic substances

NaOH(s) + H2O(l) → Na+(aq) + OH-

(aq)




B + H2O HB+ + OH−

Using similar reasoning to that used beforeIn water, the concentration of the hydroxide ion, [OH−], is related to the concentration of the hydrogen ion by Kw = [H+][OH−], therefore

Substitution of the expression for [OH−] into the expression for Kb give

BASE DISSOCIATION CONSTANTThe equilibrium constant for the dissociation of a base.

NH3(aq) + H2O NH4+

(aq) + OH-(aq)

NaOH(aq) Na+(aq) + OH-

(aq)



Analyzing Acids and Bases

Titration is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of a known reactant. Because volume measurements play a key role in titration, it is also known as volumetric analysis

Titration curve of a strong base titrating a strong acid

known volume of unknown concentration (of acid)

The EQUIVALENCE POINT, or STOICHIOMETRIC POINT, of a chemical reaction occurs during a chemical titration when the amount of titrant added is equivalent, or equal, to the amount of analyte present in the sample.



Analyzing Acids and Bases

The volume of titrant added from the buret is measured. For our example, lets assume that 18.3 mL of 0.115 M NaOH has been added to 25.00 mL of nitric acid solution. The following setup shows how the molarity of the nitric acid solution can be calculated from this data.

= or 0.0842 M HNO3

Titration curve of a strong base titrating a strong acid

ENDPOINT - The volume or amount of acid or base added to a solution to neutralize the unknown solution during a titration. When using an indicator, the endpoint occurs when enough titrant has been added to change the color of the indicator.



Analyzing Acids and BasesENDPOINT - The volume or

amount of acid or base added to a solution to neutralize the unknown solution during a titration. Indicator Low pH color Transition pH range High pH color

Gentian violet (Methyl violet) yellow 0.0–2.0 blue-violet

Thymol blue (first transition) red 1.2–2.8 yellow

Thymol blue (second transition) yellow 8.0–9.6 blue

Methyl orange red 3.1–4.4 orange

Bromocresol purple yellow 5.2–6.8 purple

Bromothymol blue yellow 6.0–7.6 blue

Phenol red yellow 6.8–8.4 red

Cresol Red yellow 7.2–8.8 reddish-purple

Phenolphthalein colorless 8.3–10.0 fuchsia

Alizarine Yellow R yellow 10.2–12.0 red

pH meter

To determine endpoint indicators can be used (paper or soluble indicator dyes) or a pH meter



Titration Calculations

M1V1 = M2V2

HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)

acid/basemolar ratio

1 : 1

1 : 2

1 : 3

30 mL of 0.10 M NaOH neutralized 25.0 mL of hydrochloric acid. Determine the concentration of the acid

NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)

0.03 L x 0.1 mol/L = 0.025 L x M2

M2 = 0.003 mol/0.035 L = 0.12 mol/L



50 mL of 0.2 mol L-1 NaOH neutralized 20 mL of sulfuric acid. Determine the concentration of the acid1.Write the balanced chemical equation for the reaction NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

2.Extract the relevant information from the question: NaOH V = 50mL, M = 0.2M H2SO4 V = 20mL, M = ?

3.Check the data for consistency NaOH V = 50 x 10-3L, M = 0.2M H2SO4 V = 20 x 10-3L, M = ?

4.Calculate moles NaOH n(NaOH) = M x V = 0.2 x 50 x 10-3 = 0.01 mol 5.From the balanced chemical equation find the mole ratio NaOH:H2SO4

2:1 6.Find moles H2SO4

NaOH: H2SO4 is 2:1

So n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10-3 moles H2SO4 at the

equivalence point 7.Calculate concentration of H2SO4: M = n ÷ V

n = 5 x 10-3 mol, V = 20 x 10-3L M(H2SO4) = 5 x 10-3 ÷ 20 x 10-3 = 0.25M or 0.25 mol L-1

Titration Calculations



Hydrolysis Reactions of Salts

Hydrolysis is a chemical reaction with water.Salts of weak acids and/or bases

hydrolyze in aqueous solutions.

CH3COONa(aq) + H2O(l) CH3COOH(aq) + Na+(aq) + OH-(aq)

NH4Cl(aq) + H2O(l) NH4OH(aq) + H3O+(aq) + Cl-(aq)

acidic pH

basic pH



BuffersBUFFER

A solution with the ability to resist changing pH when acids (H+) or bases (OH-) are added.

BUFFERS usually consist of a pair of compounds

one of which has the ability to react with H+

and the other with the ability to react with OH-.

BUFFER CAPACITY

the amount of acid (H+) that can be absorbed by a buffer without causing a significant change in pH.

CH3COO-(aq) + H+(aq) CH3COOH(aq)

CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(l)

This is how a buffer solution resists changes ion pH



Buffers

Wyeth amphojel tablets of aluminum hydroxide

CaCO3

Antacids create buffered solutions

In blood plasma, the carbonic acid and hydrogen carbonate ion equilibrium buffers the pH. In this buffer, carbonic acid (H2CO3) is the hydrogen-ion donor (acid) and hydrogen carbonate ion (HCO3

-) is the hydrogen-ion acceptor (base).

H2CO3(aq) H+(aq) + HCO3-(aq)

Additional H+ is consumed by HCO3-

and additional OH- is consumed by H2CO3.

The value of Ka for this equilibrium is 7.9 × 10-7.

The pH of arterial blood plasma is 7.40. If the pH falls below this normal value, a condition called acidosis is produced. If the pH rises above the normal value, he condition is called alkalosis.

Al(OH)3



Buffers

The equilibrium constant will be:

In 1 L of solution, there are going to be about 55 moles of water.

Kc (a constant) times the concentration of water

(another constant) on the left-hand side. The product of those is then given the name Ka.

An introduction to pKa

pKa bears exactly the same relationship to Ka as pH does to the hydrogen ion concentration:

The higher the value for pKa, the weaker the acid.



Buffers

equilibrium pKa value

H3PO4 H2PO4− + H+ pKa1 = 2.15

H2PO4− HPO4

2− + H+ pKa2 = 7.20

HPO42− PO4

3− + H+ pKa3 = 12.37

acid Ka (mol dm-3) pKa

hydrofluoric acid 5.6 x 10-4 3.3

formic acid 1.6 x 10-4 3.8

acetic acid 1.7 x 10-5 4.8

hydrogen sulfide 8.9 x 10-8 7.1

Phosphoric acid, H3PO4, is

an example of a polyprotic acid as it can lose three protons.



Buffers

pH = -log[H+] or -log[H3O+]

Henderson-Hasselbalch equationRelationship between the pH, pKa and the concentrations of acid and base in the buffer.

pH = pKa + log [base]--------------------------

[acid]

same equation

pKa = pH - log [base]--------------------------

[acid]



Buffers

If we need to make a buffer solution of a certain pH, we would usually select an acid with the pKa near the desired pH and then adjust the concentration of the acid and the conjugate base (the anion of the acid) to give the desired pH. We can assume that the amount of acid that dissociates is very small and can be neglected. This means that the buffer concentration of the acid and the anion are “equal” to made-up concentrations.



Buffers

Calculate the pH of buffers that contain the acid and conjugate base in following concentrations: a) [HPO4

2-] = 0.33 M, [PO43-] = 0.52 M

pH = pKa + log{[PO43-]/[PO4

2-]} = 12.66 + log(0.52 M/0.33 M) = 12.66 + 0.20 = 12.86

b) [CH3COOH] = 0.40 M, [CH3COO-] = 0.25

pH = pKa + log[CH3COO-]/[CH3COOH] = 4.74 + log(0.25 M/0.40 M) = 4.74 - 0.2 = 4.54

What ratio of concentrations of NaH2PO4 and Na2HPO4 in solution would give a buffer with pH = 7.65?

pH = pKa + log{[PO42-]/[PO4

-]}

7.65 = 7.21 + log{[PO42-]/[PO4

-]}

log{[PO42-]/[PO4

-]} = 0.44

[PO42-]/[PO4

-] = 100.44 = 2.75



Buffers

1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq)

(Kb = 1.8 x 10-5 mol.L-1)

a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution?

pOH

= pKb + log

[acid] --------------------------

[base]

NH4+(aq) + H2O(l) H3O

+(aq) + NH3(aq)

[acid] = [NH4+] = 1 mol/2 L = 0.5 mol/L

[base] = [NH3] = [NH4OH] = 1 mol/L

pOH = -log(1.8x10-5mol/L) + log(0.5/1 molL-1)

pOH = 4.74 - 0.30 = 4.44

pH = 14 - pOH = 9.56



Buffers

1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq)

(Kb = 1.8 x 10-5 mol.L-1)

a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution?

b. Buffer solution diluted 1:100 with pure water, ratio of conjugate acid and base remains unchanged. Therefore pH is unchanged. pH = 9.56



Buffers



Buffers1.00 mol of ammonium chloride is added to 2.00 L of 1.00 mol.L-1 NH3(aq)

(Kb = 1.8 x 10-5 mol.L-1)

a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? c. 10.0 mL of 0.100 mol.L-1 HCl(aq) is added to 100 mL of each of the above solutions. What is the final pH in each case? d. 1.00 mL of 0.100 mol.L-1 NaOH(aq) is added to fresh 100 mL portions of each solution. What is the final pH in each case? [NH4

+] = 1.00 mol/2.00 L = 0.500 mol.L-1,

[NH3] = 1.00 mol.L-1 (given). 100 mL of solution contain: amount of NH4

+ = 0.500 mol.L-1 x 0.100 L = 0.0500 mol

amount of NH3 = 1.00 mol.L-1 x 0.100 L = 0.100 mol

amount of HCl = 0.100 mol.L-1 x 0.010 L = 0.0010 mol

neutralization reaction: HCl(aq) + NH3(aq) ==> NH4

+(aq) + Cl-(aq) (complete)

i.e. 0.0010 mol of NH3 will be converted to [NH4ˆ+]

NH4+(aq) + H2O(l) <===> H3O

+(aq) + NH3(aq)

init/mol 0.0500 0.100 ch/mol +0.0010 -0.0010 0.0510 final/mol 0.099

pH = pKa - log10(a/b) = -log10{Kw/Kb} - log10{[NH4+]/[NH3]}

= 9.26 - log10{0.0051/0.099} = 9.26 + 0.29 => pH = 9.55 14 - 4.74 = 9.26




(Kb = 1.8 x 10-5 mol.L-1)

a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? c. 10.0 mL of 0.100 mol.L-1 HCl(aq) is added to 100 mL of each of the above solutions. What is the final pH in each case? d. 1.00 mL of 0.100 mol.L-1 NaOH(aq) is added to fresh 100 mL portions of each solution. What is the final pH in each case?

solution b. We are adding the same amount of HCl to a solution which is 100 times more dilute, i.e. contains 0.0100 times the amounts of NH4

+ & NH3:

amount of NH4+ = 0.0500 mol x 0.00100 = 0.000500 mol

amount of NH3 = 0.100 mol x 0.00100 = 0.00100 mol

amount of HCl = 0.100 mol.L-1 x 0.010 L = 0.0010 mol

The amounts of HCl and NH3 are equal;

we no longer have a buffer, but exact neutralization, since all the NH3 is converted

to NH4+ we have 0.00150 mol of NH4

+ in 110 mL

of solution. This is a solution of a weak acid. [NH4+] = 0.00150 mol/0.110 L = 0.0136 mol.L-1

Ka = Kw/Kb = 1.00 x 10-14/1.8 x 10-5

= 5.6 x 10-10

Since [NH4+]>>Ka we can use the approximation [NH4+]>>[H3O

+]

(sqrt = square root.) [H3O

+] = sqrt(C0.Ka) =sqrt(0.00150 x 5.6 x 10-10) mol.L-1

= 9.17 x 10-7 => pH = 6.04




(Kb = 1.8 x 10-5 mol.L-1)

a. Calculate the pH of the solution. b. 10.00 mL of the solution is placed into a 1000 mL volumetric flask and made up to the mark with pure water. What is the pH of this solution? c. 10.0 mL of 0.100 mol.L-1 HCl(aq) is added to 100 mL of each of the above solutions. What is the final pH in each case? d. 1.00 mL of 0.100 mol.L-1 NaOH(aq) is added to fresh 100 mL portions of each solution. What is the final pH in each case?

d. (Outline solution only) Similar to c, but a smaller amount of base added. The neutralisation reaction is now: NaOH(aq) + NH4

+(aq) ---> Na+(aq) + NH3(aq) + H2O(l) (complete)

i. solution a: amount of NH4+ = 0.0500 mol

amount of NH3 = 0.100 mol

amount of NaOH = 0.00010 mol NH4

+(aq) + H2O(l) <===> H3O+(aq) + NH3(aq)

init/mol 0.0500 0.100 ch/mol -0.0001 +0.0001 final/mol 0.0499 0.100 pH = 9.57

ii. solution b amount of NH4+ = 0.00050 mol

amount of NH3 = 0.00100 mol

amount of NaOH = 0.00010 mol still a buffer;similar to above final [NH4

+] = 0.00040 mol ; f

inal [NH3] =0.00110 mol pH = 9.73

acids, bases and salts chemistry 21a dr. dragan marinkovic for thousands of years people have known...

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