acids, bases and salts chemistry 21a dr. dragan marinkovic for thousands of years people have known...
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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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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?
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
Hydrochloric Acid - HClChloric 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
Show acid/base conjugate pairs.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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+
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
Perchloric Acid - HClO4
Sulfuric Acid - H2SO4
Sulfurous Acid - H2SO3
Phosphoric Acid - H3PO4
Phosphorous Acid - H3PO3
Carbonic Acid - H2CO3
Acetic Acid - HC2H3O2
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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,
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
Bases Sodium Hydroxide - NaOHPotassium Hydroxide - KOHAmmonium Hydroxide - NH4OH
Calcium Hydroxide - Ca(OH)2
Magnesium Hydroxide - Mg(OH)2
Barium Hydroxide - Ba(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
Barium Hydroxide - Ba(OH)2
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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)
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
2H2O H3O+ + OH-
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
2H2O H3O+ + OH-
at 25oCKw = [H3O
+][OH-] = 10-14
[H3O+] = [OH-] = 10-7
Keqx[H2O]2 = [H3O+][OH-]
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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 = [H3O+][OH-]
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
Chemistry 21A Dr. Dragan Marinkovic
[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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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 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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
pH = -log [H+]
pOH = -log [OH-]
Convert the following pH values in both [H+] and [OH-] values.
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
pH = -log [H+]
pOH = -log [OH-]
Convert the following pH values in both [H+] and [OH-] values.
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
pH meter
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
Sulfurous Acid - H2SO3
Phosphorous Acid - H3PO3
Carbonic Acid - H2CO3
Acetic Acid - HC2H3O2
Boric Acid - H3BO3
Silicic Acid - H2SiO3
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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?
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
Zinc Hydroxide - Zn(OH)2
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)
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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).
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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)
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
Sulfurous Acid - H2SO3
Phosphorous Acid - H3PO3
Carbonic Acid - H2CO3
Acetic Acid - HC2H3O2
Boric Acid - H3BO3
Silicic Acid - H2SiO3
WEAK ACIDSAcids that are partially ionized
(usually less than 5%) in equilibrium.
moderately weak
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
The strength of Acids and Bases
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+
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
The strength of Acids and Bases
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
The strength of Acids and 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
Zinc Hydroxide - Zn(OH)2
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)
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
The strength of Acids and Bases
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)
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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]
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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.
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
Buffers
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
Buffers
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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?
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
ACIDS, BASES and SALTS
Chemistry 21A Dr. Dragan Marinkovic
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?
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
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