ACIDS BASES and SALTS For thousands of years

ACIDS, BASES and SALTS 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 "sour". Acids taste sour, are corrosive to metals, term acere, which means 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. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Arrhenius Acids: Substances that when placed in 1884 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: Na. OH(aq) → Na+(aq) + OH-(aq) Nitric Acid - HNO 3 Chloric Acid - HCl. O 3 Perchloric Acid - HCl. O 4 How these acids and bases dissociate? Sulfuric Acid - H 2 SO 4 Phosphoric Acid - H 3 PO 4 Acetic Acid - HC 2 H 3 O 2 Swedish chemist Svante Arrhenius, received the Nobel Prize in Chemistry in 1903 One of the founders of the science of Physical Chemistry Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH) Barium Hydroxide - Ba(OH)2 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 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 There is strong evidence defined by the reaction: acid + base conjugate base + conjugate acid CH 3 CO 2 H + H 2 O CH 3 CO 2 - + H 3 O+ H 2 O + NH 3 OH- + NH 4+ 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 H 30+ forms immediately. 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 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Hydrochloric Acid - HCl Chloric Acid - HCl. O 3 Perchloric Acid - HCl. O 4 Show acid/base Sulfuric Acid - H 2 SO 4 conjugate pairs. Phosphoric Acid - H 3 PO 4 Acetic Acid - HC 2 H 3 O 2 Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH) Barium Hydroxide - Ba(OH)2 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 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 H 2 O, HSO 4 - , HPO 42 -, HSO 3 - etc. HSO 4 - + H 3 O+ H 2 SO 4 + H 2 O HSO 4 - + OH- SO 42 - + H 2 O CH 3 CO 2 H + H 2 O CH 3 CO 2 - + H 3 O+ H 2 O + NH 3 OH- + NH 4+ Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Classical acid (and salt) naming system: Anion (Salt) Acid Prefix Acid Suffix Example Prefix Suffix per hypo ate per ic acid perchloric acid (HCl. O 4) ate ic acid chloric acid (HCl. O 3) ite ous acid chlorous acid (HCl. O 2) ite hypo ous acid hypochlorous acid (HCl. O) ide hydro ic acid hydrochloric acid (HCl) sulfur sulf nitr phosphor phosp carbon h brom carbo iod n brom Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS fertilizers, explosives Nitric Acid - HNO 3 Strong Acids Nitrous Acid - HNO 2 Hydrohalic acids: HCl, HBr, HI bleach Hypochlorous Acid - HCl. O Nitric acid: HNO 3 Chlorous Acid - HCl. O 2 Sulfuric acid: H 2 SO 4 Chloric Acid - HCl. O 3 Perchloric acid: HCl. O 4 Perchloric Acid - HCl. O 4 Sulfuric Acid - H 2 SO 4 car batteries Sulfurous Acid - H 2 SO 3 involved metabolism such as adenosine diphosphate Phosphoric Acid - H 3 PO 4 (ADP) and triphosphate (ATP); DNA, RNA Phosphorous Acid - H 3 PO 3 Carbonic Acid - H 2 CO 3 soft drinks, seltzer water Acetic Acid - HC 2 H 3 O 2 vinegar, pickles Oxalic Acid - H 2 C 2 O 4 kidney and bladder stones Boric Acid - H 3 BO 3 treatment of skin irritations; insecticide Silicic Acid - H 2 Si. O 3 orthosilicic acid H 4 Si. O 4, is the form predominantly absorbed by humans and is found in numerous tissues including bone, tendons, aorta, liver and kidney. glass etching Hydrofluoric Acid - HF Hydrochloric Acid - HCl stomach acid Hydrobromic Acid - HBr Hydroiodic Acid - HI Hydrosulfuric Acid - Hrotten eggs smell 2 S Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS SULFURIC ACID World production in 2001 was 165 million tonnes Hot concentrated sulfuric acid is an oxidizing agent with an approximate value of US$8 billion. Fe(s) + 2 H 2 SO 4(conc) Fe. SO 4(aq) + 2 H 2 O(l) + SO 2(g) Concentrated sulfuric acid is very good at removing the water fro C 12 H 22 O 11(s) + n. H 2 SO 4 (l) 12 C(s) + 11 H 2 O (l) + n. H 2 SO 4(l) Making hydrogen peroxide (H 2 O 2) by reacting barium peroxide with sulfuric acid. Ba. O 2(s) + H 2 SO 4(aq) Ba. SO 4(s) + H 2 O 2(aq) Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Some important organic acids: CH 3 COOH acetic acid ascorbic acid, vitamin C Scurvy is a disease resulting from a deficiency of vitamin C, which is required for the synthesis of collagen in humans CH 3 CHOHCOOH citric acid acetylsalicylic acid lactic acid (milk acid) non-steroidal antiinflammatory drug (NSAID) Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS glycine an α-amino acid, with the amino group on the left and the carboxyl group on the right L- and D-alanine 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. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS glycylglyci ne glycine 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, A polypeptide is a chain of amino acids. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Bases Sodium Hydroxide - Na. OH lye, oven and drain cleaner Potassium Hydroxide - KOH glass cleaner Ammonium Hydroxide - NH 4 OH caustic lime, mortar, plaster laxatives, antacids, deodorants 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 an absorbent in surgical dressings Zinc Hydroxide - Zn(OH)2 Lithium Hydroxide - Li. OH Strong bases Sodium Hydroxide - Na. OH Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH)2 Barium Hydroxide - Ba(OH)2 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS CH 3 NH 2 methylamine atropine CH 3 CH 2 NH 2 ethylamine Injections of ATROPINE are used in the treatment of bradychardia (an extremely the first effective treatment for malaria low heart rate) Atropine occurs in the deadly All organic bases nightshade plant (Atropa bellado (like inorganic ones) react with acids to form salts. 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, C 17 H 19 NO 3, is the most abundant of opium’s 24 alkaloids, accounting for 9 to 14% morphine of opium-extract by mass. Named after the Roman god of dreams, Morpheus. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 2 H 2 O H 3 O+ + OH- Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 2 H 2 O H 3 O+ + OH- Keqx[H 2 O]2 = [H 3 O+][OH-] [H 2 O] = const. = 55. 5 M Keqx[H 2 O]2 = Kw KW = const. at o. C +][OH Kw = [H 3 O 25 ] = 10 -14 The product of water Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 2 H 2 O H 3 O+ + OH- Keqx[H 2 O]2 = [H 3 O+][OH-] The product of water if [H 3 O+] = [OH-] = 10 -7 THE SOLUTION IS NEUTRAL at o. C +][OH Kw = [H 3 O 25 ] = 10 -14 T (°C) Kw p. Kw neutral p. H 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 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 2 H 2 O H 3 O+ + OH- Keqx[H 2 O]2 = [H 3 O+][OH-] [H 2 O+] = const. = 55. 5 M Keqx[H 2 O]2 = Kw KW = const. at o. C 25 +][OH -] = 10 -14 Kw = [H 3 O [H 3 O+] = [OH-] = 10 -7 THE p. H CONCEPT For convenience instead of exponential numbers, negative logarithms of these numbers are used. p. H = -log[H+] or -log[H 3 O+] IN THE NEUTRAL SOLUTION at 25 o. C p. H = -log [1 x 10 -7] = -(-7. 00) = 7. 00 [H 3 O+] = 10 -p. H p. OH = -log[OH-] Chemistry 21 A Dr. Dragan Marinkovic

[H+] p. H Example 1 x 100 0 HCl 1 x 10 -1 1 Stomach acid 1 x 10 -2 Acids 1 x 10 -3 (acidic Solutions) 1 x 10 -4 2 Lemon juice 3 Vinegar 4 Soda (Coca-Cola) 1 x 10 -5 5 Rainwater 1 x 10 -6 6 Milk 1 x 10 -7 7 Pure water 1 x 10 -8 8 Egg whites 1 x 10 -9 9 Baking soda 1 x 10 -10 10 Tums® antacid Neutral Bases (basic 1 x 10 -11 Solutions) 1 x 10 -12 11 Ammonia 12 Mineral lime - Ca(OH)2 1 x 10 -13 13 Drano® 1 x 10 -14 14 Na. OH ACIDS, BASES and SALTS p. H = -log [H+] Note: concentration is commonly abbreviated by using square brackets, thus [H+] = hydrogen ion concentration. +] is in units When measuring p. H, [H The p. H of blood is of maintained within the narrow moles of H+ per liter of solution. range of 7. 35 to 7. 45. Normal urine p. H averages about 6. Saliva has a p. H between 6. 0 and 7. Tear p. H was measured in 44 normal subjects. The normal p. H range was 6. 5 to 7. 6; the mean value was 7. 0. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS p. H in living systems Compartment p. H Gastric acid 0. 7 Lisosomes 4. 5 Granules of chromaffin cells 5. 5 Urine 6. 0 Neutral H 2 O 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 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Calculate the molar concentration of H 3 O+ in water solutions with the following OH- molar concentrations: a) 6. 9 x 10 -5 b) 0. 074 Calculate the molar concentration of OH- in water solutions with the following H 3 O+ molar concentrations: a) 0. 0087 b) 9. 9 x 10 -10 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Calculate the molar concentration of H 3 O+ in water solutions with the following OH- molar concentrations: a) 6. 9 x 10 -5 b) 0. 074 Kw = [H 3 O+][OH-] = 10 -14 [H 3 O+] = 10 -14/[OH-] = 10 -14/[6. 9 x 10 -5] = 1. 449 x 10 -10 M [H 3 O+] = 10 -14/[OH-] = 10 -14/[7. 4 x 10 -2] = 1. 35 x 10 -13 M Calculate the molar concentration of OH- in water solutions with the following H 3 O+ molar concentrations: a) 0. 0087 b) 9. 9 x 10 -10 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Calculate the molar concentration of H 3 O+ in water solutions with the following OH- molar concentrations: a) 6. 9 x 10 -5 b) 0. 074 Kw = [H 3 O+][OH-] = 10 -14 [H 3 O+] = 10 -14/[OH-] = 10 -14/[6. 9 x 10 -5] = 1. 449 x 10 -10 M [H 3 O+] = 10 -14/[OH-] = 10 -14/[7. 4 x 10 -2] = 1. 35 x 10 -13 M Calculate the molar concentration of OH- in water solutions with the following OH- H 3 O+ molar concentrations: a) 0. 0087 b) 9. 9 x 10 -10 [OH-] = 10 -14 /[H 3 O+] = 10 -14/[8. 7 x 10 -3] = 1. 15 x 10 -12 M [OH-] = 10 -14 /[H 3 O+] = 10 -14/[9. 9 x 10 -10] = 1 x 10 -5 M Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Determine the p. H of water solutions a) [H+] = 7. 5 x 10 -6 with the following characteristics. b) [OH-] = 2. 5 x 10 -4 Classify each solution as acidic, basic c) [OH-] = 8. 6 x 10 -10 or neutral. a) p. H = 3. 95 Convert the following p. H values b) p. H = 4. 00 in both [H+] and [OH-] values. c) p. H = 11. 86 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Determine the p. H of water solutions a) [H+] = 7. 5 x 10 -6 with the following characteristics. b) [OH-] = 2. 5 x 10 -4 Classify each solution as acidic, basic c) [OH-] = 8. 6 x 10 -10 or neutral. p. H = -log [H+] p. OH = -log [OH-] p. H + p. OH = 14 a) p. H = -log [H+] = -log(7. 5 x 10 -6) = 5. 12 b) p. OH = -log [OH-] = -log(2. 5 x 10 -4) = 3. 6 p. H = 14 - p. OH = 14 - 3. 6 = 10. a) p. H = 3. 95 Convert the following p. H values b) p. H = 4. 00 in both [H+] and [OH-] values. c) p. H = 11. 86 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Determine the p. H of water solutions a) [H+] = 7. 5 x 10 -6 with the following characteristics. b) [OH-] = 2. 5 x 10 -4 Classify each solution as acidic, basic c) [OH-] = 8. 6 x 10 -10 or neutral. p. H = -log [H+] p. OH = -log [OH-] p. H + p. OH = 14 p. H = -log [H+] = -log(7. 5 x 10 -6) = 5. 12 p. OH = -log [OH-] = -log(2. 5 x 10 -4) = 3. 6 p. H = 14 - p. OH = 14 - 3. 6 = 10. a) p. H = 3. 95 Convert the following p. H values b) p. H = 4. 00 in both [H+] and [OH-] values. c) p. H = 11. 86 [H 3 O+] = 10 -p. H = 10 -3. 95 = 1. 12 x 10 -4 [H 3 O+] = 10 -p. H [OH-] = 10 -p. OH = 10 -10. 05 = 8. 91 x 10 -11 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS p. H meter Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Types of Acids Properties of Acids Monoprotic - a solution that produces one mole of H+ ions per mole of acid HCl , HNO 3 Diprotic - a solution that produces two moles of H+ ions per mole of acid H 2 SO 4 Triprotic - a solution that produces three moles of H+ ions per mole of acid H 3 PO 4 Polyprotic - two ore more H+ per mole of acid WEAK ACIDS STRONG ACIDS Acids that are partially ionized Acids that are essentially 100% Nitrous Acid - HNO 2 (usually less than 5%) in equilibrium. Ionized in aqueous solutions Sulfurous Acid - H 2 SO 3 Hydrohalic acids: HCl, HBr, HI Phosphorous Acid - H 3 PO 3 Nitric acid: HNO 3 Carbonic Acid - Sulfuric acid: H 2 SO 4 H 2 CO 3 Perchloric acid: HCl. O 4 Acetic Acid - HC 2 H 3 O 2 Boric Acid - H 3 BO 3 Silicic Acid - H 2 Si. O 3 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Properties of Acids react with: Metals Zn(s) + H 2 SO 4(aq) → Zn. SO 4(aq) + H 2(g Metal oxides Fe. O(s) + 2 HCl(aq) → Fe. Cl 2(aq) + H 2 O Hydroxides/bases Al(OH)3(s) + 3 CH 3 COOH(aq) → Al(CH 3 COO)3(aq) + Carbonates 3 Ca. CO 3(s) + 2 H 3 PO 4(aq) → Ca 3(PO 4)2(aq) + 3 CO 2(g) + 3 Bicarbonates KHCO 3(aq) + HNO 3(aq) → KNO 3(aq) + CO 2(g) + H Strong acids react with salts of weak acids Na 3 BO 3(aq) + 3 HBr(aq) → H 3 BO 3(aq) + 3 Na. Br The major products of al these reactions (metal compounds with acids) are SALTS. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and 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? M 1 = 3 mol/L, V 1 = 1 L, V 2 = 6 L M 1 V 1 = M 2 V 2 M 1 V 1/V 2 = M 2 = (3 mol/L x 1 L) / (6 L) = 0. 5 M Why does the formula work? Because we are equating mol to mol: M 1 V 1 = 3 mol M 2 V 2 = 3 mol Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Properties of Acids What volume of 0. 5 M HCl can be prepared from 1 L of 12 M HCl? M 1 = 12 mol/L, V 1 = 1 L, M 2 = 0. 5 L M 1 V 1 = M 2 V 2 M 1 V 1/M 2 = V 2 = (12 mol/L x 1 L) / (0. 5 L) = 24 L Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Properties of Acids 1. How many m. L of a 14 M stock solution must be used to make 250 m. L of a 1. 75 M solution? 2. You have 200 m. L of 6. 0 M HF. What concentration results if this is diluted to a total volume of 1 L? 3. 100 m. L of 6. 0 M Cu. SO 4 must be diluted to what final volume so that the resulting solution is 1. 5 M? 4. What concentration results from mixing 400 m. L of 2. 0 M HCl with 600 m. L of 3. 0 M HCl? 5. What is the concentration of Na. Cl when 3 L of 0. 5 M Na. Cl are mixed with 2 L of 0. 2 M Na. Cl? 6. What is the concentration of Na. Cl when 3 L of 0. 5 M Na. Cl 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? Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Properties of Bases Slimy or soapy feel on fingers, due to saponification of the lipids in human skin. Na. OH + H O → Na+ + OH (s) 2 (l) (aq) Concentrated or strong bases are caustic (corrosive) on organic matter and react violently with acidic substances Strong Bases Weak Bases Na. OH - Sodium Hydroxide Aluminum Hydroxide - Al(OH)3 KOH - Potassium Hydroxide Iron (II) Hydroxide - Fe(OH)2 Ca(OH)2 - Calcium Hydroxide Iron (III) Hydroxide - Fe(OH)3 Ba(OH)2 - Barium Hydroxide Zinc Hydroxide - Zn(OH)2 Bases react with: Acids Al(OH)3(s) + 3 CH 3 COOH(aq) → Al(CH 3 COO)3(aq) + “Acidic oxides” 2 Na. OH(aq) + CO 2(g) → Na 2 CO 3(aq) + H 2 O Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Salts 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 Na. Cl, about 300 m. Osm/L. Less commonly, this solution is referred to as physiological saline or isotonic saline, 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. Na. Cl(s) + H 2 O(l) → Na+(aq) + OH-(aq Na. OH HCl Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Salts According 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 (CH 3 COO-) and monoatomic (F-) as well as polyatomic ions (SO 42 -). Salt's solution in water is called electrolytes. Both, the electrolytes and molten salts conduct electricity. Salts with OH- are basic salts (Ca. OHCl, Ba. OHNO 3) and with H+ are acidic salts (Na. HSO 4). 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). Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Salts Silver electroplating. The anode is a silver bar and the cathode is an iron spoon. anode cathode + - Diagram of a Hoffman voltameter used for the electrolysis of water to produce Anode: The positive terminal of an electrical current flow. hydrogen and oxygen gases Cathode: The negative terminal of an electric current system. 2 H 2 O → 2 H 2 + O 2 Cathode (reduction): 2 H+(aq) + 2 e− → H 2 Anode (oxidation): 2 H 2 O(l) → O 2(g) + 4 H+(aq Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Salts Anode: The positive terminal of an electrical current flow. Anions – negative ions – in aqueous solutions move towards ANODE, e. g. Cl-, NO 3 -, SO 42 - CATIONS – positive (usually metal) ions, in aqueous so move towards CATHODE, K+, Al 3+, Ba 2+ Cathode: The negative terminal of an electric current sy 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 Mg. Cl 2 is 2(+) x 1 M = 2 eq Number of salt equivalents in 3. 12 x 10 -2 mol of Fe(NO 3)3 is 3(+) x 3. 12 x 10 -2 = 9. 36 x 10 -2 eq = 93. 6 meq Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Salts When crysralluized from aqueous solutions When heated, these salts lose their many salts crystallise as hydrates: crystalline water and become Cu. SO 4 • 5 H 2 O - copper (II) sulfate pentahydrate “anhydrous salts”. Co. CI 2 • 6 H 2 O - cobalt (II) chloride hexahydrate Sn. Cl 2 • 2 H 2 O - stannous (tin II) chloride dihydrate HYDRATE is a salt containing specific numb water molcules as part of solid crystalline struc Cu. SO 4 • 5 H 2 O Co. CI 2 • 6 H 2 O Cu. SO 4 WATER OF HYDRATION is water retained part of the solid crystalline structure of some s Co. C I 2 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Salts Syntheses of salts – reactions yielding salts Metal + nonmetal Fe(s) + S(s) → Fe. S(s) Metals Zn(s) + H 2 SO 4(aq) → Zn. SO 4(aq) + H 2( Metal oxides Fe. O(s) + 2 HCl(aq) → Fe. Cl 2(aq) + H 2 O Hydroxides/bases Al(OH)3(s) + 3 CH 3 COOH(aq) → Al(CH 3 COO)3(aq) + Carbonates 3 Ca. CO 3(s) + 2 H 3 PO 4(aq) → Ca 3(PO 4)2(aq) + 3 CO 2(g) + 3 Bicarbonates KHCO 3(aq) + HNO 3(aq) → KNO 3(aq) + CO 2(g) + H Strong acids react with salts of weak acids Na 3 BO 3(aq) + 3 HBr(aq) → H 3 BO 3(aq) + 3 Na. B Two soluble salts when mixed forming a new insoluble s Ba. Cl 2(aq) + K 2 SO 4(aq) → Ba. SO 4(s) + 2 KCl(aq) “Acidic oxides” 2 Na. OH(aq) + CO 2(g) → Na 2 CO 3(aq) + H 2 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Types of Acids The strength of Acids and Bases Monoprotic - a solution that produces one mole of H+ ions per mole of acid HCl , HNO 3 Diprotic - a solution that produces two moles of H+ ions per mole of acid H 2 SO 4 Triprotic - a solution that produces three moles of H+ ions per mole of acid H 3 PO 4 Polyprotic - two ore more H+ per mole of acid STRONG ACIDS Acids that are essentially 100% Ionized in aqueous solutions WEAK ACIDS Acids that are partially ionized (usually less than 5%) in equilibrium Nitrous Acid - HNO 2 % dissociation 6. 7 Sulfurous Acid - moderate H 2 SO 3 Hydrohalic acids: HCl, HBr, HI ly 34 Phosphorous Acid - weak Nitric acid: HNO 3 H PO 0. 2 3 3 Sulfuric acid: H 2 SO 4 Carbonic Acid - 1. 3 Perchloric acid: HCl. O 4 H CO 3 0. 01 2 Acetic Acid - HC 2 H 3 O 2 Boric Acid - H 3 BO 3 Silicic Acid - H 2 Si. O 3 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS The strength of Acids and Bases ACID DISSOCIATION CONSTANT The equilibrium constant for the dissociation of an acid. HA + H 2 O A− + H 3 O+ CH 3 COOH + H 2 O CH 3 COO- + H n all but the most concentrated solutions it can be assumed that the concentration of water, [H 2 O], 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 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 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 Ka 1 and the constants for dissociation of successive protons as Ka 2, etc. Phosphoric acid, H 3 PO 4, is an example of a polyprotic acid as it can lose three protons. equilibrium H 3 PO 4 H 2 PO 4− + H+ Ka 1 H 2 PO 4− HPO 42− + H+ Ka 2 HPO 42− PO 43− + H+ Ka 3 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS The strength of Acids and Bases 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 Na. OH(s) + H 2 O(l) → Na+(aq) + OH-(aq) Weak Bases Strong Bases Aluminum Hydroxide - Al(OH)3 Na. OH - Sodium Hydroxide Iron (II) Hydroxide - Fe(OH)2 KOH - Potassium Hydroxide Iron (III) Hydroxide - Fe(OH)3 Ca(OH)2 - Calcium Hydroxide Ba(OH)2 - Barium Hydroxide Zinc Hydroxide - Zn(OH)2 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS The strength of Acids and Bases NH 3(aq) + H 2 O NH 4+(aq) + OH BASE DISSOCIATION CONSTANT Na. OH(aq) Na+(aq) + OH-( The equilibrium constant for the dissociation of a base. B + H 2 O HB+ + OH− Using similar reasoning to that used before In 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 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS 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 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. known volume of unknown concentration (of acid) Titration curve of a strong base titrating a strong acid Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Analyzing Acids and Bases The volume of titrant added from the buret is measured. For our example, lets assume that 18. 3 m. L of 0. 115 M Na. OH has been added to 25. 00 m. L 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 HNO 3 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. Titration curve of a strong base titrating a strong acid Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Analyzing Acids and Bases p. H meter ENDPOINT - The volume or amount of acid or base To determine endpoint indicators can be used added to a solution to (paper or soluble indicator dyes) or a p. H meter neutralize the unknown solution during a titration. Transition p. H Indicator Low p. H color High p. H color range 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 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Titration Calculations acid/base molar ratio HNO 3(aq) + Na. OH(aq) Na. NO 3(aq) + H 2 O(l) 1 : 1 H 2 SO 4(aq) + 2 Na. OH(aq) Na 2 SO 4(aq) + 2 H 2 O 1 : 2 (l) H 3 PO 4(aq) + 3 Na. OH(aq) Na 3 PO 4(aq) + 3 H 2 O 1 : 3 (l) M 1 V 1 = M 2 V 2 30 m. L of 0. 10 M Na. OH neutralized 25. 0 m. L of hydrochloric acid. Determine the concentration of the acid Na. OH(aq) + HCl(aq) -----> Na. Cl(aq) + H 2 0. 03 L x 0. 1 mol/L = 0. 025 L x M 2 = 0. 003 mol/0. 035 L = 0. 12 mol/L Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Titration Calculations 50 m. L of 0. 2 mol L-1 Na. OH neutralized 20 m. L of sulfuric acid. Determine the concentration of the acid 1. Write the balanced chemical equation for the reaction Na. OH(aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O(l) 2. Extract the relevant information from the question: Na. OH V = 50 m. L, M = 0. 2 M H 2 SO 4 V = 20 m. L, M = ? 3. Check the data for consistency Na. OH V = 50 x 10 -3 L, M = 0. 2 M H 2 SO 4 V = 20 x 10 -3 L, M = ? 4. Calculate moles Na. OH n(Na. OH) = M x V = 0. 2 x 50 x 10 -3 = 0. 01 mol 5. From the balanced chemical equation find the mole ratio Na. OH: H 2 SO 4 2: 1 6. Find moles H 2 SO 4 Na. OH: H 2 SO 4 is 2: 1 So n(H 2 SO 4) = ½ x n(Na. OH) = ½ x 0. 01 = 5 x 10 -3 moles H 2 SO 4 at the equivalence point 7. Calculate concentration of H 2 SO 4: M = n ÷ V n = 5 x 10 -3 mol, V = 20 x 10 -3 L M(H 2 SO 4) = 5 x 10 -3 ÷ 20 x 10 -3 = 0. 25 M or 0. 25 mol L-1 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Hydrolysis Reactions of Salts Hydrolysis is a chemical reaction with water. Salts of weak acids and/or bases hydrolyze in aqueous solution CH 3 COONa(aq) + H 2 O(l) CH 3 COOH(aq) + Na+(aq) + basic p. H NH 4 Cl(aq) + H 2 O(l) NH 4 OH(aq) + H 3 O+(aq) + Cl-( acidic p. H Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers BUFFER A solution with the ability to resist changing p. H 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 O CH 3 COOH(aq) + OH-(aq) CH 3 COO-(aq) + H 2 O(l) CH 3 COO-(aq) + H+(aq) CH 3 COOH(aq) BUFFER CAPACITY the amount of acid (H+) that can be absorbed b This is how a buffer solution a buffer without causin significant change in p resists changes ion p. H Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers Antacids create buffered solutions In blood plasma, the carbonic acid and hydrogen carbonate ion equilibrium buffers the In this buffer, carbonic acid (H 2 CO 3) is the hydrogen-ion donor (acid) and hydrogen carbonate ion (HCO 3 -) is the hydrogen-ion acceptor (base). H 2 CO 3(aq) H+(aq) + HCO 3 -(aq) Al(OH)3 Wyeth amphojel tablets of aluminum hydroxide Ca. CO 3 Additional H+ is consumed by HCO 3 - and additional OH- is consumed by H 2 CO 3. The value of Ka for this equilibrium is 7. 9 × 10 -7. The p. H of arterial blood plasma is 7. 40. If the p. H falls below this normal value, a condition called acidosis is produced. If the p. H rises above the normal value, he condition is called alkalosis. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers The equilibrium constant will be: In 1 L of solution, there are going to be about 55 moles of wate 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 p. Ka bears exactly the same relationship to Ka as p. H does to the hydrogen ion concentration: The higher the value for p. Ka, the weaker the acid. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers acid Ka (mol dm-3) p. Ka 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, H 3 PO 4, is an example of a polyprotic acid as it can lose three protons. equilibrium p. Ka value H 3 PO 4 H 2 PO 4− + H+ p. Ka 1 = 2. 15 H 2 PO 4− HPO 42− + H+ p. Ka 2 = 7. 20 HPO 42− PO 43− + H+ p. Ka 3 = 12. 37 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers p. Ka = p. H - log [base] ------- [acid] p. H = -log[H+] or -log[H 3 O+] same equation Henderson-Hasselbalch equation Relationship between the p. H, p. Ka and the concentrations of acid and base in the buffer. p. H = p. Ka + log [base] ------- [acid] Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers If we need to make a buffer solution of a certain p. H, we would usually select an acid with the p. Ka near the desired p. H and then adjust the concentration of the acid and the conjugate base (the anion of the acid) to give the desired p. H. 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. Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers Calculate the p. H of buffers that contain the acid and conjugate base in following c a) [HPO 42 -] = 0. 33 M, [PO 43 -] = 0. 52 M p. H = p. Ka + log{[PO 43 -]/[PO 42 -]} = 12. 66 + log(0. 52 M/0. 33 M) = 12. 66 + 0. 20 = b) [CH 3 COOH] = 0. 40 M, [CH 3 COO-] = 0. 25 p. H = p. Ka + log[CH 3 COO-]/[CH 3 COOH] = 4. 74 + log(0. 25 M/0. 40 M) = 4. 74 - 0. 2 What ratio of concentrations of Na. H 2 PO 4 and Na 2 HPO 4 in solution would give a buffer with p. H = 7. 65? p. H = p. Ka + log{[PO 42 -]/[PO 4 -]} 7. 65 = 7. 21 + log{[PO 42 -]/[PO 4 -]} = 0. 44 [PO 42 -]/[PO 4 -] = 100. 44 = 2. 75 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers 1. 00 mol of ammonium chloride is added to 2. 00 L of 1. 00 mol. L -1 NH 3(aq) (Kb = 1. 8 x 10 -5 mol. L-1) a. Calculate the p. H of the solution. b. 10. 00 m. L of the solution is placed into a 1000 m. L volumetric flask and made up to the mark with pure water. What is the p. H of this solution? [acid] p. OH = p. Kb + log ------- [base] NH 4+(aq) + H 2 O(l) H 3 O+(aq) + NH 3(aq) [acid] = [NH 4+] = 1 mol/2 L = 0. 5 mol/ [base] = [NH 3] = [NH 4 OH] = 1 mol/L p. OH = -log(1. 8 x 10 -5 mol/L) + log(0. 5/1 p. OH = 4. 74 - 0. 30 = 4. 44 p. H = 14 - p. OH = 9. 56 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers 1. 00 mol of ammonium chloride is added to 2. 00 L of 1. 00 mol. L -1 NH 3(aq) (Kb = 1. 8 x 10 -5 mol. L-1) a. Calculate the p. H of the solution. b. 10. 00 m. L of the solution is placed into a 1000 m. L volumetric flask and made up to the mark with pure water. What is the p. H of this solution? b. Buffer solution diluted 1: 100 with pure water, ratio of conjugate acid and base remains unchanged. Therefore p. H is unchanged. p. H = 9. 56 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers 1. 00 mol of ammonium chloride is added to 2. 00 L of 1. 00 mol. L -1 NH 3(aq) (Kb = 1. 8 x 10 -5 mol. L-1) a. Calculate the p. H of the solution. b. 10. 00 m. L of the solution is placed into a 1000 m. L volumetric flask and made up to the mark with pure water. What is the p. H of this solution? c. 10. 0 m. L of 0. 100 mol. L-1 HCl(aq) is added to 100 m. L of each of the above solutions. What is the final p. H in each case? d. 1. 00 m. L of 0. 100 mol. L-1 Na. OH(aq) is added to fresh 100 m. L portions of each solution. What is the final p. H in each case? [NH 4+] = 1. 00 mol/2. 00 L = 0. 500 mol. L-1, [NH 3] = 1. 00 mol. L-1 (given). neutralization reaction: HCl(aq) + NH 3(aq) ==> NH 4+(aq) + Cl-(aq) (complete) i. e. 0. 0010 mol of NH 3 will be converted to [NH 4ˆ+] 100 m. L of solution contain: amount of NH 4+ = 0. 500 mol. L-1 x 0. 100 L = 0. 0500 mol NH 4+(aq) + H 2 O(l) <===> H 3 O+(aq) + NH 3(aq) amount of NH 3 = 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 init/mol 0. 0500 0. 100 ch/mol +0. 0010 -0. 0010 0. 0510 final/mol 0. 099 p. H = p. Ka - log 10(a/b) = -log 10{Kw/Kb} - log 10{[NH 4+]/[NH 3]} = 9. 26 - log 10{0. 0051/0. 099} = 9. 26 + 0. 29 => p. H = 9. 55 14 - 4. 74 = 9. 26 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers 1. 00 mol of ammonium chloride is added to 2. 00 L of 1. 00 mol. L -1 NH 3(aq) (Kb = 1. 8 x 10 -5 mol. L-1) a. Calculate the p. H of the solution. b. 10. 00 m. L of the solution is placed into a 1000 m. L volumetric flask and made up to the mark with pure water. What is the p. H of this solution? c. 10. 0 m. L of 0. 100 mol. L-1 HCl(aq) is added to 100 m. L of each of the above solutions. What is the final p. H in each case? d. 1. 00 m. L of 0. 100 mol. L-1 Na. OH(aq) is added to fresh 100 m. L portions of each solution. What is the final p. H 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 NH 4+ & NH 3: amount of NH 4+ = 0. 0500 mol x 0. 00100 = 0. 000500 mol The amounts of HCl and NH 3 are equal; we no longer have a buffer, but exact amount of NH 3 = 0. 100 mol x 0. 00100 = 0. 00100 mol amount of HCl = 0. 100 mol. L-1 x 0. 010 L = 0. 0010 mol neutralization, since all the NH 3 is converted to NH 4+ we have 0. 00150 mol of NH 4+ in 110 m. L [NH 4+] = 0. 00150 mol/0. 110 L = 0. 0136 mol. L-1 of solution. This is a solution of a weak acid. Ka = Kw/Kb = 1. 00 x 10 -14/1. 8 x 10 -5 = 5. 6 x 10 -10 Since [NH 4+]>>Ka we can use the approximation [NH 4+]>>[H 3 O+] (sqrt = square root. ) [H 3 O+] = sqrt(C 0. Ka) =sqrt(0. 00150 x 5. 6 x 10 -10) mol. L-1 = 9. 17 x 10 -7 => p. H = 6. 04 Chemistry 21 A Dr. Dragan Marinkovic

ACIDS, BASES and SALTS Buffers 1. 00 mol of ammonium chloride is added to 2. 00 L of 1. 00 mol. L -1 NH 3(aq) (Kb = 1. 8 x 10 -5 mol. L-1) a. Calculate the p. H of the solution. b. 10. 00 m. L of the solution is placed into a 1000 m. L volumetric flask and made up to the mark with pure water. What is the p. H of this solution? c. 10. 0 m. L of 0. 100 mol. L-1 HCl(aq) is added to 100 m. L of each of the above solutions. What is the final p. H in each case? d. 1. 00 m. L of 0. 100 mol. L-1 Na. OH(aq) is added to fresh 100 m. L portions of each solution. What is the final p. H in each case? d. (Outline solution only) Similar to c, but a smaller amount of base added. The neutralisation reaction is now: Na. OH(aq) + NH 4+(aq) ---> Na+(aq) + NH 3(aq) + H 2 O(l) (complete) i. solution a: amount of NH 4+ = 0. 0500 mol amount of NH 3 = 0. 100 mol amount of Na. OH = 0. 00010 mol ii. solution b amount of NH 4+ = 0. 00050 mol + + NH 4 (aq) + H 2 O(l) <===> H 3 O (aq) + NH 3(aq) amount of NH 3 = 0. 00100 mol init/mol 0. 0500 0. 100 ch/mol -0. 0001 +0. 0001 amount of Na. OH = 0. 00010 mol still a buffer; final/mol 0. 0499 0. 100 p. H = 9. 57 similar to above final [NH +] = 0. 00040 mol ; f 4 inal [NH 3] =0. 00110 mol p. H = 9. 73 Chemistry 21 A Dr. Dragan Marinkovic