1. 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 l itmus (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.

2. 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 - HNO 3 Chloric Acid - HClO 3 Perchloric Acid - HClO 4 Sulfuric Acid - H 2 SO 4 Phosphoric Acid - H 3 PO 4 Acetic Acid - HC 2 H 3 O 2 Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH) 2 Barium Hydroxide - Ba(OH) 2 How these acids and bases dissociate?

3. 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 CH 3 CO 2 H + H 2 O CH 3 CO 2 - + H 3 O + H 2 O + NH 3 OH - + NH 4 + 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 H 3 0 + forms immediately.

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

5. 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 H 2 O, HSO 4 -, HPO 4 2-, HSO 3 - etc. HSO 4 - + H 3 O + H 2 SO 4 + H 2 O HSO 4 - + OH - SO 4 2- + 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 +

6. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Classical acid (and salt) naming system: Anion (Salt) Prefix Anion (Salt) Suffix Acid PrefixAcid SuffixExample perateperic acidperchloric acid (HClO 4 ) ateic acidchloric acid (HClO 3 ) iteous acidchlorous acid (HClO 2 ) hypoitehypoous acid hypochlorous acid (HClO) idehydroic acidhydrochloric acid (HCl) sulf nitr phosph carbon brom iod sulfur nitr phosphor carbon brom iod

7. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Hydrofluoric Acid - HF Hydrochloric Acid - HCl Hydrobromic Acid - HBr Hydroiodic Acid - HI Hydrosulfuric Acid - H 2 S Nitric Acid - HNO 3 Nitrous Acid - HNO 2 Hypochlorous Acid - HClO Chlorous Acid - HClO 2 Chloric Acid - HClO 3 Perchloric Acid - HClO 4 Sulfuric Acid - H 2 SO 4 Sulfurous Acid - H 2 SO 3 Phosphoric Acid - H 3 PO 4 Phosphorous Acid - H 3 PO 3 Carbonic Acid - H 2 CO 3 Acetic Acid - HC 2 H 3 O 2 Oxalic Acid - H 2 C 2 O 4 Boric Acid - H 3 BO 3 Silicic Acid - H 2 SiO 3 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 H 4 SiO 4, 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: HNO 3 Sulfuric acid: H 2 SO 4 Perchloric acid: HClO 4

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

9. 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 CH 3 COOH CH 3 CHOHCOOH

10. 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

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

12. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Bases Sodium Hydroxide - NaOH Potassium Hydroxide - KOH Ammonium Hydroxide - NH 4 OH 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 bases Sodium Hydroxide - NaOH Potassium Hydroxide – KOH Calcium Hydroxide - Ca(OH) 2 Barium Hydroxide - Ba(OH) 2

13. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic CH 3 CH 2 NH 2 CH 3 NH 2 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, C 17 H 19 NO 3, 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 )

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

15. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic 2H 2 O H 3 O + + OH - at 25 o C K W = const. [H 2 O] = const. = 55.5 M K w = [H 3 O + ][OH - ] = 10 -14 K eq x[H 2 O] 2 = [H 3 O + ][OH - ] K eq x[H 2 O] 2 = K w The product of water

16. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic 2H 2 O H 3 O + + OH - at 25 o C K w = [H 3 O + ][OH - ] = 10 -14 [H 3 O + ] = [OH - ] = 10 -7 K eq x[H 2 O] 2 = [H 3 O + ][OH - ] T (°C) Kw pKw neutral pH 0 0.114 × 10 -14 14.947.47 5 0.186 × 10 -14 14.737.37 20 0.681 × 10 -14 14.177.08 7015.85 × 10 -14 12.806.40 100 51.3 × 10 -14 12.296.14 The product of water if THE SOLUTION IS NEUTRAL

17. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic 2H 2 O H 3 O + + OH - pH = -log[H + ] or -log[H 3 O + ] pH = -log [1 x 10 -7 ] = -(-7.00) = 7.00 pOH = -log[OH - ] [H 3 O + ] = 10 -pH at 25 o C K W = const. [H 2 O + ] = const. = 55.5 M K w = [H 3 O + ][OH - ] = 10 -14 [H 3 O + ] = [OH - ] = 10 -7 K eq x[H 2 O] 2 = [H 3 O + ][OH - ] K eq x[H 2 O] 2 = K w THE pH CONCEPT IN THE NEUTRAL SOLUTION For convenience instead of exponential numbers, negative logarithms of these numbers are used.

18. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic [H + ] pH Example Acids (acidic Solutions) 1 x 10 0 0HCl 1 x 10 -1 1Stomach acid 1 x 10 -2 2Lemon juice 1 x 10 -3 3Vinegar 1 x 10 -4 4Soda (Coca-Cola) 1 x 10 -5 5Rainwater 1 x 10 -6 6Milk Neutral1 x 10 -7 7Pure water Bases (basic Solutions) 1 x 10 -8 8Egg whites 1 x 10 -9 9Baking soda 1 x 10 -10 10Tums ® antacid 1 x 10 -11 11Ammonia 1 x 10 -12 12Mineral lime - Ca(OH) 2 1 x 10 -13 13Drano ® 1 x 10 -14 14NaOH 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.

19. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic pH in living systems CompartmentpH Gastric acid0.7 Lisosomes4.5 Granules of chromaffin cells5.5 Urine6.0 Neutral H 2 O at 37 °C6.81 Cytosol7.2 Cerebrospinal fluid (CSF)7.3 Blood7.34 – 7.45 Mitochondrial matrix7.5 Pancreas secretions8.1

20. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Calculate the molar concentration of H 3 O + in water solutions with the following OH - molar concentrations: Calculate the molar concentration of OH - in water solutions with the following H 3 O + molar concentrations: a)6.9x10 -5 b)0.074 a)0.0087 b)9.9x10 -10

21. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Calculate the molar concentration of H 3 O + in water solutions with the following OH - molar concentrations: Calculate the molar concentration of OH - in water solutions with the following H 3 O + molar concentrations: a)6.9x10 -5 b)0.074 a)0.0087 b)9.9x10 -10 K w = [H 3 O + ][OH - ] = 10 -14 [H 3 O + ] = 10 -14 /[OH - ] = 10 -14 /[6.9x10 -5 ] = 1.449x10 -10 M [H 3 O + ] = 10 -14 /[OH - ] = 10 -14 /[7.4x10 -2 ] = 1.35x10 -13 M

22. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Calculate the molar concentration of H 3 O + in water solutions with the following OH - molar concentrations: Calculate the molar concentration of OH - in water solutions with the following OH - H 3 O + molar concentrations: a)6.9x10 -5 b)0.074 a)0.0087 b)9.9x10 -10 K w = [H 3 O + ][OH - ] = 10 -14 [H 3 O + ] = 10 -14 /[OH - ] = 10 -14 /[6.9x10 -5 ] = 1.449x10 -10 M [H 3 O + ] = 10 -14 /[OH - ] = 10 -14 /[7.4x10 -2 ] = 1.35x10 -13 M [OH - ] = 10 -14 /[H 3 O + ] = 10 -14 /[8.7x10 -3 ] = 1.15x10 -12 M [OH - ] = 10 -14 /[H 3 O + ] = 10 -14 /[9.9x10 -10 ] = 1x10 -5 M

23. 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 = 3.95 b) pH = 4.00 c) pH = 11.86

24. 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 = 3.95 b) pH = 4.00 c) pH = 11.86 a) pH = -log [H + ] = -log(7.5x10 -6 ) = 5.12 pH + pOH = 14 b) pOH = -log [OH - ] = -log(2.5x10 -4 ) = 3.6pH = 14 - pOH = 14 - 3.6 = 10.4

25. 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 = 3.95 b) pH = 4.00 c) pH = 11.86 pH = -log [H + ] = -log(7.5x10 -6 ) = 5.12 pH + pOH = 14 pOH = -log [OH - ] = -log(2.5x10 -4 ) = 3.6pH = 14 - pOH = 14 - 3.6 = 10.4 [H 3 O + ] = 10 -pH [H 3 O + ] = 10 -pH = 10 -3.95 = 1.12x10 -4 [OH - ] = 10 -pOH [OH - ] = 10 -pOH = 10 -10.05 = 8.91x10 -11

26. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic pH meter

27. 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, 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 Hydrohalic acids: HCl, HBr, HI Nitric acid: HNO 3 Sulfuric acid: H 2 SO 4 Perchloric acid: HClO 4 WEAK ACIDS Acids that are partially ionized (usually less than 5%) in equilibrium. Properties of Acids Nitrous Acid - HNO 2 Sulfurous Acid - H 2 SO 3 Phosphorous Acid - H 3 PO 3 Carbonic Acid - H 2 CO 3 Acetic Acid - HC 2 H 3 O 2 Boric Acid - H 3 BO 3 Silicic Acid - H 2 SiO 3

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

29. 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? 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 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

30. 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? 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 V 2 = (12 mol/L x 1 L) / (0.5 L) = 24 L

31. 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 CuSO 4 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?

32. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Properties of Bases Strong Bases NaOH - Sodium Hydroxide KOH - Potassium Hydroxide Ca(OH) 2 - Calcium Hydroxide Ba(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) + 3CH 3 COOH (aq) → Al(CH 3 COO) 3(aq) + 3H 2 O ( l ) “Acidic oxides” 2NaOH (aq) + CO 2(g) → Na 2 CO 3(aq) + H 2 O ( 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) + H 2 O ( l ) → Na + (aq) + OH - (aq)

33. 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) + H 2 O ( l ) → Na + (aq) + OH - (aq) NaOHHCl

34. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic 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 4 2- ). Salt's solution in water is called electrolytes. Both, the electrolytes and molten salts conduct electricity. Salts with OH - are basic salts (CaOHCl, BaOHNO 3 ) and with H + are acidic salts (NaHSO 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).

35. 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 + - 2H 2 O → 2H 2 + O 2 Cathode (reduction): 2H + ( aq ) + 2e − → H 2 ( g ) Anode (oxidation): 2H 2 O( l ) → O 2 ( g ) + 4H + ( aq ) + 4e Silver electroplating. The anode is a silver bar and the cathode is an iron spoon. anode cathode

36. 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 -, NO 3 -, SO 4 2- CATIONS – positive (usually metal) ions, in aqueous solutions move towards CATHODE, K +, Al 3+, Ba 2+ 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 MgCl 2 is 2(+) x 1 M = 2 eq Number of salt equivalents in 3.12x10 -2 mol of Fe(NO 3 ) 3 is 3(+) x 3.12x10 -2 = 9.36x10 -2 eq = 93.6 meq

37. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Salts When crysralluized from aqueous solutions many salts crystallise as hydrates : CuSO 4 5H 2 O - copper (II) sulfate pentahydrate CoCI 2 6H 2 O - cobalt (II) chloride hexahydrate SnCl 2 2H 2 O - 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. CuSO 4 5H 2 O CoCI 2 CoCI 2 6H 2 O CuSO 4

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

39. 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, 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 Hydrohalic acids: HCl, HBr, HI Nitric acid: HNO 3 Sulfuric acid: H 2 SO 4 Perchloric acid: HClO 4 The strength of Acids and Bases % dissociation 6.7 34 0.2 1.3 0.01 Nitrous Acid - HNO 2 Sulfurous Acid - H 2 SO 3 Phosphorous Acid - H 3 PO 3 Carbonic Acid - H 2 CO 3 Acetic Acid - HC 2 H 3 O 2 Boric Acid - H 3 BO 3 Silicic Acid - H 2 SiO 3 WEAK ACIDS Acids that are partially ionized (usually less than 5%) in equilibrium. moderately weak

40. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic 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 + 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 CH 3 COOH + H 2 O CH 3 COO - + H 3 O +

41. 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 K a1 and the constants for dissociation of successive protons as K a2, 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 + K a1 H 2 PO 4 − HPO 4 2− + H + K a2 HPO 4 2− PO 4 3− + H + K a3

42. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic The strength of Acids and Bases Strong Bases NaOH - Sodium Hydroxide KOH - Potassium Hydroxide Ca(OH) 2 - Calcium Hydroxide Ba(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) + H 2 O ( l ) → Na + (aq) + OH - (aq)

43. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic The strength of Acids and Bases 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 K w = [H + ][OH − ], therefore Substitution of the expression for [OH − ] into the expression for K b give BASE DISSOCIATION CONSTANT The equilibrium constant for the dissociation of a base. NH 3(aq) + H 2 O NH 4 + (aq) + OH - (aq) NaOH (aq) Na + (aq) + OH - (aq)

44. 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.

45. 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 HNO 3 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.

46. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Analyzing Acids and Bases ENDPOINT - The volume or amount of acid or base added to a solution to neutralize the unknown solution during a titration. IndicatorLow pH color Transition pH range High pH color Gentian violet (Methyl violet)yellow0.0–2.0blue-violet Thymol blue (first transition)red1.2–2.8yellow Thymol blue (second transition)yellow8.0–9.6blue Methyl orangered3.1–4.4orange Bromocresol purpleyellow5.2–6.8purple Bromothymol blueyellow6.0–7.6blue Phenol redyellow6.8–8.4red Cresol Redyellow7.2–8.8reddish-purple Phenolphthaleincolorless8.3–10.0fuchsia Alizarine Yellow Ryellow10.2–12.0red pH meter To determine endpoint indicators can be used (paper or soluble indicator dyes) or a pH meter

47. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Titration Calculations M 1 V 1 = M 2 V 2 HNO 3(aq) + NaOH (aq) NaNO 3(aq) + H 2 O ( l ) H 2 SO 4(aq) + 2NaOH (aq) Na 2 SO 4(aq) + 2H 2 O ( l ) H 3 PO 4(aq) + 3NaOH (aq) Na 3 PO 4(aq) + 3H 2 O ( l ) acid/base molar 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) + H 2 O(l) 0.03 L x 0.1 mol/L = 0.025 L x M 2 M 2 = 0.003 mol/0.035 L = 0.12 mol/L

48. 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 acid 1.Write the balanced chemical equation for the reaction NaOH(aq) + H 2 SO 4 (aq) -----> Na 2 SO 4 (aq) + 2H 2 O(l) 2.Extract the relevant information from the question: NaOH V = 50mL, M = 0.2M H 2 SO 4 V = 20mL, M = ? 3.Check the data for consistency NaOH V = 50 x 10 -3 L, M = 0.2M H 2 SO 4 V = 20 x 10 -3 L, 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:H 2 SO 4 2:1 6.Find moles H 2 SO 4 NaOH: H 2 SO 4 is 2:1 So n(H 2 SO 4 ) = ½ x n(NaOH) = 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 = 0.25M or 0.25 mol L -1 Titration Calculations

49. 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. CH 3 COONa(aq) + H 2 O( l ) CH 3 COOH(aq) + Na + (aq) + OH - (aq) NH 4 Cl(aq) + H 2 O( l ) NH 4 OH(aq) + H 3 O + (aq) + Cl - (aq) acidic pH basic pH

50. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Buffers BUFFER 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. CH 3 COO - (aq) + H + (aq) CH 3 COOH(aq) CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq ) + H 2 O( l ) This is how a buffer solution resists changes ion pH

51. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Buffers Wyeth amphojel tablets of aluminum hydroxide CaCO 3 Antacids create buffered solutions In blood plasma, the carbonic acid and hydrogen carbonate ion equilibrium buffers the pH. 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) Additional H + is consumed by HCO 3 - and additional OH - is consumed by H 2 CO 3. The value of K a 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

52. 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. K c (a constant) times the concentration of water (another constant) on the left-hand side. The product of those is then given the name K a. An introduction to pK a pK a bears exactly the same relationship to K a as pH does to the hydrogen ion concentration: The higher the value for pK a, the weaker the acid.

53. ACIDS, BASES and SALTS Chemistry 21A Dr. Dragan Marinkovic Buffers equilibriumpK a value H 3 PO 4 H 2 PO 4 − + H + pK a1 = 2.15 H 2 PO 4 − HPO 4 2− + H + pK a2 = 7.20 HPO 4 2− PO 4 3− + H + pK a3 = 12.37 acidK a (mol dm -3 )pK a hydrofluoric acid5.6 x 10 -4 3.3 formic acid1.6 x 10 -4 3.8 acetic acid1.7 x 10 -5 4.8 hydrogen sulfide8.9 x 10 -8 7.1 Phosphoric acid, H 3 PO 4, is an example of a polyprotic acid as it can lose three protons.

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

55. 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 pK a 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.

56. 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) [HPO 4 2- ] = 0.33 M, [PO 4 3- ] = 0.52 M pH = pK a + log{[PO 4 3- ]/[PO 4 2- ]} = 12.66 + log(0.52 M/0.33 M) = 12.66 + 0.20 = 12.86 b) [CH 3 COOH] = 0.40 M, [CH 3 COO - ] = 0.25 pH = pK a + log[CH 3 COO-]/[CH 3 COOH] = 4.74 + log(0.25 M/0.40 M) = 4.74 - 0.2 = 4.54 What ratio of concentrations of NaH 2 PO 4 and Na 2 HPO 4 in solution would give a buffer with pH = 7.65? pH = pK a + log{[PO 4 2- ]/[PO 4 - ]} 7.65 = 7.21 + log{[PO 4 2- ]/[PO 4 - ]} log{[PO 4 2- ]/[PO 4 - ]} = 0.44 [PO 4 2- ]/[PO 4 - ] = 10 0.44 = 2.75

57. 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 NH 3 (aq) (K b = 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 = pKbpKb + log [acid] ------------- ------------- [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/L [base] = [NH 3 ] = [NH 4 OH] = 1 mol/L pOH = -log(1.8x10 -5 mol/L) + log(0.5/1 molL -1 ) pOH = 4.74 - 0.30 = 4.44 pH = 14 - pOH = 9.56

58. 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 NH 3 (aq) (K b = 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

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

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

61. 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 NH 3 (aq) (K b = 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? [NH 4 + ] = 1.00 mol/2.00 L = 0.500 mol.L -1, [NH 3 ] = 1.00 mol.L -1 (given). 100 mL of solution contain: amount of NH 4 + = 0.500 mol.L -1 x 0.100 L = 0.0500 mol 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 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 4 ˆ + ] NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) init/mol 0.0500 0.100 ch/mol +0.0010 -0.0010 0.0510 final/mol 0.099 pH = pK a - log 10 (a/b) = -log 10 {K w /K b } - log 10 {[NH 4 + ]/[NH 3 ]} = 9.26 - log 10 {0.0051/0.099} = 9.26 + 0.29 => pH = 9.55 14 - 4.74 = 9.26

62. 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 NH 3 (aq) (K b = 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 NH 4 + & NH 3 : amount of NH 4 + = 0.0500 mol x 0.00100 = 0.000500 mol 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 The amounts of HCl and NH 3 are equal; we no longer have a buffer, but exact neutralization, since all the NH 3 is converted to NH 4 + we have 0.00150 mol of NH 4 + in 110 mL of solution. This is a solution of a weak acid. [NH 4 + ] = 0.00150 mol/0.110 L = 0.0136 mol.L -1 K a = K w /K b = 1.00 x 10 -14 /1.8 x 10 -5 = 5.6 x 10 -10 Since [NH 4 +]>>K a we can use the approximation [NH 4 + ]>>[H 3 O + ] (sqrt = square root.) [H 3 O + ] = sqrt(C 0.K a ) =sqrt(0.00150 x 5.6 x 10 -10 ) mol.L -1 = 9.1 7 x 10 -7 => pH = 6.04

63. 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 NH 3 (aq) (K b = 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) + 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 NaOH = 0.00010 mol NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (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 NH 4 + = 0.00050 mol amount of NH 3 = 0.00100 mol amount of NaOH = 0.00010 mol still a buffer; similar to above final [NH 4 + ] = 0.00040 mol ; f inal [NH 3 ] =0.00110 mol pH = 9.73