Unit 2 ACIDS AND BASES

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1 Unit 2 ACIDS AND BASES Operational Definition One that is used to classify a substance Acids are aqueous solutions that: 1. turn blue litmus red (ph less than 7 at 25 C) 2. neutralize basic solutions 3. taste sour 4. react with active metals to produce hydrogen gas 5. react with carbonates to produce carbon dioxide gas 6. conduct electricity (increased concentration increases conductivity)

2 Bases are aqueous solutions that: 1. turn red litmus blue (ph greater than 7 at 25 C) 2. neutralize acid solutions 3. taste bitter 4. do not react with active metals or carbonates 5. Conduct electricity (increased concentration increases conductivity)

3 Theoretical Definition One that is used to explain the properties of a substance Arrhenius definition: (theoretical) Acid: substance that dissolves in water to increase hydrogen ion (H + ) concentration Ex: HCl(g) H + (aq) + Cl (aq)

4 Base: substance that dissolves in water to increase hydroxide ion (OH ) concentration Ex: NaOH(s) Na + (aq) + OH (aq) This theory can also explain some difficult species such as CO 2 (acid) and NH 3 (base) by first reacting with water then dissociating into H + and OH ions. Ex: Ex: CO 2 as acid CO 2 + H 2 O H 2 CO 3 H + + HCO 3 NH 3 as base NH 3 + H 2 O NH 4 OH NH 4+ + OH

5 Limitations 1. Cannot explain why species such as HCO 3 act as bases operationally when, according to Arrhenius, it is an acid. Ex: HCO 3 H + + CO Proposes the existance of certain species for which there is no empirical evidence (NH 4 OH cannot be isolated)

6 Modified Arrhenius Theory Acid: substance that reacts with water to increase hydronium (H 3 O + ) ion concentration Ex:1. HCl + H 2 O H 3 O + + Cl 2. CO 2 + H 2 O H 2 CO 3 + H 2 O H 3 O + +HCO 3 Base: substance that reacts with water to increase the hydroxide (OH ) ion concentration (only necessary for molecular, not ionic bases) Ex: NH 3 + H 2 O OH + NH 4 +

7 Advantages 1. Can explain why species such as HCO 3 act as bases Ex: HCO 3 + H 2 O OH + H 2 CO 3 2. Also explains why NH 3 is a base without postulating the existance of NH 4 OH Ex: NH 3 + H 2 O NH 4+ + OH

8 Limitations Can explain species such as HCO 3 acting as a base or an acid but cannot predict whether a solution of HCO 3 will be basic or acidic. Ex: HCO 3 as base HCO 3 + H 2 O OH + H 2 CO 3 Ex: HCO 3 as acid HCO 3 + H 2 O H 3 O + + CO 3 2

9 Bronsted-Lowry Theory (proton transfer) Acid: a chemical species that loses or donates a proton in a chemical reaction. (proton donor) acid Ex: HCl + H 2 O H 3 O + + Cl H + Base: a chemical species that gains or accepts a proton in a chemical reaction. (proton acceptor) base Ex: NH 3 + H 2 O OH + NH 4 + H +

10 Advantages 1. Can be used to predict when substances such as HCO 3 can act as an acid or base. Base when placed in contact with a stronger acid. Acid when placed in contact with a weaker acid. See data table to classify acid/base strength. Ex. As base HCO 3 + H 3 O + H 2 O + H 2 CO 3 base + stronger acid As acid HCO 3 + OH H 2 O + CO 2 3 acid + weaker acid

11 2. Allows for explanation of non-aqueous acid/base reactions. Ex. HCl(g) + NH 3 (g) NH 4 Cl(s) (DEMO) 3. Explains acids and bases in terms of only reactions rather than aqueous solutions and reactions. (Broader more inclusive definition) Disadvantages 1. Not good for acid/base reactions where the main concern is identification of the acid or base. We must rely on the operational definition for this 2. Not good for determining relative acidity of a solution. (ph, [OH ] or [H 3 O + ]) (Complete # s 1,2,3,4 pgs )

12 NOTE: Bronsted-Lowry acid/base reactions produce a new acid and base and are equilibrium systems Ex: CH 3 COOH + H 2 O H 3 O + + CH 3 COO acid base acid base CH 3 COOH and CH 3 COO acid-base conjugate pairs. This is a pair of substances in an acid/base reaction that differ by a proton H + Ex: H 3 O + and H 2 O are also conjugate acid-base pairs

13 Acid-Base Reactions - a chemical reaction in which a proton is transferred from the strongest acid present to the strongest base present to form a new acid and base. May operate in reverse (equilibrium) - acid-base classification is for a specific reaction. ie. A substance may be an acid in some reactions and a base in others (amphoteric) Acid strength is a measure of an acids tendency to donate a proton ie. Strong acid - large tendency to donate a proton. Strongest acids give up a proton 100% Base strength is a measure of a bases tendency to accept a proton ie. Strongest bases accept protons 100% See acid/base data table for acid/base strength classification

14 Strong vs Weak Acids There are 6 main strong acids as listed at the top of the table. These acids ionize 100% in water therefore reactions involving these 6 acids are non equilibrium systems. Monoprotic acids Those that contain only one proton to donate Ex HCl + H 2 O H 3 O + + Cl HNO 3 + H 2 O H 3 O + + NO 3

15 Polyprotic acids Those that contain more than one proton These acids react multiple times with H 2 O, once for each proton. As each proton is removed, the remaining acid becomes weaker. Ex H 2 SO 4 (strong diprotic acid) 1. H 2 SO 4 + H 2 O H 3 O + + HSO 4 SA WA 2. HSO 4 + H 2 O H 3 O + + SO 4 2

16 Ex H 3 PO 4 (Weak Triprotic Acid) 1. H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 SA WA 2. H 2 PO 4 + H 2 O H 3 O + + HPO 4 2 SA WA 3. HPO H 2 O H 3 O + + PO 4 3

17 Predicting the most likely acid/base reaction and the extent of the equilibrium, if present. Ex: 1. Spilled oven cleaner containing aqueous sodium hydroxide can be neutralized using vinegar (acetic acid) Step 1: List all species initially present. Rules: i. Soluble ionic compounds- list as ions ii. Insoluble ionic compounds- list in pure state iii. Strong acids- list as H 3 O + and balancing ion iv. Weak acids- list as molecules v. Water is always present Na + (aq), OH (aq), CH 3 COOH(aq), H 2 O(l)

18 Step 2: Classify each species as acid/base or both A SA A Na + (aq), OH (aq), CH 3 COOH(aq), H 2 O(l) SB B Step 3: Identify the strongest acid and base present using the data table. Step 4: Write a reaction with the strongest acid and base as the reactants. Transfer a proton from the acid to the base to predict the products. Equilibrium if acid is weak. CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) H +

19 Step 5: Select the strongest acid present in the equilibrium. This will donate protons more than the other acid CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) SA WA Since the strongest acid is on the reactants side, the equilibrium favours the formation of products. Ex 2: What happens when bleach (sodium hypochlorite) and vinegar are mixed? SA CH 3 COOH(aq), Na +, ClO (aq), H 2 O(l) SB A B CH 3 COOH(aq) + ClO (aq) CH 3 COO (aq) + HClO(aq) SA Products are favoured WA

20 Water Equilibrium Water is slightly conductive. This means there must be ions present in pure H 2 O. This is explained as follows: (Arrhenius) 10 6 % H 2 O(l) H + (aq) + OH (aq) or Bronsted-Lowry 10 6 % H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) The equilibrium highly favours the H 2 O side so this equilibrium is not important in most aqueous reactions. However, since acids and bases produce H + and OH this equilibrium is extremely important.

21 Equilibrium Constant for Water Applying the equilibrium constant to this equilibrium system gives: K = [H + ] [OH ] [H 2 O] Since the [H 2 O(l)] is large and does not change much in this equilibrium, we incorporate it with the equilibrium constant K ie. K w = [H + ] [OH ] = ionization constant for water = 1.00 x This value is for water at 25 C Therefore: [H + ] [OH ] = 1.00 x [H + ] = 1.00 x [OH ] [OH ] = 1.00 x [H + ]

22 Using this equilibrium we can calculate [H + ] and [OH ] in acid and base solutions. NOTE: In neutral solutions [H + ] = [OH ] since these are produced in a 1:1 ratio. This means [H + ] = 1.00 x 10-7 mol/l and [OH ] = 1.00 x 10-7 mol/l If acid,h +, is added to a neutral solution, the equilibrium will shift to favour more water thereby using up the OH such that [H + ] [OH ] = 1.00 x at 25 C We can use this idea to calculate [H + ] and [OH ] in acid or base solutions.

23 ph Determination The ph scale was developed to describe the [H + ] in a simple manner. The [H + ] can be used to indicate how acidic or basic a solution is but the range is too large to be of use. In 1909 a Danish biochemist, Soren Sorensen, described this range using the numerical system of logarithms In this system the logarithm of a number is the power of 10 that equals that number. Ex: logarithm of 10 is 1 because 10 1 = 10 logarithm of 100 is 2 because 10 2 = 100 ph = -log[h + ]

24 From ph of a solution we may quickly determine whether it is acidic or basic and by how much. See ph scale on page 570 in the text. ph problems 1. What is the ph of a solution with [H + ] of 4.7 x 10-11? ph = - log [H + ] = - log (4.7 x ) 2 sig.figs. = decimal places 2. What is the ph of a solution that has [H + ] of 2.0 x 10-3? ph = - log [H + ] = - log (2.0 x 10-3 ) 2 sig.figs. = decimal places

25 3. What is the ph of a solution formed by dissolving 2.65 g of strontium hydroxide (Sr(OH) 2 ) in enough water to form ml of solution Given m Sr(OH) 2 = 2.65 g v = ml M Sr(OH) 2 = g/mol c Sr(OH) 2 = 2.65 g x mol x 1 x 1000 ml g ml 1L = x 10-2 mol/l Since Sr(OH) 2 is a strong base the [OH ] is the stoichiometric amount of the [Sr(OH) 2 ].

26 Sr(OH) 2 (s) Sr 2+ (aq) + 2OH (aq) Mole ratio = 2 mol OH (aq) 1 mol Sr(OH) 2 (s) [OH ] = x 10 2 mol/l Sr(OH) 2 (s) x 2 mol OH (aq) = x 10-2 mol/l 1 mol Sr(OH) 2 (s) [H + ] = Kw [OH ] = 1.00 x x 10-2 = x mol/l

27 ph = - log [H + ] = - log (1.148 x ) 3 sig.figs. = decimal places The ph of the solution is and the solution is basic. Complete text questions 12 to 15 pg 566 and 16 to 19 pg 569

28 Determining [H + ] and [OH ] from ph [H + ] can be determined by obtaining the antilog of ph [H + ] = 10 -ph Ex 1. What is the [H + ] of a solution with a ph of 4.58? [H + ] = 10 - ph = = 2.6 x 10 5 mol/l

29 2. What is the [OH ] of a solution with a ph of 5.22? [H + ] = 10 -ph = = 6.0 x 10 6 mol/l [OH ] = Kw [H + ] = 1.00 x x 10-6 = 1.7 x 10-9 mol/l

30 [OH ] and poh Follows the same pattern as ph except replace [H + ] with [OH ] and poh = - log [OH ] [OH ] = 10 - poh Relationship between ph and poh ph + poh = This allow for convenient conversion between ph and poh

31 Ex 1. What is the ph of a 0.25 mol/l solution of barium hydroxide? (Ba(OH) 2 ) Ba(OH) 2 (s) Ba 2+ (aq) + 2OH (aq) [OH ] = 0.25 mol/l Ba(OH) 2 = 0.50 mol/l x 2 mol OH 1 mol Ba(OH) 2 poh = - log [OH ] = - log (0.50) = 0.30 ph = = 13.70

32 Acids/Bases and Dilution Since acids and bases are solutions, we can apply the principles of dilution to change ph or [H + ] / poh or [OH ] in these solutions. Dilution: c i v i = c f v f Acids: c i = initial [H + ] c f = final [H + ] Bases: c i = initial [OH ] c f = final [OH ] Remember that if ph is given you must convert to [H + ] and if poh is given you must convert to [OH ].

33 Ex 1. In preparing a solution a chemist takes stock HCl solution of ph 1.50 and dilutes 125 ml of this solution to 495 ml. What is the [H + ] in the dilute solution? Given ph = 1.50 v i v f [H + ] f? = 125 ml = 495 ml To use the dilution formula we need [H + ] i [H + ] i = 10 - ph = = mol/l

34 c i v i = c f v f c f = c i v i v f c f = mol/l x 125 ml 495 ml = mol/l The [H + ] in the dilute acid solution is mol/l. You could also be asked to calculate the ph of the dilute solution. Complete # s 26 to 29 pg 547

35 Revisiting Acid/Base Strength Strong Acid: completely ionized in aqueous solution (100% reaction with water) Weak Acid: slightly ionized in aqueous solution (<100% reaction with water) For strong acids, the concentration of H + or H 3 O + is the stoichiometric equivalent amount of the concentration of the acid. Ex: HCl - strong monoprotic acid HCl + H 2 O H 3 O + + Cl [H 3 O + ] = [HCl] H 2 SO 4 - strong diprotic acid H 2 SO 4 + H 2 O H 3 O + + HSO 4 [H 3 O + ] = [H 2 SO 4 ]

36 For weak acids with a concentration of 0.10 mol/l, the concentration of H + or H 3 O + is A) [H 3 O + ] = % reaction x [acid] Ex: What is the [H 3 O + ] of a 0.10 mol/l acetic (ethanoic) acid solution? CH 3 COOH + H 2 O H 3 O + + CH 3 COO % rxn = 1.3% [H 3 O + ] = 1.3% x [CH 3 COOH] = 1.3% x 0.10 mol/l 100% = mol/l

37 B) Use of ionization constant Ka for [acid] == 0.10 mol/l CH 3 COOH + H 2 O H 3 O + CH 3 COO Keq = [H 3 O + ] [CH 3 COO ] [CH 3 COOH] [H 2 O] Since the value of [H 2 O] changes so little, it can be incorporated into Keq to give Ka. Keq [H 2 O] = [H 3 O + ] [CH 3 COO ] [CH 3 COOH] Ka = [H 3 O + ] 2 because [H 3 O + ] = [CH 3 COO ] [CH 3 COOH]

38 In General: For any weak acid. Ka = [H 3 O + ] 2 eq [acid] eq Weak acid and base solutions are equilibrium systems. To solve many of these problems remember equilibrium problems from Ch. 13. Steps. 1. Write a reaction equation. 2. Determine initial values. 3. Setup an ICE table. 4. Use the ICE table to determine the required information. 5. If the initial concentration of the weak acid and Ka is known you can use the 500 rule to avoid using a quadratic. [weak acid]i/ka > 500 initial change in initial concentration, x, is negligible

39 ph of a weak acid eg. Find ph of mol/l HF(aq). Step #1: Write a balanced equation HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Step #2,3,4: Equilibrium Concentrations using ICE table I C -x +x +x E x x x Using 500 rule = 152 cannot ignore drop in x 6.6 x 10-4

40 Step #5: Substitute into Ka expression x 10 [x][x] [ x] x 2 = 6.6 x x 10-4 x x x 10-4 x x 10-5 = 0 a = 1 b = 6.6 x 10-4 c = -6.6 x 10-5 Using Quadratic formula x mol/l Ignor negative values

41 For weak bases we use the base dissociation constant Kb. Same relationship to bases as Ka is to acids. ie: NH 3 + H 2 O NH 4+ + OH Kb = [OH ] 2 [NH 3 ] Relationship between Ka and Kb Ka x Kb = Kw

42 Excess Acid or Base To calculate the ph of a solution produced by mixing an acid with a base: 1. write the B-L equation (NIE) or balanced rxn equation 2. calculate the moles of H 3 O + and OH - 3. subtract to determine the moles of excess H 3 O + or OH - 4. divide by total volume to get concentration 5. calculate ph

43 eg ml of M Ca(OH) 2 (aq) is mixed with 10.0 ml of M HCl(aq). Determine the ph of the resulting solution. ANSWER: Species present: Ca 2+ (aq) OH - (aq) H 3 O + (aq) Cl - (aq) H 2 O(l) SB SA NIE: OH [OH ] - (aq) = 2 x + [Ca(OH) H 3 O + (aq) 2 ] 2 H 2 O(l) n OH- = cv n H3O+ = cv = mol/L x L = mol/L x L = 4.00 x 10-4 mol (excess) = 5.00 x 10-5 mol(limiting) n OH- remaining = 4.00 x 10-4 mol x 10-5 mol = 3.50 x 10-4 mol

44 C OH n = 3.50 x 10-4 mol/ L [OH-] remaining = mol/l poh = - log ( ) = ph = = remaining V total

45 Acid-Base Stoichiometry Same as solution stoichiometry 1. Write a balanced acid/base reaction equation. 2. List the givens and required under the appropriate reactant or product. 3. Convert given to moles. 4. Use a mole ratio to determine moles of required. 5. Convert to required information.

46 Ex ml of mol/l H 2 SO 4 (aq) was used to neutralize 36.5 ml of NaOH(aq). Calculate the molar concentration of the NaOH solution. H 2 SO 4 (aq) + 2 NaOH(aq) 2 HOH(l) + Na 2 SO 4 (aq) v= 25.0mL v= 36.5 ml c= mol/l c=? c NaOH = 25.0 ml H 2 SO 4 x mol x 2 mol NaOH x 1 L 1 mol H 2 SO ml = mol/l

47 Buffers A mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Ex: acetic acid (CH 3 COOH) and a solution of sodium acetate (NaCH 3 COO). CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) acid conjugate base If acid (H 3 O + ) is added to this equilibrium the equilibrium shifts to the left to reduce the amount of H 3 O +. This can only occur if there is excess CH 3 COO - to react with. Because of this process, the ph of the solution changes very little.

48 Indicators An indicator is a weak acid that changes color with changes in ph To choose an indicator for a titration, the ph of the endpoint must be within the ph range over which the indicator changes color HIn(aq) + H 2 O(l) H 3 O + (aq) + In (aq Colour 1 Colour 2 HIn is the acid form of the indicator. Adding OH removes the H 3 O + and shifts the equilibrium to the right causing a colour change

49 Methyl Orange: ph range ( ) HMo(aq) + H 2 O(l) H 3 O + (aq) + Mo (aq) red yellow Bromothymol Blue: ph range ( ) HBb(aq) + H 2 O(l) H 3 O + (aq) + Bb (aq) yellow blue

50 Acid-Base Titration (p. 603 ) A titration is a lab technique used to determine an unknown solution concentration. A standard solution is added to a known volume of solution until the endpoint of the titration is reached.

51 Acid-Base Titration The endpoint occurs when there is a sharp change in colour. The equivalence point occurs when the moles of hydronium equals the moles of hydroxide. The colour change is caused by the indicator added to the titration flask.

52 Acid-Base Titration An indicator is a chemical that changes colour over a given ph range. (See indicator table) A buret is used to deliver the standard solution.

53 Acid-Base Titration standard solution - solution of known concentration. primary standard - a standard solution which can be made by direct weighing of a stable chemical. Titration Lab

54 Multi-Step Titrations (p ) Polyprotic acids do not donate all their protons at the same time when they react with a base. eg. Write the equations for the steps that occur when H 3 PO 4 (aq) is titrated with NaOH(aq) H 3 PO 4 (aq) + OH (aq) H 2 PO 4 (aq) + H 2 O(l) H 2 PO 4 (aq) + OH (aq) HPO 42 (aq) + H 2 O(l) HPO 42 (aq) + OH (aq) PO 43 (aq) + H 2 O(l)

55 Multi-Step Titrations Overall equation calculation H 3 PO 4 (aq) + OH-(aq) H 2 PO 4 (aq) + H 2 O(l) H 2 PO 4 (aq) + OH (aq) HPO 42 (aq) + H 2 O(l) HPO 42 (aq) + OH (aq) PO 43 (aq) + H 2 O(l) H 3 PO 4 (aq) + 3 OH (aq) PO 43 (aq) + 3 H 2 O(l)

56 Acid base titration curves A graph of ph vs volume of titrant (the material in the burette) as acid is added to base or base is added to acid is called a titration curve. These curves have characteristic shapes as the following examples will show. The curve starts with a gentle slope which gets greater as the endpoint approaches and decreases after the endpoint. The endpoint (ph) depends on the type of acid base titration.

57 Strong Base TITRANT With Strong Acid

58 Strong Acid TITRANT With Strong Base

59 Strong Acid TITRANT With Weak Base

60 Strong Base TITRANT With Weak Acid

61 Weak Base TITRANT With Weak Acid

62 Weak Acid TITRANT With Weak Base

63 Strong Base TITRANT With a Diprotic Acid

64 Strong Acid TITRANT With a Diprotic Base

65 Strong Base TITRANT with Triprotic Acid

66 Strong Acid (HCl) With Weak Base (NH 3 )

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