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2 The Common Ion Effect Consider a solution that contains a week acid, CH 3 COOH, and a soluble salt of that acid e.g. CH 3 COONa. The equilibrium equation for CH 3 COOH (a weak electrolyte) is CH 3 COONa is an ionic compound and therefore a strong electrolyte: It dissociates completely. CH 3 COONa(aq) Na + (aq) + CH 3 COO - Both substances dissociate into CH 3 COO - (which is a common ion to both). What happens when we add CH 3 COONa to CH 3 COOH? 2

3 The Common Ion Effect At equilibrium H + and CH 3 COO - are constantly moving into and out of solution, but the concentrations of ions is constant and equal. Adding CH 3 COO - from CH 3 COONa causes equilibrium to shift to the (recall LeChâtelier s Principle). Adding CH 3 COO - also causes a in the equil concentration of H + to counteract the increase in CH 3 COO - 3

4 The Common Ion Effect Whenever a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it normally would if it were alone in solution. o This Observation is called the common ion effect. o The common ion effect is important in buffer solutions. 4

5 The Common Ion Effect Example Calculate the fluoride ion concentration and ph of a solution that is 0.20M in HF and 0.10 M in HCl. K a for HF is HF is a weak acid: HF(aq) H + (aq) + F - (aq) Initial conc. 0.20M 0.10M 0 HCl is a strong acid / strong electrolyte: HCl(aq) H + (aq) + Cl - (aq) conc. 0.10M 0.10M 0.10M 5

6 The Common Ion Effect Example In solution we have HF, H + (the common ion), Cl - (spectator ion it is the conjugate base of a strong acid) Problem wants [F - ]: HF(aq) H + (aq) + F - (aq) K [H + ] [F ] a = = [HF] Because HCl, a strong acid, is also present, the initial [H + ] is not 0, but rather 0.10 M. 6

7 HF(aq) H + (aq) + F - (aq) [HF], M [H + ], M [F ], M Initially Change x +x +x Equilibrium 0.20 x x x K [H + ] [F ] a = = [HF] (0.10+x) (x) (0.20-x) = If we assume that x much smaller, relative to 0.10 M or 0.20 M then: 7

8 The Common Ion Effect (0.10) (x) (0.20) = x = 1.4 x 10-3 M Therefore, [F ] = x = M [H + ] = ( x) = M ph = log (0.10) and ph = 1.00 Percentage dissociation of HF = [H + ] x 100 HF = x 100 = 0.70% 0.20 Note that the dissociation of 0.1 M HF is 2.7% thus clearly showing that the presence of the F - ions from the dissolved HCl greatly inhibits the dissociation of HF. 8

9 Buffered Solutions These are solutions of a weak conjugate acid-base pair. They are particularly resistant to ph changes, e.g. when small amounts of acid strong acid or base is added. Human blood is an example. 9

10 Buffered Solutions Composition and Action of Buffered Solutions A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X - ): HX(aq) H + (aq) + X - (aq) The K a expression is K a [H ][X [HX] - ] Thus [H ] K a [HX] - [X ] 10

11 Buffered Solutions The ph is determined by the value of K a for the weak acid component and the ratio of the concentrations of the acid-base pair, [HX]/[X-]. When OH - is added to the buffer, the OH - reacts with HX to produce X - and water. But, the [HX]/[X - ] ratio remains more or less constant, so the ph is not significantly changed. OH - (aq) + HX H 2 O(l) + X - (aq) When H + is added to the buffer, X - is consumed to produce HX. Once again, the [HX]/[X - ] ratio is more or less constant, so the ph does not change significantly. H + (aq) + X - (aq) HX(aq) 11

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13 Buffered Solutions Calculating the ph of a Buffer Buffers share a common ion (they are conjugate acidbase pairs). We can therefore using the same procedure we used for such solutions to calculate the ph of a buffered solution. Alternatively we can use another approach based on the equation: K a [H ] [H ][X [HX] K a - ] [HX] - [X ] 13

14 Buffered Solutions Calculating the ph of a Buffer Taking the negative log of both sides of [HX] [HX] -log[h ] log(k a ) logk log - a [X ] [X ] [HX] [H ] Ka - [X ] But... -log[h ] ph,and log K a pk a The equation becomes ph pk a - [X ] log [HX] Or ph pk a [base] log [acid] This is the Henderson Hasselbalch equation. 14

15 Buffered Solutions Calculating the ph of a Buffer Example: What is the ph of a buffer that is 0.12M in lactic acid, HC 3 H 5 O 3, and 0.10M in sodium lactate? K a for lactic acid is Answer: ph = pk a + log ph = log (1.4 [base] [acid] 10 4 ) +log (0.10) (0.12) ph = (-0.08) ph =

16 Buffered Solutions Buffer Capacity and ph Range Buffer capacity is the amount of acid or base the buffer can neutralize before there is a significant change in ph. Buffer capacity depends on the composition of the buffer i.e. amounts of acid and base. The greater the amounts of conjugate acid-base pair, the greater the buffer capacity. The ph range of the buffer depends on K a. 16

17 Buffered Solutions Buffer Capacity and ph Range The ph range is the range of ph values over which a buffer system works effectively. ph pk a [base] log [acid] We can see from this equation that when [base] and [acid] are equal, ph = pk a This gives the optimum ph for any buffer. It is best to choose an a buffer whose acid form has pk a close to the desired ph. 17

18 Buffered Solutions Addition of Strong Acids or Bases to Buffers When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction. The amount of strong acid or base added results in a neutralization reaction: X - + H 3 O + HX + H 2 O HX + OH - X - + H 2 O. By knowing how much H 3 O + or OH - was added (stoichiometry) we know how much HX or X - is formed. With the concentrations of HX and X - (note the change in volume of solution) we can calculate the ph from the 18 Henderson-Hasselbalch equation.

19 Buffered Solutions Addition of Strong Acids or Bases to Buffers The calculation is done two parts: stoichiometric and equilibrium. 1. Consider the acid-base neutralization reaction, and determine its effect on [HX] and [X - ]. 2. Use K a and the new concentrations of HX and X - from step 1 to calculate [H + ]. 19

20 Example: Calculating ph changes in buffers A buffer is made by adding mol CH 3 COOH and mol CH 3 COONa to enough water to make 1.00 L of solution. The ph of the buffer is (a) (b) Buffered Solutions Addition of Strong Acids or Bases to Buffers Calculate the ph of this solution after mol of NaOH is added. Calculate the ph if mol of NaOH were added to 1L of pure water (compare with ph in a). Solution: Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 ph = pk a = log ( ) =

21 (a) Solving this problem requires two steps: 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution (stoichiometric calculation) 2. Use the Henderson Hasselbalch equation to determine the new ph of the solution (Equilibrium calculation). 21

22 Buffered Solutions 1. Stoichiometry Calculation - Prior to the neutralization there are mol each of CH 3 COO - and CH 3 COOH - The mol NaOH will react with mol of the acetic acid: CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) Buffer before reaction mol mol Addition mol - Buffer after addition mol mol 22

23 2. Equilibrium Calculation - We focus on the equilibrium that will determine the ph of the buffer: CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) - Now use the Henderson Hasselbalch equation to calculate the new ph: ph = log ph = ph = 4.80 Buffered Solutions (0.320 mol/l) (0.200 mol/l) We could use moles instead of conc since volumes of acid/base are equal 23

24 Buffered Solutions (b) A solution made by adding mol NaOH to 1.00L of pure water can be calculated by first calculation poh and then subtracting from 14. ph = 14 (-log0.020) = Note that addition of a small amount of NaOH changes the ph of water significantly. On the other hand addition of a small amount of NaOH causes very little change to the ph of the buffer. 24

25 Buffered Solutions Exercise: A buffer solution contains 0.50 M sodium lactate (CH 3 CH 2 COONa) and 0.50 M lactic acid (CH 3 CH 2 COOH). a) What is the ph of the buffer solution? b) What is the ph after adding 0.1 mol HCl to one litre of buffer? c) What is the ph after adding 0.1 mol KOH to one litre of buffer? (K a lactic acid = 1.4 x 10-4 ) 25

26 Buffered Solutions In an NH 4+ /NH 3 buffer, what concentration changes will occur if a small amount of OH - is added? [NH 4+ ] [NH 3 ] ph 1. Increase Increase Increase 2. Decrease Increase Increase 3. Increase Decrease Increase 4. Increase Decrease Decrease 5. Decrease Increase Decrease NH 4+ (aq) + OH - NH 3 (aq) + H 2 O(l) 26

27 Buffered Blood The major buffer system to control ph in blood is the carbonic acid bicarbonate system The important equilibria is this buffer system is The principal organs that control blood ph are lungs and kidneys. Buffered Solutions H + (aq) + HCO 3- (aq) H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) Blood is a buffered solution whose ph is maintained between 7.35 and

28 Acid-Base Titrations A known concentration of base (or acid) is slowly added to a solution of acid (or base). 28

29 Acid-Base Titrations A ph meter or indicators are used to determine when the solution has reached the equivalence point, i.e. the point at which the stoichiometric amount of acid equals that of base. A plot of ph versus volume of acid (or base) added is called a titration curve. 29

30 Acid-Base Titrations Strong Acid-Strong Base Titrations When a strong acid is added to a strong base, the general shape of the titration curve looks like this. 30

31 Acid-Base Titrations Strong Acid-Strong Base Titrations Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl). The ph can be calculated at various stages of the titration which can be separated into 4 regions in the titration curve. 31

32 Acid-Base Titrations Strong Acid-Strong Base Titrations 1. The initial ph: Before the addition of the base, the ph of the solution is equal to the ph of the acid. From the start of the titration to near the equivalence point, the ph is low and goes up slowly. 32

33 Acid-Base Titrations Strong Acid-Strong Base Titrations 2. Between the initial ph and the equivalence point: Just before and after the equivalence point, the ph increases rapidly. ph of solution is determined by the remaining concentration of the acid. 33

34 Acid-Base Titrations Strong Acid-Strong Base Titrations 3. Equivalence point: At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. ph of solution is

35 Acid-Base Titrations Strong Acid-Strong Base Titrations 4. After the equivalence point: As more base is added, the rapid increase in ph eventually flattens. The ph of solution is determined by concentration of the excess NaOH. 35

36 Acid-Base Titrations Strong Acid-Strong Base Titrations The equivalence point in a titration is the point at which the acid and base are present in stoichiometric quantities. The end point in a titration is the observed point. The difference between equivalence point and end point is called the titration error. 36

37 Acid-Base Titrations Strong Base-Strong Acid Titrations Titration of a strong base with a strong acid yields a similar but analogous curve of ph vs added acid. At first, the strong base is in excess, so the ph > 7. 37

38 Acid-Base Titrations Weak Acid-Strong base titrations Consider the titration of acetic acid, HC 2 H 3 O 2 and NaOH. 38

39 Acid-Base Titrations Weak Acid-Strong base titrations Initially, the solution contains only weak acid. Therefore, ph is given by the equilibrium calculation. Before the equivalence point, i.e. as strong base is added, the strong base consumes a stoichiometric quantity of weak acid: CH 3 COOH(aq) + NaOH(aq) CH 3 COO - (aq) + H 2 O(l) There is an excess of acid before the equivalence point. 39

40 Acid-Base Titrations Weak Acid-Strong base titrations Therefore, we have a mixture of weak acid and its conjugate base. The ph is given by the buffer calculation. First the amount of C 2 H 3 O 2 - generated is calculated, as well as the amount of HC 2 H 3 O 2 consumed. (Stoichiometry.) Then the ph is calculated using equilibrium conditions. (Henderson-Hasselbalch.) 40

41 Acid-Base Titrations Weak Acid-Strong base titrations At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C 2 H 3 O 2 - has been generated. Therefore, the ph is given by the C 2 H 3 O 2 - solution. This means ph > 7. Note that for a weak acid-strong base titration the ph 7. After the equivalence point, the ph is given by the strong base in excess. The equivalence point is determined by K a of the acid. 41

42 Acid-Base Titrations Weak Acid-Strong base titrations Effect of K a on titration curves Things to note: The initial ph is higher for weak acids. The ph change before the equivalence point is smaller for weak acids. The ph at the equivalence point is above 7.00 for weak acids. 42

43 Acid-Base Titrations Weak Base-Strong Acid titrations The ph at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice. 43

44 Acid-Base Titrations Weak Base-Strong Acid titrations Which indicator would best determine the endpoint of the titration of NaC 2 H 3 O 2 with HNO 3? Methyl violet Methyl orange Bromthymol blue Phenolphthalein Alizarin yellow R 44

45 Acid-Base Titrations Titrations of Polyprotic Acids In polyprotic acids, each ionizable proton dissociates in steps. Therefore, in a titration there are n equivalence points corresponding to each ionizable proton. 45

46 Solubility Equilibria So far we have considered homogeneous acid-base equilibria (all species in the same phase). We will now consider heterogeneous equilibria, which involve the dissolution or precipitation of ionic compounds. The compounds are usually slightly soluble. 46

47 Solubility Equilibria Previously we used solubility rules to predict whether a substance was going to be soluble in water or whether a precipitate would form. Now we use solubility equilibria to predict quantitatively the amount of a given compound that will or will not dissolve. 47

48 Solubility Equilibria The Solubility Product Constant, K sp Consider the equilibrium for saturated BaSO 4 BaSO 4 (s) Ba 2+ (aq) + SO 2-4 (aq) The solubility product expression is: K sp K sp is the solubility product. [Ba ][SO It explains how soluble the solid is in water. BaSO 4 is ignored because it is a pure solid so its concentration is constant ] 48

49 Solubility Equilibria The Solubility Product Constant, K sp K sp is the product of the molar concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium reaction. K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/l) or 100 ml (g/ml) of solution, or in mol/l (M). 49

50 Solubility Equilibria The Solubility Product Constant, K sp To convert solubility to K sp solubility needs to be converted into molar solubility (via molar mass); molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation), K sp is the product of equilibrium concentration of ions. 50

51 Solubility Equilibria The Solubility Product Constant, K sp The magnitude of K sp is a measure of how much of the solid dissolves to form a saturated solution. It is a unitless number. 51

52 Solubility Equilibria The Solubility Product Constant, K sp Example: The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of K sp. ALWAYS: Write a balanced chemical reaction Solution: The balanced chemical reaction is: AgCl(s) Ag + (aq) + Cl - K sp = [Ag + ][Cl - ] Since the mole ratio of of AgCl to both Ag + and Cl - is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. Therefore, K sp = (1.3 x 10-5 )(1.3 x 10-5 ) = 1.7 x

53 Solubility Equilibria The Solubility Product Constant, K sp Example: Calculate the solubility of CaF 2 in g/l given that K sp = 4.0 x 10-8 Solution: CaF 2 Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][F - ] 2 Since for every mole of Ca 2+ formed, 2 moles of F - are formed. These can be assigned as: [Ca 2+ ] = x and [F - ] = 2x K sp = (x)(2x) 2 = 4.0 x 10-8 x = 2.2 x 10-3 M 1 mole of CaF 2 is 78.1 g; therefore 2.2 x 10-3 moles is g and therefore the solubility is g/l 53

54 Solubility Equilibria The Solubility Product Constant, K sp Example: Calculate the molar solubility of Mg(OH) 2 buffered at ph = K sp for Mg(OH) 2 is 1.8 x Important: write balanced chemical equation. Solution: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) 1 mole of Mg(OH) 2 furnishes 1 mole of Mg 2+ and 2 moles of OH - poh = = 5.00, so [OH - ] = 1 x 10-5 M K sp = [Mg 2+ ][OH - ] 2 = 1.8 x [Mg 2+ ](1 x 10-5 ) 2 = 1.8 x [Mg 2+ ] = 0.18 M 54

55 Factors that Affect Solubility The solubility of a substance is affected by: Temperature and presence of other solutes Presence of common ions ph of the solution Presence of complexing agents 55

56 Factors That Affect Solubility The Common-Ion Effect In general the solubility of a slightly soluble salt decreases in the presence of a common ion. Solubility and ph The solubility of slightly soluble salts containing basic anions increases as [H + ] increases i.e. as the ph is lowered. 56

57 Factors That Affect Solubility 57

58 ph, solubility and sinkholes (a) The sequence A B C shows how a sinkhole forms. An underground void develops as limestone, CaCO 3 (s), dissolves. Collapse of the overlying ground into an underground cavity causes sinkhole formation. (b) The large sinkhole shown here has destroyed several buildings and part of a highway. 58

59 Factors That Affect Solubility Solubility and ph Tooth Decay and Flouridation Tooth cavities (or dental caries) are caused when acids, from the action of bacteria (e.g. Streptococus mutans) on mainly sugars, dissolve tooth enamel. Tooth enamel consists of mainly a mineral called hydroxyapatite, Ca 10 (PO 4 ) 6 (OH) 2. Ca 10 (PO 4 ) 6 (OH) 2 (s) + 8H + (aq) 10Ca 2+ (aq) + 6HPO 4 2- (aq) +2H 2 O(l) 59

60 Factors That Affect Solubility Solubility and ph Tooth Decay and Flouridation Fluoride ion, present in drinking water, toothpaste and other sources, can react with hydroxyapatite to form fluoroapatite, Ca 10 (PO 4 ) 6 F 2. This mineral, in which F - has replaced OH -, is much more resistant to attack by acids because it is less soluble in acid than hydroxyapatite. In this way tooth decay is reduced. 60

61 Factors That Affect Solubility Examples on the Common Ion effect The K sp for CaF 2 is 3.9 x at 25 o C. What are the concentrations of Ca 2+ and F - ions if we dissolve CaF 2 in a 1.00 M solution of KF? 61

62 Factors That Affect Solubility Examples on ph of solution Calculate the molar solubility of Fe(OH) 2 when buffered at ph 7.0. (K sp for Fe(OH) 2 = 7.9 x ) 62

63 Precipitation and the Solubility Product If we have the concentration of substances in a reaction mixture, can we predict whether a precipitate will form or not? We can make use of the reaction quotient or ion product, Q and compare with K sp. Ion product, Q = *initial ion+ x [initial ion] y The ion product expression has the form of the K sp expression each concentration is raised to the power of the number of ions in the ionic formula. However, the concentrations are the initial concentrations not the equilibrium concentrations like in K sp. 63

64 Precipitation and the Solubility Product In summary: Q < K sp the solution is unsaturated Solid dissolves until Q = K sp Q = K sp the solution is saturated. Equilibrium exists. Therefore, no precipitation occurs. Q > K sp the solution is supersaturated and precipitation occurs until Q = K sp 64

65 Precipitation and the Solubility Product Example: Will a precipitate of lead(ii) chloride form when you add a solution of Pb(NO 3 ) 2 and NaCl to water to give a solution with 0.050M Pb 2+ and 0.10 M Cl -? (K sp (PbCl 2 ) = 1.6 x 10-5 ) Solution: Pb(NO 3 ) 2 (aq) + 2NaCl(aq) PbCl 2 (s) + 2NaNO 3 (aq) PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) Q = [Pb 2+ ][Cl - ] 2 = (0.050)(0.1) 2 = 5 x 10-4 Because Q < K sp, precipitation will occur until Q = K sp and equilibrium is reached. 65

66 End of Chapter Additional Aspects of Aqueous Equilibria 66