Chapter FifteenPrentice-Hall ©2002Slide 1 of our slides 1

Chapter FifteenPrentice-Hall ©2002Slide 2 of our slides 2 Ionic equilibriums(1) (Mark=1.5)

Equilibrium In Solutions Of Weak Acids And Weak Bases weak acid:HA + H 2 O  H 3 O + + A - [H 3 O + ][A - ] K a = [HA] weak base:B + H 2 O  HB + + OH - [HB + ][OH - ] K b = [B] our slides 3

Some Acid-Base Equilibrium Calculations cHA≈[HA] [H 3 O + ][A - ] [H 3 O + ][A - ] K a = = cHA –[H 3 O + ] cHA cHA > [HA] Analytical C> Equilibrium C - the calculations can be simplified. - When M acid /K a or M base /K b > 100, - When K a or K b <1×10 -4 (In usual Conc.) our slides 4

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An Example 1. Determine the concentrations of H 3 O +, CH 3 COOH and CH 3 COO -, and the pH of 1.00 M CH 3 COOH solution. K a = 1.8 x What is the pH of a solution that is M in methylamine, CH 3 NH 2 ? K b = 4.2 x our slides 6

Are Salts Neutral, Acidic or Basic? Salts are ionic compounds formed in the reaction between an acid and a base. 1. NaCl Na + is from NaOH, a strong base Cl - is from HCl, a strong acid H 2 O NaCl (s) → Na + (aq) + Cl - (aq) Na + and Cl - ions do not react with water. The solution is neutral our slides 7

Are Salts Neutral, Acidic or Basic ? 2.KCN K + is from KOH, a strong base CN - is from HCN, a weak acid H 2 O KCN (s) → K + (aq) + CN - (aq) K + ions do not react with water, but CN - ions do. CN - + H 2 O  HCN + OH - hydrolysis The OH - ions are produced, so the solution is basic our slides 8

Are Salts Neutral, Acidic or Basic? 3.NH 4 Cl NH 4 + is from NH 3, a weak base Cl - is from HCl, a strong acid H 2 O NH 4 Cl (s) → NH 4 + (aq) + Cl - (aq) Cl - ions do not react with water, but NH 4 + ions do. NH H 2 O  H 3 O + + NH 3 hydrolysis The H 3 O + ions are produced, so the solution is acdic our slides 9

Are Salts Neutral, Acidic or Basic? 3.NH 4 CN NH 4 + is from NH 3, a weak base CN - is from HCN, a weak acid H 2 O NH 4 CN (s) → NH 4 + (aq) + CN - (aq) NH H 2 O  H 3 O + + NH 3 K a hydrolysis CN - + H 2 O  HCN + OH - K b hydrolysis (K a >K b,Acidic)’’’(K a < K b,Basic)‘’’ (K a = K b,Nutral) our slides 10

Ions As Acids And Bases Certain and anion ions can cause an aqueous solution to become acidic or basic due to hydrolysis. Salts of strong acids and strong bases form neutral solutions. Salts of weak acids and strong bases form basic solutions. Salts of strong acids and weak bases form acidic solutions. Salts of weak acids and weak bases form solutions that are acidic in some cases, neutral or basic in others our slides 11

Strong Acids And Strong Bases Strong acids: HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 Strong bases: Group IA and IIA hydroxides our slides 12

An Example Indicate whether the solutions (a) Na 2 S and (b) KClO 4 are acidic, basic or neutral our slides 13

The pH Of a Salt Solution What is the pH of 0.1M NaCN solution? What is the pH of 0.1M NH 4 Cl solution? What is the pH of 0.1M NH 4 CN solution? K a of HCN=1.0× K b for NH3=1.0×10 -5 K a x K b = K w so, K b = K w /K a our slides 14

Chapter FifteenPrentice-Hall ©2002Slide 15 of our slides 15

Common Ion Effect Illustrated ((1.00 M CH 3 COOH)) ((1.00 M CH 3 COOH M CH 3 COONa)) yellow: pH < 3.0 blue-violet: pH > our slides 16 CH 3 COOH  CH 3 COO - + H +

The Common Ion Effect Calculate the pH of 0.10 M CH 3 COOH solution. K a of CH 3 COOH=1.0×10 -5 Calculate the pH of 0.10 M CH 3 COONa solution. Calculate the pH of 0.10 M CH 3 COOH/ 0.10 M CH 3 COONa solution our slides 17

Depicting Buffer Action our slides 18

Buffer Solutions A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. A buffer contains CH 3 COOH  CH 3 COO - Acidic buffer NH 3  NH 4 + Alkalin buffer our slides 19

How A Buffer Solution Works The acid component of the buffer can neutralize small added amounts of OH -, and the basic component can neutralize small added amounts of H 3 O +. CH 3 COOH  CH 3 COO - + H our slides 20

Chapter FifteenPrentice-Hall ©2002Slide 21 of 31 Ionization constant of an acid Taking log of the equation on both sides, our slides 21

Chapter FifteenPrentice-Hall ©2002Slide 22 of 31 Ionization constant of an acid Multiplying both sides of the equation by our slides 22 Henderson-Hasselbach equation

Henderson-Hasselbalch Equation For Buffer Solutions [conjugate base] pH = pK a + log [conjugate acid] If [conjugate acid] = [conjugate base], pH = pK a Requirement: -[B] / [A] between 0.10 and our slides 23

Buffer Capacity There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed. In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal. [B]=[A] [Buffer]=[Acid]+[Base] our slides 24

Buffer Capacity [Buffer]=[Acid]+[Base] [Acid] ↑ & [Base] ↑ Capacity ↑ In equimolar buffersis is important Capacity ↑ our slides 25

Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? our slides 26

Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? (b)If 5 mmol NaOH is added to L of this solution, what will be the pH? our slides 27

Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? (b)If 5 mmol NaOH (50ml * 0.1M) is added to L of this solution, what will be the pH? (c)If 5 mmol HCl is added to L of this solution, what will be the pH? our slides 28

Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? (b)If 5 mmol NaOH is added to L of this solution, what will be the pH? (c)If 5 mmol HCl is added to L of this solution, what will be the pH? (d)If 5 mmol NH 4 Cl is added to L of this solution, what will be the pH? our slides 29

Chapter FifteenPrentice-Hall ©2002Slide 30 of 31 Calculations in Buffer Solutions 2) What concentration of acetate ion in 500 ml of M CH 3 COOH (pK a =5) produces a buffer solution with pH = 4.00? our slides 30

Calculations in Buffer Solutions 2) What concentration of acetate ion in500 ml of M CH 3 COOH (pK a =5) produces a buffer solution with pH = 4.00? our slides 31 How many mg?

Acid-Base Indicators An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless. HIn + H 2 O  H 3 O + + In - Acid-base indicators are often used for applications in which a precise pH reading isn’t necessary. A common indicator used in chemistry laboratories is Phenolphetalein our slides 32

Chapter FifteenPrentice-Hall ©2002Slide 33 of our slides 33

Chapter FifteenPrentice-Hall ©2002Slide 34 of our slides 34

Neutralization Reactions Neutralization is the reaction of an acid and a base. Titration is a common technique for conducting a neutralization. At the equivalence point in a titration, the acid and base have been brought together in exact stoichiometric proportions. The point in the titration at which the indicator changes color is called the end point. The indicator endpoint and the equivalence point for a neutralization reaction can be best matched by plotting a titration curve, a graph of pH versus volume of titrant. In a typical titration, 50 mL or less of titrant that is 1 M or less is used our slides 35

Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the some points and draw the curve. 4 essential points. 1)initial point 2)equivalence point 3)before the equivalence point 4)beyond the equivalence point Ml تیترانت pH محیط our slides 36

Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? our slides 37

Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of mL of M HCl with M NaOH. (a)initial pH. (Before the addition of any NaOH). Answer Q1. There are:HCl & H2O Answer Q2. HCl Answer Q3. [HCl] Answer Q4. pH=-log[H + ] our slides 38

Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of mL of M HCl with M NaOH. b)equivalence point. Answer Q1. There are:NaCl & H 2 O Answer Q2. H 2 O Answer Q3. Answer Q4. pH= our slides 39

Concentrations before equivalente point. HCl + NaOH → NaCl +H 2 O our slides N1V1 N2V2 N1V1-N2V2 N2V2

Concentrations at equivalente point. HCl + NaOH → NaCl +H 2 O our slides =100 0 N1V1 N2V2 N1V1-N2V2 N2V2=N1V1

Concentrations after equivalente point. HCl + NaOH → NaCl +H 2 O our slides N1V1 N2V2 N2V2-N1V1 N2V2

Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of mL of M HCl with M NaOH. c)before the equivalence point. Answer Q1. There are:HCl,NaCl & H 2 O Answer Q2. HCl Answer Q3. Answer Q4. [H + ]=N pH=-log[H + ] our slides 43

Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of mL of M HCl with M NaOH. d)after the equivalence point. Answer Q1. There are:NaOH,NaCl & H 2 O Answer Q2. NaOH Answer Q3. Answer Q4. [OH - ]=N pOH=-log[OH - ] pH=14-pOH our slides 44

Titration Curve For Strong Acid - Strong Base pH is low at the beginning. pH changes slowly until just before equivalence point. pH changes sharply around equivalence point. pH = 7.0 at equivalence point. Further beyond equivalence point, pH changes slowly. Any indicator whose color changes in pH range of 4 – 10 can be used in titration our slides 45

Chapter FifteenPrentice-Hall ©2002Slide 46 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the some points and draw the curve. K a =1× essential points. 1)initial point 2)equivalence point 3)beyond the initial point 4)before the equivalence point 5)beyond the equivalence point our slides 46

Chapter FifteenPrentice-Hall ©2002Slide 47 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? our slides 47

Chapter FifteenPrentice-Hall ©2002Slide 48 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of mL of M CH 3 COOH with M NaOH. (a)initial pH. (Before the addition of any NaOH). Answer Q1. There are: CH 3 COOH & H 2 O Answer Q2. CH 3 OOH Answer Q3. CH 3 OOH Answer Q4. pH=-log[H + ] our slides 48

Chapter FifteenPrentice-Hall ©2002Slide 49 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of mL of M CH 3 COOH with M NaOH. b)equivalence point. Answer Q1. There are: CH3COO -, Na + & H 2 O Answer Q2. CH3COO - Answer Q3. Answer Q4. pOH=-log[OH - ] K a ×K b =K w our slides 49

Chapter FifteenPrentice-Hall ©2002Slide 50 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of mL of M CH 3 COOH with M NaOH. c)beyond the initial point. Answer Q1. There are: CH 3 COOH, CH3COO -,Na + & H 2 O Answer Q2. CH 3 COOH, CH3COO - Answer Q3. Answer Q our slides 50

Chapter FifteenPrentice-Hall ©2002Slide 51 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of mL of M CH 3 COOH with M NaOH. d)before the equivalence point. Answer Q1. There are: CH 3 COOH, CH3COO -,Na + & H 2 O Answer Q2. CH 3 COOH, CH3COO - Answer Q3. Answer Q our slides 51

Chapter FifteenPrentice-Hall ©2002Slide 52 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of mL of M CH 3 COOH with M NaOH. e)after the equivalence point. Answer Q1. There are:NaOH, CH3COO -, Na + & H 2 O Answer Q2. NaOH Answer Q3. Answer Q4. [OH - ]=N pOH=-log[OH - ] pH=14-pOH our slides 52

Titration Curve For Weak Acid - Strong Base The initial pH is higher because weak acid is partially ionized. At the half-neutralization point, pH = pKa. pH >7 at equivalence point because the anion of the weak acid hydrolyzes. The steep portion of titration curve around equivalence point has a smaller pH range. The choice of indicators for the titration is more limited our slides 53

Chapter FifteenPrentice-Hall ©2002Slide 54 of our slides 54

Chapter FifteenPrentice-Hall ©2002Slide 55 of 31 Application of K a The K a of nicotinic acid, HNic, is 1.4e-5. A solution containing 0.22 M HNic. What is its pH? What is the degree of ionization? Solution: HNic  H + + Nic – 0.22-x x x x 2 K a = ———— = 1.4e – x(use approximation, small indeed) x =  (0.22*1.4e-5) = pH = – log (0.0018) = 2.76 Degree of ionization = / 0.22 = = 0.79% our slides 55

Chapter FifteenPrentice-Hall ©2002Slide 56 of 31 Determine K a and percent ionization Nicotinic acid, HNic, is a monoprotic acid. A solution containing M HNic, has a pH of What is its K a ? What is the percent of ionization? Solution: HNic  H + + Nic – x x x x = [H + ] = 10 –3.39 = 4.1e-4 [HNic] = – = (4.1e-4) 2 K a = ————— = 1.4e Degree of ionization = / = = 3.4% our slides 56

Chapter FifteenPrentice-Hall ©2002Slide 57 of 31 Using the quadratic formula The K a of nicotinic acid, HNic, is 1.4e-5. A solution containing M HNic. What is its pH? What is the degree of ionization? Solution: HNic  H + + Nic – x x x x 2 K a = —————— = 1.4e-5x e-5 x – 1.4e-8 = – x –1.4e–5 +  (1.4e–5) 2 + 4*1.4e-8 x = —————————————————— = M 2 pH = – log ( ) = 3.95 Degree of ionization = / = = 11.1% our slides 57

Chapter FifteenPrentice-Hall ©2002Slide 58 of 31 Degree of or percent ionization The degree or percent of ionization of a weak acid always decreases as its concentration increases, as shown from the table given earlier. Concentration of acid % ionization our slides 58 Deg.’f ioniz’n % % %

Chapter FifteenPrentice-Hall ©2002Slide 59 of 31 Polyprotic acids Polyprotic acids such as sulfuric and carbonic acids have more than one hydrogen to donate. H 2 SO 4 → H + + HSO 4 – K a1 very large completely ionized HSO 4 –  H + + SO 4 2– K a2 = H 2 CO 3  H + + HCO 3 – K a1 = 4.3e-7 HCO 3 –  H + + CO 3 2– K a2 = 4.8e-11 Ascorbic acid (vitamin C) is a diprotic acid, abundant in citrus fruit. Others: H 2 S, H 2 SO 3, H 3 PO 4, H 2 C 2 O 4 (oxalic acid) … our slides 59

Chapter FifteenPrentice-Hall ©2002Slide 60 of 31 Species concentrations of diprotic acids Evaluate concentrations of species in a 0.10 M H 2 SO 4 solution. Solution: H 2 SO 4 → H + + HSO 4 – completely ionized (0.1–0.1) HSO 4 –  H + + SO 4 2– K a2 = –y 0.10+y yAssume y = [SO 4 2– ] (0.10+y) y ————— = (0.10-y) [SO 4 2– ] = y = 0.01M [H+] = = 0.11 M; [HSO 4 – ] = = 0.09 M our slides 60

Chapter FifteenPrentice-Hall ©2002Slide 61 of 31 Species concentrations of weak diprotic acids Evaluate concentrations of species in a 0.10 M H 2 S solution. Solution: H 2 S = H + + HS – K a1 = 1.02e-7 (0.10–x) x+y x-yAssume x = [HS – ] HS – = H + + S 2– K a2 = 1.0e-13 x–y x+y yAssume y = [S 2– ] (x+y) (x-y) (x+y) y ————— = 1.02e-7 ———— = 1.0e-13 (0.10-x)(x-y) [H 2 S] = 0.10 – x = 0.10 M [HS – ] = [H + ] = x  y = 1.0e–4 M; [S 2– ] = y = 1.0e-13 M 0.1>> x >> y: x+ y = x-y = x x =  0.1*1.02e-7 = 1.00e-4 y = 1e our slides 61