Ch 16 Aqueous Ionic Equilibrium

Transcription

1 Ch 16 Aqueous Ionic Equilibrium (Buffers and Acid/BaseTitration) buret Erlenmeyer flask Modified By Dr. Cheng-Yu Lai

2 Review- Reaction of Weak Bases/Acids with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O BH + + OH - K b [ BH ][ OH ] [ B] How about Weak acids with water? What is Ka? HA + H 2 O A - + H 3 O +

3 Buffer Solutions Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in ph. Weak acid + Conjugate base For Example: CH 3 CO 2 H + CH 3 CO 2 1- HF + F 1- NH NH 3 H 2 PO HPO 4 2- Buffered solutions contain either: A weak acid and its salt (conjugate base) A weak base and its salt (conjugate acid)

4 Buffer Solution Preparation CH 3 CO 2 H(aq) + H 2 O(l) Weak acid H 3 O 1+ (aq) + CH 3 CO 2 1- (aq) Conjugate base (NaCH 3 CO 2 ) Addition of OH 1- to a buffer: CH H 2 O(l) + CH 3 CO 1-3 CO 2 H(aq) + OH 1- (aq) 100% 2 (aq) Addition of H 3 O 1+ to a buffer: CH 3 CO 1-2 (aq) + H 3 O 1+ (aq) 100% H 2 O(l) + CH 3 CO 2 H(aq)

5 Action of a Buffer Figure 16.3 pg Pearson Education, Inc.

6 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (d) HCl/NaCl (a) HF is a weak acid and F - is its conjugate base (b) HBr is a strong acid ; same as d (c) CO 3 2- is a weak base and HCO 3 - is its conjugate acid

7 Why buffer works? Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium. The ph of 0.10 M acetic acid is Calculate the ph of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. CH 3 CO 2 H(aq) + H 2 O(l) H 3 O 1+ (aq) + CH 3 CO 2 1- (aq) I C -x +x +x E x x x K a = [H 3 O 1+ ][CH 3 CO 2 1- ] [CH 3 CO 2 H] = 1.8 x 10-5

8 The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left. The Common-Ion Effect 1.8 x 10-5 = (x)( x) ( x) x(0.10) 0.10 x = [H 3 O 1+ ] = 1.8 x 10-5 M ph =-log([h 3 O 1+ ]) = -log(1.8 x 10-5 ) = 4.74 Why is there a difference in ph? Le Châtelier s Principle CH H 3 O 1+ (aq) + CH 3 CO 1-3 CO 2 H(aq) + H 2 O(l) 2 (aq)

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10 Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left. HA (aq) + H 2 O (l) A (aq) + H 3 O + (aq) This causes the ph to be higher than the ph of the acid solution, lowering the H 3 O + ion concentration.

11 Example 16.2 Calculating the ph of a Buffer Solution as an Equilibrium Problem and with the Henderson Hasselbalch Equation Calculate the ph of a buffer solution that is M in benzoic acid (HC 7 H 5 O 2 ) and M in sodium benzoate (NaC 7 H 5 O 2 ). For benzoic acid, K a = Solution Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table. Large Ka has greatest percent ionization Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

12 The Henderson-Hasselbach Equation Generic weak acid derivation HA + H 2 O H 3 O + + A - K a = [H 3 O + ] [A - ]/[HA] pk a = - log K a = - log ([H 3 O + ] [A - ]/[HA]) Log xy = log x + log y pk a = - log [H 3 O + ] - log ([A - ]/[HA]) Conjugate base pk a = ph - log ([A - ]/[HA]) ph = pk a + log ([A - ]/[HA]) ph = pk a + log [Conjugate Base] [Acid]

13 What is the ph of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Ka = 1.8E-4 Mixture of weak acid and conjugate base! HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) x +x x Equilibrium (M) x x x Common ion effect X small 0.30 x x 0.52 HCOOH pk a = 3.77 ph = pk a + log [HCOO- ] [HCOOH] ph = log [0.52] [0.30] = EXAM

14 Buffer Capacity and Range 1. A solution that resists a change in ph when either hydroxide ions or protons are added. the effective ph range of a buffer is pk a ± 1. A buffer will be effective when 0.1 < [base]:[acid] < A buffer is most effective when ph = pk. At this ph [acid] = [conjugate base]. The buffer is equally effective against added acid or added base. 3. The buffer loses effectiveness when added acid/base exceeds the [buffer]. When choosing an acid to make a buffer, choose one whose pk a is closest to the ph of the buffer.

15 ph Titration Curves Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. buret standard solution (known concentration) Erlenmeyer flask unknown concentration solution An indicator is added which changes color once the reaction is complete Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/15

16 Titration, Titration Curves and Indicators Acid / conjugate base and indicator or Base/conjugated acid and indicator 3. If we observe the resulting ph while titrating either a strong or a weak acid with a strong base, we can plot that data to get a titration curve. 4. The titration curve is sigmoidal in shape, with a steep rise in ph around the neutralization or equivalence point.

17 Selection of Indicators Buffer Range 4±6 6±8 8±10

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21 Strong Acid-Strong Base Titrations

22 Strong Acid-Strong Base Titrations Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

23 Strong Acid-Strong Base Titrations 1. Before Addition of Any NaOH M HCl Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/23

24 Strong Acid-Strong Base Titrations 2. Before the Equivalence Point = ½ of [HCl] left 100% H 3 O 1+ (aq) + OH 1- (aq) 2H 2 O(l) Excess H 3 O 1+ Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/24

25 Strong Acid-Strong Base Titrations 3. At the Equivalence Point The quantity of H 3 O 1+ is equal to the quantity of added OH 1-. ph = 7 Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/25

26 Strong Acid-Strong Base Titrations 4. Beyond the Equivalence Point Excess OH 1- Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/26

27 a. at equivalent point b. V OH added = ml acid mole = M1V1 = M2V2= base mole V OH = ml c. V OH added = ml

28 ph Strong Base titrated with strong Acid Half Neutralization Equivalence point amount acid added = amount of base. M1V1=M2V2 ph L acid added Begin with 0.1 L of 0.1M NaOH - titrate with 0.1M HCl

29 Weak Acid-Strong Base Titrations

30 Weak Acid-Strong Base Titrations 1. Before Addition of Any NaOH CH 3 CO 2 H(aq) + H 2 O(l) H 3 O 1+ (aq) + CH 3 CO 2 1- (aq) M CH 3 CO 2 H Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/30

31 Weak Acid-Strong Base Titrations 2. Before the Equivalence Point CH H 2 O(l) + CH 3 CO 1-3 CO 2 H(aq) + OH 1- (aq) 100% 2 (aq) Excess CH 3 CO 2 H Half Neutralization Buffer region ph=pka Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/31

32 Weak Acid-Strong Base Titrations 3. At the Equivalence Point CH 3 CO 2 1- (aq) + H 2 O(l) OH 1- (aq) + CH 3 CO 2 H(aq) ph > 7 Notice how the ph at the equivalence point is shifted up versus the strong acid titration. Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/32

33 Weak Acid-Strong Base Titrations 4. Beyond the Equivalence Point Excess OH 1- Notice that the strong and weak acid titration curves correspond after the equivalence point. Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/33

34 Which indicator(s) would you use for a titration of HNO 2 with KOH? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, ph > 7 Use cresol red or phenolphthalein = Salt KNO 2 is basic 17.5

35 Weak Acid-Strong Base Titrations weak acid 5 acids have the same equivalent Strong acid

36 Titration of A Weak Polyprotic Acid 7. more 36

37 Polyprotic Acid-Strong Base Titrations

38 Weak Acid-Strong Base Titration Calculation CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) At equivalence point (ph > 7): why? CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) 14 ml

39 25.0 ml of 0.10M Acetic acid K a =1.8 x Half Neutralization ph? 3. 4 XS base [NaOAC]+ OH - 1. Starting ph=? Equivalence point ph? 0.10 M NaOH 39

40 Example 16.7 Weak Acid Strong Base Titration ph Curve A 40.0 ml sample of M HNO 2 is titrated with M KOH. Calculate: a. the volume required to reach the equivalence point b. the ph after adding 5.00 ml of KOH c. the ph at one-half the equivalence point Solution a. The equivalence point occurs when the amount (in moles) of added base equals the amount (in moles) of acid initially in the solution. Begin by calculating the amount (in moles) of acid initially in the solution. The amount (in moles) of KOH that must be added is equal to the amount of the weak acid. Calculate the volume of KOH required from the number of moles of KOH and the molarity. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

41 Example 16.7 Weak Acid Strong Base Titration ph Curve Continued b. Use the concentration of the KOH solution to calculate the amount (in moles) of OH in 5.00 ml of the solution. Prepare a table showing the amounts of HNO 2 and NO 2 before and after the addition of 5.00 ml KOH. The addition of the KOH stoichiometrically reduces the concentration of HNO 2 and increases the concentration of NO 2. Since the solution now contains significant amounts of a weak acid and its conjugate base, use the Henderson Hasselbalch equation and pk a for HNO 2 (which is 3.34) to calculate the ph of the solution. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

42 Example 16.7 Weak Acid Strong Base Titration ph Curve Continued c. At one-half the equivalence point, the amount of added base is exactly one-half the initial amount of acid. The base converts exactly half of the HNO 2 into NO 2, resulting in equal amounts of the weak acid and its conjugate base. The ph is therefore equal to pk a. For Practice 16.7 Determine the ph at the equivalence point for the titration of HNO 2 and KOH in Example Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

43 Weak Base-Strong Acid Titrations 1. Before Addition of Any HCl 0.10 M NH 3 K b = 1.8 x 10-5 (pk a = 9.26) M NH 3 Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/43

44 Weak Base-Strong Acid Titrations 2. Before the Equivalence Point NH NH 1+ 3 (aq) + H 3 O 1+ (aq) 100% 4 (aq) + H 2 O(l) Excess NH 3 Buffer region ph = pk a + log ([NH 3 ]/[NH 4+ ]) Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/44

45 Weak Base-Strong Acid Titrations 3. At the Equivalence Point NH4Cl salt (WB+SA), ph 7 NH 4 1+ (aq) + H 2 O(l) H 3 O 1+ (aq) + NH 3 (aq) ph < 7 Mole of acid = moles of OH - added. Equivalence pt.

46 Weak Base-Strong Acid Titrations 4. Beyond the Equivalence Point Excess H 3 O 1+ Copyright 2008 Pearson Prentice Hall, Inc. Chapter 15/46

47 Calculate the ph of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the ph after the addition of 20.0 ml of M NaOH to 80.0 ml of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) ph = pk a + log [NH 3] [NH 4+ ] pk a = 9.25 ph = log [0.30] [0.36] = 9.17 start (moles) end (moles) NH + 4 (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) final volume = 80.0 ml ml = 100 ml [NH 4+ ] = [NH 3 ] = ph = log [0.25] [0.28] = 9.20

48 Strong Acid-Weak Base Titration Calculation HCl (aq) + NH 3 (aq) NH 4 Cl (aq) H + (aq) + NH 3 (aq) NH 4 Cl (aq) At equivalence point (ph < 7): NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq)

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50 25.0 ml of 0.10M Acetic acid K a =1.8 x Half Neutralization ph? 3. 4 XS base [NaOAC]+ OH - 1. Starting ph=? Equivalence point ph? 0.10 M NaOH 50

51 Region 1 : Calculate the ph of 25.0 ml of 0.10M acetic acid (HOAc). The K a of HOAc = 1.8 x 10-5 Region 2- Start Titration Problem: Calculate the ph of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5 ). Region 3. Equivalent Point NaOAC salt, ph > 7 Region 3 problem: Calculate the ph at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH K a (HOAc)=1.8 x Region 4 problem: Calculate the ph of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 40.00mL of 0.080M NaOH. K a (HOAc)=1.8 x 10-5.

52 Region 1 : Calculate the ph of 25.0 ml of 0.10M acetic acid (HOAc). The K a of HOAc = 1.8 x 10-5 I C E HOAc H + + OAC x 0 +x 0 +x 0.10-x x x K a = [H+][OAc-] [HOAc] K a = [x][x] 0.10-x Small,drop, WHY? K a = x x = = [H + ] ph = -log =

53 Region 2- Start Titration Problem: Calculate the ph of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5 ). Region 2, we have : HOAC and NaOAC I C E HOAc + NaOH H2O + NaOAC- 2.5 mmole mmole +1.12mmole +1.12mmole +1.12mmole 1.38 mmole +1.12mmole

54 Problem: Calculate the ph of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5 ). 1. initial M of Acid: HOAc + OH - H 2 O + OAc - 1:1 ratio mole/0.039l = M acid after RXN =INITIAL 2. initial M of OAC - anion from M of OH mole/0.039l = M OAc - after RXN =INITIAL ** 1.8 x 10 K a - 5 [H ][A [HA] [x][ x] [ x] ] [HOAc] [H + ] + [OAc - ] I ** C -x +x +x E x x x Can we drop these? x = =[H + ] ph = - Log [ ] = 4.65 Common Ion Effect lower [H+], higher ph 54

55 25.0 ml of 0.10M Acetic acid K a =1.8 x Half Neutralization ph Equivalence point [NaOAC] 1.pH= M NaOH 55

56 Region 3. Equivalent Point. Region 3 problem: Calculate the ph at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH K a (HOAc)=1.8 x Region 3, we have : NaOAC and it hydrolyzes I C E OAC - + H2O HOAc + OH mmole mmole-x 0 x x 2.5 mmole-x 0 x x

57 Region 3 problem: Calculate the ph at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH K a (HOAc)=1.8 x Find moles of Acid (which = s moles of base at Eq pt.) 25.00mL 0.10mol HOAc = mol HOac 1000mL 2. Find mol of OH mL 0.080mol NaOH 1molOH - = mol OH mL 1mol NaOH 3. Since the # of moles are the same, this is at the Eq. Pt. and therefore the initial moles of OAc - ion = mol 57

58 moles of Acid = moles of base = = Eq. Pt. New Volume = L..and [OAc - ] = / =0.0444M Question: What equilibrium(ia) gives rise to the ph? K b K b [OH ][HOAc] [OAc ] 2 [x ] [ x] 1. hydrolysis of the acetate anion: OAc - + H 2 O HOAc + OH- I C E [OAc - ] [HOAc] [OH - ] x +x +x x x x 58

59 moles of Acid = moles of base = = Eq. Pt. New Volume = L... and [OAc - ] =0.0444M Question: What equilibrium(ia) gives rise to the ph? 1. hydrolysis of the acetate anion: OAc - + H 2 O HOAc + OH - [OH ][HOAc] OAc - HOAc OH - [OAc ] I [x ] C -x +x +x K b [ x] E x x x Can we drop this x? How do we find K b? K b Also: K a K b =K w =1.0 x o C so K b = 5.56 x

60 moles of Acid = moles of base = = Eq. Pt. New Volume = L... and [OAc - ] =0.0444M Question: What equilibrium(ia) gives rise to the ph? K b 5.56x 10 [OH ][HOAc] [OAc ] [x ] [ x] 1. hydrolysis of the acetate anion: OAc - + H 2 O HOAc + OH - OAc - HOAc OH - I C E x +x +x x x x x = 4.97 x 10-6 = [OH - ] poh = 5.30 ph = =

61 Region 4 problem: Calculate the ph of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 40.00mL of 0.080M NaOH. K a (HOAc)=1.8 x Find moles of acid: 25.00mL 0.10mol HOAc = mol HOAc 1000mL 2. Find moles of OH - : 40.00mL 0.080molOH - = mol OH mL 3. Subtract to find XS: = mole XS OH - 4. Find OH - concentration, poh, and ph: poh = -Log[0.0007/ ] = 1.97 and ph =

62 Region 4 tips : All the acid has been neutralized and XS base has been added HA + OH - H 2 O + A - + XS OH - What is the dominating factor that controls the ph? The XS strong base (OH - ) Calculations: Find moles of XS OH - and then use the new volume to find [OH - ] and then the poh and ph. 62

63 [H ][A [HA] [H + ] [A - ] K a 2 1 ] [OH ][HA] [H ][A ] K b [A ] [HA] [H + ]=[A - ] [OH - ]=[HA] K a Exam 3 4 poh=-log[xs OH - ] 63