Chapter 15. Acid-Base Equilibria

Chapter 15 Acid-Base Equilibria

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion already involved in the equilibrium reaction. An application of Le Châtelier s principle. 2

The Common Ion Effect Consider a solution of acetic acid (HC 2 H 3 O 2 ), in which you have the following equilibrium: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 (aq) If we were to add NaC 2 H 3 O 2 to this solution, it would provide additional C 2 H 3 O 2 ions which are present on the right side of the equilibrium. 3

The Common Ion Effect Consider a solution of acetic acid (HC 2 H 3 O 2 ), in which you have the following equilibrium: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 (aq) NaC 2 H 3 O 2(s) Na + (aq) + C 2 H 3 O 2 (aq) common ion The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased. 4

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion already involved in the equilibrium reaction. Addition of a common ion suppresses the ionization of a weak acid or a weak base. 5

A Problem To Consider 15.1. An aqueous solution is 0.025 M in formic acid, HCO 2 H and 0.018 M in sodium formate, NaHCO 2. What is the ph of the solution. The K a for formic acid is 1.7 10-4. Consider the equilibrium below: HCO 2 H (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) K a + [H3O ][HCO 2] 1.710 [HCO H] 2 4 6

15.1. (Continued) A Problem To Consider HCO 2 H (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) Starting 0.025 0 0.018 Change -x +x +x Equilibrium 0.025-x x 0.018+x K a + [H3O ][HCO 2] x(0.018 x) 1.710 [HCO H] (0.025 x) 2 4 7

15.1. (Continued) A Problem To Consider HCO 2 H (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) K a + [H3O ][HCO 2] x(0.018 x) 1.710 [HCO H] (0.025 x) 2 4 Assume that x is small compared with 0.018 and 0.025: (0.018 x) 0.018 and (0.025 x) 0.025 x(0.018) 0.025 1.710 and x (1.710 ) 2.410 (0.025) 0.018 4 4 4 8 M

15.1. (Continued) A Problem To Consider HCO 2 H (aq) + H 2 O (l) H 3 O + (aq) + HCO 2 (aq) x [H O ] 2.410 3 + 4 M Note that x is much smaller than 0.018 or 0.025 (5 % rule). + 4 ph log[h3o ] log(2.410 ) 3.63 For comparison, the ph of 0.025 M formic acid is 2.69. 9

Buffered Solutions Buffered solution is a solution characterized by the ability to resist changes in ph when limited amounts of acid or base are added to it. A buffer solution contains either: 1) a weak acid and its conjugate base or 2) a weak base and its conjugate acid Both must be present! 10

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3- is its conjugate acid buffer solution 11

Buffers To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 12

Buffering: How Does It Work? Consider an equal molar mixture of CH 3 COOH (weak acid) and CH 3 COONa: Add strong acid: H + (aq) + CH 3 COO (aq) CH 3 COOH (aq) Add strong base: OH (aq) + CH 3 COOH (aq) CH 3 COO (aq) + H 2 O (l) 13

ph of a Buffered Solution Consider mixture of a weak acid HA and salt NaA: HA (aq) H + (aq) + A (aq) NaA (s) Na + (aq) + A (aq) K a + [H ][A ] [HA] + K [HA] [H ] a [A ] [HA] log [H ] lg o K a log [A ] log [H ] lg o K a [A ] log [HA] 14

ph of a Buffered Solution Consider mixture of a weak acid HA and salt NaA: HA (aq) H + (aq) + A (aq) + ph log[h ] NaA (s) Na + (aq) + A (aq) pk a log K a log [H ] lg o K a [A ] log [HA] ph p K a log [A ] p [HA] K a [conjugate base] log [weak acd i ] Henderson-Hasselbalch Equation 15

Henderson-Hasselbalch Equation ph p K a log [A ] [HA] Used to determine ph of a buffered solution For a particular buffering system (conjugate acid base pair), all solutions that have the same ratio [A ] / [HA] will have the same ph. If concentrations of a weak acid and its conjugate base are the same, ph of that buffer is equal to pk a of the acid. 16

A Problem To Consider 15.2. What is the ph of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? pk a of HCOOH is 3.77. Initial (M) HCOOH (aq) H + (aq) + HCOO (aq) 0.30 0.00 0.52 Change (M) -x +x +x Equilibrium (M) 0.30 - x x 0.52 + x Common ion effect: presence of HCOO suppresses the ionization of HCOOH. Therefore: 0.30 x 0.30 0.52 + x 0.52 17

15.2. (Continued) A Problem To Consider Initial (M) Change (M) Equilibrium (M) HCOOH (aq) H + (aq) + HCOO (aq) 0.30 0.00 0.52 -x +x +x 0.30 - x 0.30 x 0.52 + x 0.52 Mixture of weak acid HCOOH and its conjugate base HCOOK. This will be a buffer solution. Applying Henderson-Hasselbalch equation: [HCOO ] ph pk a log [HCOOH] 0.52 3.77 log 4.01 0.30 18

Buffer Capacity Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base. The ph in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. Buffer capacity depends on the amount of acid and conjugate base present in the solution. H 3 O + (aq) + A (aq) HA (aq) + H 2 O (l) OH (aq) + HA (aq) A (aq) + H 2 O (l) 19

Buffer Capacity As long as the ratio of the concentrations of the weak acid and weak base remains virtually constant, the ph will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A or B and BH + ) are large compared with amounts of H + or OH added. H 3 O + (aq) + A (aq) HA (aq) + H 2 O (l) OH (aq) + HA (aq) A (aq) + H 2 O (l) 20

Buffer Capacity Buffer capacity is defined as the amount of protons (H + ) or hydroxide ions (OH ) the buffer can absorb without a significant change in ph. Buffer capacity is determined by the magnitudes of [HA] and [A ]. A buffer with large capacity contains large concentrations of the buffering components. Optimal buffering occurs when [HA] is equal to [A ]. It is for this condition that the ratio [A ] / [HA] is most resistant to change when H + or OH is added to the buffered solution. 21

Preparing a buffer of given ph: A buffer must be prepared from a conjugate acid-base pair in which the K a of the acid is approximately equal to the desired H 3 O + concentration (or pk a of the acid is approximately equal to the desired ph level). log [A ph pk ] a [HA] For example, we need a buffer with ph 4.90. So, we need a conjugate acid-base pair with a pk a close to the desired ph = 4.90. The K a for acetic acid is 1.7 10-5, and its pk a is 4.77. We can use a mixture of acetic acid (HC 2 H 3 O 2 ), and sodium acetate (NaC 2 H 3 O 2 ). To get a buffer ph to 4.90 we need to increase the ratio of [NaC 2 H 3 O 2 ] /[HC 2 H 3 O 2 ]. 22

Solving Problems with Buffered Solutions 23

A Problem To Consider 15.3. Calculate the ph of the 0.20 M NH 3 /0.20 M NH 4 Cl buffer. What is the ph of the buffer after the addition of 10.0 ml of 0.10 M HCl to 65.0 ml of the buffer? 1) Use Henderson-Hasselbalch equation to calculate the ph. The weak acid in this buffer is NH 4+, so we need K a for ammonium ion (5.6 10-10 ): ph p K a log [NH ] log(5.610 ) log 0.20 9.25 [ NH ] 0.20 3 10 4 24

A Problem To Consider 15.3. Calculate the ph of the 0.20 M NH 3 /0.20 M NH 4 Cl buffer. What is the ph of the buffer after the addition of 10.0 ml of 0.10 M HCl to 65.0 ml of the buffer? 2) Initial amounts of NH 3 and NH 4 Cl in the buffer: mol NH initial 0.0650 L 0.20M 0.013 mol 4 mol NH initial 0.0650 L 0.20M 0.013 mol 3 Amount of HCL added to the buffer: mol HCl added 0.0100 L 0.10M 0.0010 mol 25

15.3. (Continued) A Problem To Consider mol NH initial 0.0650 L 0.20M 0.013 mol 4 mol NH initial 0.0650 L 0.20M 0.013 mol 3 mol HCl added 0.0100 L 0.10M 0.0010 mol When HCl is added the following acid-base reaction occurs: NH 3(aq) + H + (aq) NH 4 + (aq) Initial (mol) 0.013 0.0010 0.013 Change (mol) -0.0010-0.0010 +0.0010 After rxn is complete (mol) 0.012 0 0.014 26

A Problem To Consider 15.3. (Continued) When HCl is added the following acid-base reaction occurs: NH 3(aq) + H + (aq) NH + 4 (aq) After rxn is complete (mol) 0.012 0 0.014 Note: [NH 3 ] [NH 4 + ] = mol NH 3 mol NH 4 + To find new ph: [NH ] 0.012 ph pk a log log(5.610 ) log 9.18 [NH ] 0. 014 3 10 4 27

A Problem To Consider 15.4. 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 0.050 M NaOH to 80.0 ml of the buffer solution? 1) Use Henderson-Hasselbalch equation to calculate the ph. The weak acid in this buffer is NH 4+, so we need K a for ammonium ion (5.6 10-10 ): ph p K a log [NH ] log(5.610 ) log 0.30 9.17 [ NH ] 0.36 3 10 4 28

A Problem To Consider 15.4. 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 0.050 M NaOH to 80.0 ml of the buffer solution? 2) Initial amounts of NH 3 and NH 4 Cl in the buffer: mol NH initial 0.0800 L 0.36M 0.029 mol 4 mol NH initial 0.0800 L 0.30M 0.024 mol 3 Amount of NaOH added to the buffer: mol NaOH added 0.0200 L 0.050M 0.0010 mol 29

A Problem To Consider 15.4. (Continued) mol NH initial 0.0800 L 0.36M 0.029 mol 4 mol NH initial 0.0800 L 0.30M 0.024 mol 3 mol NaOH added 0.0200 L 0.050M 0.0010 mol When NaOH is added the following acid-base reaction occurs: NH 4 + (aq) + OH (aq) NH 3(aq) + H 2 O (l) Initial (mol) Change (mol) After rxn is complete (mol) 0.029 0.0010 0.024-0.0010-0.0010 +0.0010 0.028 0 0.025 30

15.4. (Continued) A Problem To Consider When NaOH is added the following acid-base reaction occurs: NH 4 + (aq) + OH (aq) NH 3(aq) + H 2 O (l) After rxn is complete (mol) Note: To find new ph: [NH 3 ] [NH 4 + ] = mol NH 3 mol NH 4 + 0.028 0 0.025 [NH ] 0.025 ph pk a log log(5.610 ) log 9.22 [NH ] 0. 028 3 10 4 31

Neutralization of a Strong Acid with a Strong Base To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 32

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 33

Acid-Base Titration Curves An acid-base titration curve is a plot of the ph of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. Equivalence (Stoichiometric) Point point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. You can use titration curves to choose an appropriate indicator that will show when the titration is complete. 34

Strong Acid Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH (aq) + H + (aq) H 2 O (l) Note that the ph changes slowly until the titration approaches the equivalence point. At the equivalence point, the ph of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze. 35

Strong Acid Strong Base Titrations The ph Curve for the titration of 100.0 ml of 0.50 M NaOH with 1.0 M HCI 36

A Problem To Consider 15.5. Calculate the ph of a solution in which 10.0 ml of 0.100 M NaOH is added to 25.0 ml of 0.100 M HCl. Because the reactants are a strong acid and a strong base, the reaction is essentially complete: H 3 O + (aq) + OH (aq) H 2 O (l) + H 2 O (l) We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities: mol H 3 O + = 0.0250 L 0.100 M = 0.00250 mol mol OH = 0.0100 L 0.100 M = 0.00100 mol 37

A Problem To Consider 15.5. (Continued). Initial (mol) Change (mol) After completion (mol) H 3 O + (aq) + OH (aq) H 2 O (l) + H 2 O (l) 0.00250 0.00100-0.00100-0.00100 0.0015 0 Excess H 3 O + = 0.0150 mol moles 0.00150 mol [H 3 O ] Volume in L (0.0100 0.0250) L 0.0429 M ph log[h3o ] log(0.0429) 1.368 38

Weak Acid Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) The ph range of these titrations is shorter. The equivalence point will be on the basic side (ph > 7) since the salt produced contains the anion of a weak acid: CH 3 COO (aq) + H 2 O (l) OH (aq) + CH 3 COOH (aq) 39

A Problem To Consider 15.6. Calculate the ph of the solution at the equivalence point when 25.0 ml of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 10-5. At the equivalence point, equal molar amounts of acetic acid and sodium hydroxide react to give sodium acetate: CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) mol CH 3 COONa = 0.0250 L 0.10 M = 0.0025 mol Volume of the solution: 25.0 ml + 25.0 ml = 50.0 ml = 0.0500 L - mol CH3COONa [CH COO ] Volume in L 0.0025 mol 0.0500 L 3 0.050 M 40

A Problem To Consider 15.6. (Continued). Sodium acetate will hydrolyze according to the equation: CH 3 COO (aq) + H 2 O (l) OH (aq) + CH 3 COOH (aq) Initial (M) 0.050 0 0 Change (M) -x +x +x Equilibrium (M) 0.050-x x x K b - (CH COO ) [OH ][CH3COOH] - [CH COO ] 3 3 2 x 0.050 x 2 x 0.050 K b K 1.010 ( CH -14 - w -10 3COO ) 5.9 10-5 K a(ch3cooh) 1.710 41

A Problem To Consider 15.6. (Continued). Sodium acetate will hydrolyze according to the equation: CH 3 COO (aq) + H 2 O (l) OH (aq) + CH 3 COOH (aq) Equilibrium (M) 0.050-x x x 2 x 0.050 5.910-10 - [ OH ] x 5.410-6 M ( 5% rule : x 5% of 0.050 M ) poh log(5.410-6 ) 5.27 ph 14.00 poh 14.00 5.27 8.73 42

A Problem To Consider 15.7. Exactly 100. ml of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the ph at the equivalence point? The K a for nitrous acid is 4.4 10-4. At the equivalence point, equal molar amounts of nitrous acid and sodium hydroxide react to give sodium nitrite: HNO 2(aq) + NaOH (aq) NaNO 2(aq) + H 2 O (l) mol NaNO 2 = 0.100 L 0.10 M = 0.010 mol Volume of the solution: 100. ml + 100. ml = 200. ml = 0.200 L - mol NaNO2 [NO ] Volume in L 0.010 mol 0.200 L 2 0.050 M 43

A Problem To Consider 15.7. (Continued). Sodium nitrite will hydrolyze according to the equation: NO 2 (aq) + H 2 O (l) OH (aq) + HNO 2(aq) Initial (M) 0.050 0 0 Change (M) -x +x +x Equilibrium (M) 0.050-x x x K b 2 - [OH ][HNO2] x (NO2 ) - [NO ] 0.050 x 2 2 x 0.050 K b -14 - K w 1.010 ( NO2 ) 2.310-4 K (HNO ) 4.410 a 2-11 44

A Problem To Consider 15.7. (Continued). Sodium nitrite will hydrolyze according to the equation: NO 2 (aq) + H 2 O (l) OH (aq) + HNO 2(aq) Equilibrium (M) 0.050-x x x 2 x 0.050 2.310-11 - [ OH ] x 1.110-6 M ( 5% rule : x 5% of 0.050 M ) poh log(5.410-6 ) 5.97 ph 14.00 poh 14.00 5.97 8.03 45

Strong Acid Weak Base Titrations HCl (aq) + NH 3(aq) NH 4 Cl (aq) The equivalence point will be on the basic side (ph < 7) since the salt produced contains the anion of a weak acid: NH 4 + (aq) + H 2 O (l) NH 3(aq) + H 3 O + (aq) 46

Strong Acid Weak Base Titrations In this case, the ph declines slowly at first, then falls abruptly from about ph 7 to ph 3 Methyl red, which changes color from yellow at ph 6 to red at ph 4.8, is a possible indicator. 47

Acid-Base Indicators Marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point (but ideally they should be as close as possible). 48

ph 49

The Acid and Base Forms of the Indicator Phenolphthalein 50

The Methyl Orange Indicator is Yellow in Basic Solution and Red in Acidic Solution 51

Acid-Base Indicators [HIn] [In ] [HIn] [In ] 10 0.1 HIn (aq) H + (aq) + In (aq) Color of acid (HIn) predominates Color of conjugate base (In - ) predominates 52

The titration curve of a strong acid with a strong base. 53

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 54

Useful ph Ranges for Several Common Indicators 55