Chapter 15 Acid-Base Equilibria
Section 15.1 Solutions of Acids or Bases Containing a Common Ion Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application of Le Châtelier s principle. Copyright Cengage Learning. All rights reserved 2
Section 15.1 Solutions of Acids or Bases Containing a Common Ion Example HCN(aq) + H 2 O(l) H 3 O + (aq) + CN - (aq) Addition of NaCN will shift the equilibrium to the left because of the addition of CN -, which is already involved in the equilibrium reaction. A solution of HCN and NaCN is less acidic than a solution of HCN alone.
Section 15.2 Buffered Solutions Key Points about Buffered Solutions Buffered Solution resists a change in ph. They are weak acids or bases containing a common ion. After addition of strong acid or base, deal with stoichiometry first, then the equilibrium. Copyright Cengage Learning. All rights reserved 4
Section 15.2 Buffered Solutions Adding an Acid to a Buffer 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 Copyright Cengage Learning. All rights reserved 5
Section 15.2 Buffered Solutions 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 Copyright Cengage Learning. All rights reserved 6
Section 15.2 Buffered Solutions Solving Problems with Buffered Solutions Copyright Cengage Learning. All rights reserved 7
Section 15.2 Buffered Solutions Buffering: How Does It Work? Copyright Cengage Learning. All rights reserved 8
Section 15.2 Buffered Solutions Buffering: How Does It Work? Copyright Cengage Learning. All rights reserved 9
Section 15.2 Buffered Solutions Henderson Hasselbalch Equation A ph = p K + log HA a For a particular buffering system (conjugate acid base pair), all solutions that have the same ratio [A ] / [HA] will have the same ph. Copyright Cengage Learning. All rights reserved 10
Section 15.2 Buffered Solutions EXERCISE! What is the ph of a buffer solution that is 0.45 M acetic acid (HC 2 H 3 O 2 ) and 0.85 M sodium acetate (NaC 2 H 3 O 2 )? The K a for acetic acid is 1.8 10 5. ph = 5.02 Copyright Cengage Learning. All rights reserved 11
Section 15.2 Buffered Solutions Copyright Cengage Learning. All rights reserved 12
Section 15.2 Buffered Solutions Buffered Solution Characteristics Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base. Added H + reacts to completion with the weak base. Added OH - reacts to completion with the weak acid. Copyright Cengage Learning. All rights reserved 13
Section 15.2 Buffered Solutions Buffered Solution Characteristics The ph in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio 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. Copyright Cengage Learning. All rights reserved 14
Section 15.3 Buffering Capacity The amount of protons or hydroxide ions the buffer can absorb without a significant change in ph. Determined by the magnitudes of [HA] and [A ]. A buffer with large capacity contains large concentrations of the buffering components. Copyright Cengage Learning. All rights reserved 15
Section 15.3 Buffering Capacity 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. Copyright Cengage Learning. All rights reserved 16
Section 15.3 Buffering Capacity Choosing a Buffer pk a of the weak acid to be used in the buffer should be as close as possible to the desired ph. Copyright Cengage Learning. All rights reserved 17
Titration Curve Plotting the ph of the solution being analyzed as a function of the amount of titrant added. Equivalence (Stoichiometric) Point point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. Copyright Cengage Learning. All rights reserved 18
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 Copyright Cengage Learning. All rights reserved 19
The ph Curve for the Titration of 50.0 ml of 0.200 M HNO 3 with 0.100 M NaOH Copyright Cengage Learning. All rights reserved 20
The ph Curve for the Titration of 100.0 ml of 0.50 M NaOH with 1.0 M HCI Copyright Cengage Learning. All rights reserved 21
Weak Acid Strong Base Titration Step 1: Step 2: A stoichiometry problem (reaction is assumed to run to completion) then determine concentration of acid remaining and conjugate base formed. An equilibrium problem (determine position of weak acid equilibrium and calculate ph). Copyright Cengage Learning. All rights reserved 22
CONCEPT CHECK! Consider a solution made by mixing 0.10 mol of HCN (K a = 6.2 10 10 ) with 0.040 mol NaOH in 1.0 L of aqueous solution. What are the major species immediately upon mixing (that is, before a reaction)? HCN, Na +, OH, H 2 O Copyright Cengage Learning. All rights reserved 23
Let s Think About It Why isn t NaOH a major species? Why aren t H + and CN major species? List all possibilities for the dominant reaction. Copyright Cengage Learning. All rights reserved 24
Let s Think About It The possibilities for the dominant reaction are: 1. H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) 2. HCN(aq) + H 2 O(l) H 3 O + (aq) + CN (aq) 3. HCN(aq) + OH (aq) CN (aq) + H 2 O(l) 4. Na + (aq) + OH (aq) NaOH 5. Na + (aq) + H 2 O(l) NaOH + H + (aq) Copyright Cengage Learning. All rights reserved 25
Let s Think About It How do we decide which reaction controls the ph? H 2 O(l) + H 2 O(l) HCN(aq) + H 2 O(l) HCN(aq) + OH (aq) H 3 O + (aq) + OH (aq) H 3 O + (aq) + CN (aq) CN (aq) + H 2 O(l)
Let s Think About It HCN(aq) + OH (aq) CN (aq) + H 2 O(l) What are the major species after this reaction occurs? HCN, CN, H 2 O, Na + Copyright Cengage Learning. All rights reserved 27
Let s Think About It Now you can treat this situation as before. List the possibilities for the dominant reaction. Determine which controls the ph. Copyright Cengage Learning. All rights reserved 28
CONCEPT CHECK! Calculate the ph of a solution made by mixing 0.20 mol HC 2 H 3 O 2 (K a = 1.8 10 5 ) with 0.030 mol NaOH in 1.0 L of aqueous solution. Copyright Cengage Learning. All rights reserved 29
Let s Think About It What are the major species in solution? Na +, OH, HC 2 H 3 O 2, H 2 O Why isn t NaOH a major species? Why aren t H + and C 2 H 3 O 2 major species? Copyright Cengage Learning. All rights reserved 30
Let s Think About It What are the possibilities for the dominant reaction? 1. H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) 2. HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) 3. HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + H 2 O(l) 4. Na + (aq) + OH (aq) NaOH(aq) 5. Na + (aq) + H 2 O(l) NaOH + H + (aq) Which of these reactions really occur? Copyright Cengage Learning. All rights reserved 31
Let s Think About It Which reaction controls the ph? H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + H 2 O(l) How do you know? Copyright Cengage Learning. All rights reserved 32
Let s Think About It HC 2 H 3 O 2 (aq) + OH C 2 H 3 O 2 (aq) + H 2 O Before 0.20 mol 0.030 mol 0 Change 0.030 mol 0.030 mol +0.030 mol After 0.17 mol 0 0.030 mol K = 1.8 10 9 Copyright Cengage Learning. All rights reserved 33
Steps Toward Solving for ph HC 2 H 3 O 2 (aq) + H 2 O H 3 O + + C 2 H 3 O 2- (aq) Initial 0.170 M ~0 0.030 M Change x +x +x Equilibrium 0.170 x x 0.030 + x K a = 1.8 10 5 ph = 3.99 Copyright Cengage Learning. All rights reserved 34
EXERCISE! Calculate the ph of a 100.0 ml solution of 0.100 M acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 10 5. ph = 2.87 Copyright Cengage Learning. All rights reserved 35
CONCEPT CHECK! Calculate the ph of a solution made by mixing 100.0 ml of a 0.100 M solution of acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 10 5, and 50.0 ml of a 0.10 M NaOH solution. ph = 4.74 Copyright Cengage Learning. All rights reserved 36
CONCEPT CHECK! Calculate the ph of a solution at the equivalence point when 100.0 ml of a 0.100 M solution of acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 10 5, is titrated with a 0.10 M NaOH solution. ph = 8.72 Copyright Cengage Learning. All rights reserved 37
The ph Curve for the Titration of 50.0 ml of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH Copyright Cengage Learning. All rights reserved 38
The ph Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various K a Values with 0.10 M NaOH Copyright Cengage Learning. All rights reserved 39
The ph Curve for the Titration of 100.0 ml of 0.050 M NH 3 with 0.10 M HCl Copyright Cengage Learning. All rights reserved 40
Section 15.5 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 they are ideally as close as possible). Copyright Cengage Learning. All rights reserved 41
Section 15.5 Acid-Base Indicators The Acid and Base Forms of the Indicator Phenolphthalein Copyright Cengage Learning. All rights reserved 42
Section 15.5 Acid-Base Indicators The Methyl Orange Indicator is Yellow in Basic Solution and Red in Acidic Solution Copyright Cengage Learning. All rights reserved 43
Section 15.5 Acid-Base Indicators Useful ph Ranges for Several Common Indicators Copyright Cengage Learning. All rights reserved 44