LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution.

LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution. (Sec ) LO 3.3 The student is able to use stoichiometric calculations to predict the results of performing a reaction in the laboratory and/or to analyze deviations from the expected results. (Sec 15.4) LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec ) LO 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration. (Sec 15.4)

LO 6.13 The student can interpret titration data for monoprotic or polyprotic acids involving titration of a weak or strong acid by a strong base (or a weak or strong base by a strong acid) to determine the concentration of the titrant and the pKa for a weak acid, or the pKb for a weak base. (Sec 15.4) LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution. (Sec 15.4) LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. (Sec ) LO 6.17 The student can, given an arbitrary mixture of weak and strong acids and bases (including polyprotic systems), determine which species will react strongly with one another (i.e. with K >1) and what species will be present in large concentration at equilibrium. (Sec )

LO 6.18 The student can design a buffer solution with a target pH and buffer capacity by selecting an appropriate conjugate acid-base pair and estimating the concentrations needed to achieve the desired capacity. (Sec 15.2) LO 6.19 The student can relate the predominant form of a chemical species involving a labile proton (i.e. protonated/deprotonated form of a weak acid) to the pH of a solution and the pKa associated with the labile proton. (Sec 15.2) LO 6.20 The student can identify a solution as being a buffer solution and explain the buffer mechanism in terms of the reactions that would occur on addition of acid or base. (Sec )

AP Learning Objectives, Margin Notes and References
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. LO 6.17 The student can, given an arbitrary mixture of weak and strong acids and bases (including polyprotic systems), determine which species will react strongly with one another (i.e. with K >1) and what species will be present in large concentration at equilibrium.

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

Example HCN(aq) + H2O(l) H3O+(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.

LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. LO 6.17 The student can, given an arbitrary mixture of weak and strong acids and bases (including polyprotic systems), determine which species will react strongly with one another (i.e. with K >1) and what species will be present in large concentration at equilibrium. LO 6.18 The student can design a buffer solution with a target pH and buffer capacity by selecting an appropriate conjugate acid-base pair and estimating the concentrations needed to achieve the desired capacity. LO 6.19 The student can relate the predominant form of a chemical species involving a labile proton (i.e. protonated/deprotonated form of a weak acid) to the pH of a solution and the pKa associated with the labile proton. LO 6.20 The student can identify a solution as being a buffer solution and explain the buffer mechanism in terms of the reactions that would occur on addition of acid or base. Additional AP References LO 6.18 (see APEC 15, “Buffers and Buffered Systems”)

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

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

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

Solving Problems with Buffered Solutions
Copyright © Cengage Learning. All rights reserved

Buffering: How Does It Work?

Henderson–Hasselbalch Equation
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

EXERCISE! What is the pH of a buffer solution that is 0.45 M acetic acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)? The Ka for acetic acid is 1.8 × 10–5. pH = 5.02 pH = –logKa + log([C2H3O2–] / [HC2H3O2]) = –log(1.8 × 10–5) + log(0.85 M / 0.45 M) = 5.02 Copyright © Cengage Learning. All rights reserved

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

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

LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.20 The student can identify a solution as being a buffer solution and explain the buffer mechanism in terms of the reactions that would occur on addition of acid or base. Additional AP References LO 6.20 (see APEC 16, “Acids, Bases and Buffers”)

Determined by the magnitudes of [HA] and [A–].
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

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

LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution. LO 3.3 The student is able to use stoichiometric calculations to predict the results of performing a reaction in the laboratory and/or to analyze deviations from the expected results. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration. LO 6.13 The student can interpret titration data for monoprotic or polyprotic acids involving titration of a weak or strong acid by a strong base (or a weak or strong base by a strong acid) to determine the concentration of the titrant and the pKa for a weak acid, or the pKb for a weak base. LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution.

Additional AP References LO 1.20 (see APEC 4, “Analysis of Vinegar”) LO 6.13 (see APEC 14, “Titration Curves”) LO 6.13 (see Appendix 7.10, “Polyprotic Acid Titrations”)

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

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

The pH Curve for the Titration of 50. 0 mL of 0. 200 M HNO3 with 0
The pH Curve for the Titration of 50.0 mL of M HNO3 with M NaOH Copyright © Cengage Learning. All rights reserved

The pH Curve for the Titration of 100. 0 mL of 0. 50 M NaOH with 1
The pH Curve for the Titration of mL of 0.50 M NaOH with 1.0 M HCI Copyright © Cengage Learning. All rights reserved

Weak Acid–Strong Base Titration
Step 1: A stoichiometry problem (reaction is assumed to run to completion) then determine concentration of acid remaining and conjugate base formed. Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate pH). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider a solution made by mixing 0.10 mol of HCN (Ka = 6.2 × 10–10) with 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–, H2O Major Species: HCN, Na+, OH-, H2O. Copyright © Cengage Learning. All rights reserved

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. NaOH is a strong base and completely dissociates. HCN is a weak acid and does not dissociate very much. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… The possibilities for the dominant reaction are:
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) HCN(aq) + OH–(aq) CN–(aq) + H2O(l) Na+(aq) + OH–(aq) NaOH Na+(aq) + H2O(l) NaOH + H+(aq) Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How do we decide which reaction controls the pH?
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) HCN(aq) + OH–(aq) CN–(aq) + H2O(l) In general, the best acid (HCN in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka / Kw = 62,000. We can assume the reaction goes to completion. In this problem, OH- is limiting.

Let’s Think About It… HCN(aq) + OH–(aq) CN–(aq) + H2O(l)
What are the major species after this reaction occurs? HCN, CN–, H2O, Na+ After the reaction takes place, the major species are: HCN, CN-, H2O, Na+ Copyright © Cengage Learning. All rights reserved

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

CONCEPT CHECK! Calculate the pH of a solution made by mixing 0.20 mol HC2H3O2 (Ka = 1.8 × 10–5) with mol NaOH in 1.0 L of aqueous solution. Major Species: HC2H3O2, Na+, OH-, H2O Possibilities for reactions: 1) HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) 2) H2O + H2O  H3O+(aq) + OH-(aq) 3) HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) Reaction 3 is most likely. In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka/ Kw = 1.8 x 109. We can assume the reaction goes to completion. In this problem, OH- is limiting. After the reaction takes place, the major species are: HC2H3O2, C2H3O2-, H2O, Na+ The primary reaction could be: 3) C2H3O2-(aq) + H2O  HC2H3O2(aq) + OH-(aq) We will see that reactions 1 and 3 will give us the same answer (as long as "x" is negligible in calculations, or solved for exactly). In this case, though, Ka (reaction 1) is greater than Kb (reaction 3) and should be used. When students solve this, make sure they include the initial concentration of the conjugate base (they tend to forget and call it zero). pH = 3.99 Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are the major species in solution?
Na+, OH–, HC2H3O2, H2O Why isn’t NaOH a major species? Why aren’t H+ and C2H3O2– major species? NaOH is a strong base and completely dissociates. HC2H3O2 is a weak acid and does not dissociate very much. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are the possibilities for the dominant reaction? H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq) HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l) Na+(aq) + OH–(aq) NaOH(aq) Na+(aq) + H2O(l) NaOH + H+(aq) Which of these reactions really occur? See Slide 12. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Which reaction controls the pH?
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq) HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l) How do you know? See Slide 12. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… K = 1.8 × 109 HC2H3O2(aq) + OH– C2H3O2–(aq)
Before 0.20 mol mol Change –0.030 mol –0.030 mol mol After 0.17 mol 0.030 mol K = 1.8 × 109 Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH
HC2H3O2(aq) + H2O H3O+ + C2H3O2-(aq) Initial 0.170 M ~0 0.030 M Change –x +x Equilibrium 0.170 – x x x Ka = 1.8 × 10–5 pH = 3.99 Copyright © Cengage Learning. All rights reserved

EXERCISE! Calculate the pH of a mL solution of M acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5. pH = 2.87 pH = 2.87 Major Species: HC2H3O2, H2O Dominant reaction: HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) The mL is not necessary to know for this problem. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a solution made by mixing mL of a M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5, and 50.0 mL of a 0.10 M NaOH solution. pH = 4.74 pH = 4.74 Major Species: HC2H3O2, Na+, OH-, H2O Dominant reaction: HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka/Kw = 1.8 x 109. We can assume the reaction goes to completion. In this problem, OH- is limiting. After the reaction takes place, the major species are: HC2H3O2, C2H3O2-, H2O Primary reaction is: HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) When students solve this, make sure they include the initial concentration of the conjugate base (they tend to forget and call it zero). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a solution at the equivalence point when mL of a M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5, is titrated with a 0.10 M NaOH solution. pH = 8.72 pH = 8.72 Major Species: HC2H3O2, Na+, OH-, H2O Dominant reaction: HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). At the equivalence point, there are the same number of moles of HC2H3O2(aq) + OH-(aq) to exactly react due to the 1:1 ratio in the balanced equation. This turns out to be mol of both the acid and base in this case. This means that mL of NaOH had to be added. After the reaction takes place, the major species are: C2H3O2-, H2O, Na+ Primary reaction is: C2H3O2-(aq) + H2O  OH-(aq) + HC2H3O2(aq) Now use an ICE chart to determine the equilibrium concentrations and finally the pH. Take note that the initial concentration of C2H3O2-(aq) is M (0.010 mol / L). Copyright © Cengage Learning. All rights reserved

The pH Curves for the Titrations of 50. 0-mL Samples of 0
The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various Ka Values with 0.10 M NaOH Copyright © Cengage Learning. All rights reserved

LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. Additional AP References LO 1.20 (see APEC 4, “Analysis of Vinegar”)

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

The Acid and Base Forms of the Indicator Phenolphthalein

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

Useful pH Ranges for Several Common Indicators

LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution.

Similar presentations

Presentation on theme: "LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution.

Similar presentations

Presentation on theme: "LO 1.20 The student can design, and/or interpret data from an experiment that uses titration to determine the concentration of an analyte in a solution."— Presentation transcript:

Similar presentations

About project

Feedback