chapter 17chapter 17 complex aqueous equilibriacourses.chem.psu.edu/chem112/summer/lecture...

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Chapter 17 Chapter 17 Chapter 17 Chapter 17 Complex Aqueous Equilibria Complex Aqueous Equilibria Common Ion Effect Buffers Titrations Predict using LeChatelier’s Principle Predict using LeChatelier’s Principle CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Consider the equilibrium involving the weak acid at 298K shown above. What will happen to the pH if some solid CH COONa is added while the temperature is held CH 3 COONa is added while the temperature is held constant? 1. The pH will increase 2. The pH will decrease 3 The pH will not be effected 3. The pH will not be effected. 4. There is not enough information to determine the effect on the pH on the pH. Chem 112 2 MJ Bojan

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Page 1: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Chapter 17Chapter 17Chapter 17Chapter 17Complex Aqueous EquilibriaComplex Aqueous Equilibria

Common Ion Effect

Buffers

Titrations

Predict using LeChatelier’s PrinciplePredict using LeChatelier’s Principle

CH3COOH(aq) H+(aq) + CH3COO−(aq)

Consider the equilibrium involving the weak acid at 298K shown above. What will happen to the pH if some solid CH COONa is added while the temperature is heldCH3COONa is added while the temperature is held constant?

1. The pH will increasep

2. The pH will decrease

3 The pH will not be effected3. The pH will not be effected.

4. There is not enough information to determine the effect on the pHon the pH.

Chem 112 2MJ Bojan

Page 2: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Test the prediction by doing an experiment.

DEMO:CH3COOH(aq) H+(aq) + CH3COO−(aq)

pH =

Add CH3COONa(s): CH COONa (aq) Na+(aq) + CH COO−(aq)CH3COONa (aq) → Na+(aq) + CH3COO−(aq)

pH =

What happened to[CH3COO−]?

Common Ion Effect: if a system is at equilibrium and it is disturbed by adding a strong electrolyte with

[CH3COOH]?

[H+]?

by adding a strong electrolyte with a common ion, the ionization of the weak electrolyte will be

d d[H+]? reduced.Chem 112 3MJ Bojan

What happens when a strong acid is added to a weak acid and to a mixture of the acid and its conjugate base?

CH COOH( ) H+( ) CH COO ( )CH3COOH(aq) H+(aq) + CH3COO−(aq)

pH0 1M CH COOH(aq) Mixture pH0.1M CH3COOH(aq) Mixture pH

0.1M CH3COOH0.1M CH3COONa

Add HClAdd HCl

Add NaOH Add NaOH

How does the addition of strong acid or base effect the pH of the weak acid?

Example of the Common Ion effect

of the mixture?

Page 3: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

A Buffer is a solution that “resists” a change in pHg p

E g blood contains substancesE.g. blood contains substances that keep its pH fixed at 7.3 -important for life functions

Buffer solutions consist of either:

Chem 112 5MJ Bojan

http://chemcollective.org/activities/tutorials/buffers/buffers3

How do Buffers work? When strong acid or base is added, the ratio of weak acid: weak base does not change much.

If a small amount of hydroxide is

Buffer with[HA] = [A]

Buffer afteradding OH

Buffer afteradding H+

yadded to an equimolarsolution of HA in NaA, the HF

Similarly, if acid is added, the A− reacts with it to form HA and water

OH H+

,reacts with the OH− to make A− and water.

HA and water.

[HA][HA] [A] [HA] [A][A]

6http://chemcollective.org/activities/tutorials/buffers/buffers3

Page 4: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

QuantitativelyC l l t th H f l ti th t i 0 1M CH COOHCalculate the pH of a solution that is 0.1M CH3COOH

and 0.1M CH3COONa

Chem 112 7MJ Bojan

We can derive an expression that will enable us to calculate the pH of a buffer solution.p

Consider the equilibrium constant expression for the dissociation of a generic acid, HX:

HX(aq) H+(aq) + X−(aq)

This is the Henderson Hasselbalch equationThis is the Henderson–Hasselbalch equation.Chem 112 8MJ Bojan

Page 5: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Calculating the pH of a BufferWhat is the pH of a buffer that is 0.12 M in lactic acid,

CH3CH(OH)COOH, and 0.10 M in sodium lactate?

K f l ti id i 1 4 × 10 4Ka for lactic acid is 1.4 × 10−4.

Chem 112 9MJ Bojan

Calculating the pH of a Buffer

Calculate the pH of a solution that is 0.1M NH3 and 0.2M NH4Cl.

Kb for NH3 is 1.8 × 10−5.b 3

Chem 112 10MJ Bojan

Page 6: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

When Strong Acids or Bases Are Added to a Buffer reaction occursto a Buffer… reaction occurs.

Follow a 2-step process to determine pH1. neutralization2. pH determination

neutralizationA(aq) + H+(aq) → HA(aq)

BufferHA + A

Calculate[HA] and [A]

pH determination

neutralizationHA(aq) + OH(aq) → A(aq) + H2O()

Stoichiometry calculationEquilibriumcalculation

Chem 112 MJ Bojan 11

Stoichiometry calculation calculation

Finding the pH after adding Strong Acid or Base to a Bufferg

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Add 0 020 mol OH Add 0 020 mol H+Add 0.020 mol OH Add 0.020 mol H

Buffer: 0.300 M CH3COOH0.300 M CH3COONa

pH =??? pH =???

Chem 112 12

MJ Bojan

pH = 4.74

Page 7: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Calculating pH changes in Buffers

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of thisThe pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

Chem 112 13MJ Bojan

Finding the pH after adding Strong Acid or Base to a Bufferg

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Add 0 020 mol OH Add 0 020 mol H+Add 0.020 mol OH Add 0.020 mol H

Buffer: 0.300 M CH3COOH0.300 M CH3COONapH = 4 74

Buffer: 0.280 M CH3COOH0.320 M CH3COONapH = 4 80

Buffer: 0.320 M CH3COOH0.280 M CH3COONapH = 4 68

Chem 112 14MJ Bojan

pH = 4.74pH = 4.80 pH = 4.68

Page 8: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Choosing a BufferChoosing a Buffer

pH range of a buffer: range of pH over which the buffer acts effectivelybuffer acts effectively.

optimal pH: Buffer region

Chem 112 15MJ Bojan

Buffer Capacity: volume of acid or base that can be added before the pH of the buffercan be added before the pH of the buffer

changes appreciably.

Whi h l ti ill h th t t b ff it ?Which solution will have the greatest buffer capacity?

1M CH COOH/CH COONa1M CH3COOH/CH3COONa

0.1M CH3COOH/CH3COONa0.1M CH3COOH/CH3COONa

Chem 112 16MJ Bojan

Page 9: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Titration

A pH meter or indicators are used to determine when the solution has reached the equivalence point:

In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).

Chem 112 17MJ Bojan

The titration of a strong acid with a strong base can be described by a plot of pH vs. amount of base added.y p p

Chem 112 18MJ Bojan

Page 10: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Quantitative determination of the pH for an acid Quantitative determination of the pH for an acid AcidAcid BaseBase Titration is a two step processTitration is a two step processAcid Acid Base Base Titration is a two step processTitration is a two step process

1. Neutralization

What is in solution after neutralization?

2. Find the pHA) strong acid/baseB) weak/acid baseB) weak/acid base

equilibrium calculationbuffer solutionbuffer solution

Chem 112 19MJ Bojan

Sample calculation of pH during the titration of a Strong Acid with a Strong Baseof a Strong Acid with a Strong Base

Titrate 0.1M HCl with 0.2M NaOH starting with 50 mL of HCl solution. Calculate pH when the following quantities of NaOH are added:

a. 0 mL NaOH, b. 20 mL NaOH, c. 25 mL NaOH, d. 30 mL NaOH

a. Initial concentration of strong acid: (0mL of NaOH added)

Chem 112 20MJ Bojan

Page 11: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Sample calculation of pH during the titration of a Strong Acid with a Strong Baseof a Strong Acid with a Strong Base

Titrate 0.1M HCl with 0.2M NaOH starting with 50 mL of HCl solution. Calculate pH when the following quantities of NaOH are added:

a. 0 mL NaOH, b. 20 mL NaOH, c. 25 mL NaOH, d. 30 mL NaOH

b. concentration of solution after adding 20mL of NaOH:

Chem 112 21MJ Bojan

Sample calculation of pH during the titration of a Strong Acid with a Strong Baseof a Strong Acid with a Strong Base

Titrate 0.1M HCl with 0.2M NaOH starting with 50 mL of HCl solution. Calculate pH when 25 mL of NaOH is added.

c. concentration of solution after adding 25mL of NaOH :

Equivalence point is where moles of acid = moles of base

Chem 112 22MJ Bojan

Page 12: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Sample calculation of pH during the titration of a Strong Acid with a Strong Baseof a Strong Acid with a Strong Base

Titrate 0.1M HCl with 0.2M NaOH starting with 50 mL of HCl solution. Calculate pH when the following quantities of NaOH are added:

a. 0 mL NaOH, b. 20 mL NaOH, c. 25 mL NaOH, d. 30 mL NaOH

d. concentration of solution after adding 30mL of NaOH:

Chem 112 23MJ Bojan

The titration of a weak acid with a strong base can be described by a plot of pH vs. amount of base added.y p p

Chem 112 24MJ Bojan

Page 13: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Calculating the pH during the Titration of a Weak Acid with a Strong Base is still a two step process.g p p

Sol’n withHA and A

neutralizationHA(aq) + OH(aq) → A(aq) + H2O() 

Determine [HA], [A]

pH determination (buffer)

Equilibrium  calculation

HA and A HA(aq)   OH (aq) → A (aq)   H2O()  [HA], [A ] (buffer)

Stoichiometry calculation

At each point below the equivalence point, the pH of the solution during titration is determined from thethe solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

Buffer solutions are encountered during titrations!titrations!

Chem 112 25

MJ Bojan

Sample calculation of pH during the titration of a Weak Acid with a Strong Base

Titrate 0.1M CH3COOH (Ka = 1.8 x 10−5) with 0.2M NaOH starting with 50 mL of CH3COOH solution and calculate pH when the following quantities of NaOH are added: a. 0 mL NaOH, b. 12.5 mL NaOH, c. 25mL d. 30 mL NaOH

a. Initial concentration of weak acid: (0mL of NaOH added)

Chem 112 26MJ Bojan

Page 14: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Sample calculation of pH during the titration of a Weak Acid with a Strong Base

Titrate 0.1M CH3COOH (Ka = 1.8 x 10−5) with 0.2M NaOH starting with 50 mL of CH3COOH solution and calculate pH when the following quantities of NaOH are added: a. 0 mL NaOH, b. 12.5 mL NaOH, c. 25mL d. 30 mL NaOH

b. Half way to equivalence (12.5mL of NaOH added)

Chem 112 27MJ Bojan

Sample calculation of pH during the titration of a Weak Acid with a Strong Base

Titrate 0.1M CH3COOH (Ka = 1.8 x 10−5) with 0.2M NaOH starting with 50 mL of CH3COOH solution and calculate pH when the following quantities of NaOHare added: a. 0 mL NaOH, b. 12.5 mL NaOH, c. 25 mL NaOH d. 30 mL

c. At the equivalence point (25mL of NaOH added)

Chem 112 28MJ Bojan

Page 15: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Sample calculation of pH during the titration of a Weak Acid with a Strong Base

Titrate 0.1M CH3COOH (Ka = 1.8 x 10−5) with 0.2M NaOH starting with 50 mL of CH3COOH solution and calculate pH when the following quantities of NaOHare added: a. 0 mL NaOH, b. 12.5 mL, c. 25mL d. 30 mL NaOH

d. Beyond the equivalence point; excess strong base (30 mL of NaOH added)

Chem 112 29MJ Bojan

The initial pH of a weak acid and pH at equivalence after titration ith t b d d th l f K f th k idwith a strong base depends on the value of Ka of the weak acid.

Chem 112 30MJ Bojan

Page 16: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

INDICATORS are organic molecules that have INDICATORS are organic molecules that have different colors for the acid and conjugate base.different colors for the acid and conjugate base.

InH+(aq) In(aq) + H+(aq)Some of the more common acid-base indicators

Name pH interval forColor change

Acid color

Base color

Methyl violet 1 - 2 Yellow Violety

Methyl yellow 1.5 - 2.5 Red Yellow

Methyl orange 2.5 - 4.0 Red Yellow

Methyl red 4 2 6 3 Red YellowMethyl red 4.2 - 6.3 Red Yellow

Bromthymol blue 6.0 - 7.6 Yellow Blue

Thymol blue 8.0 - 9.6 Yellow Blue

Phenolphthalein 8.5 - 10 Colorless Pink

Alizarin yellow G 10.1-12.0 Yellow Red

Chem 112 31MJ Bojan

Choose an indicator that will change in Choose an indicator that will change in the range of the equivalence point.the range of the equivalence point.the range of the equivalence point. the range of the equivalence point.

Chem 112 32MJ Bojan

Page 17: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Titration curve of a Weak Base with a Strong Acid shows how pH changes with bases added.p g

Strong acid Strong acid -- strong base:strong base:

The pH at the equivalenceThe pH at the equivalence point

indicator of choice is:

Strong acid Strong acid -- weak baseweak base::

The pH at the equivalence pointpoint

indicator of choice is:

Chem 112 33MJ Bojan

Summary of expectations for pH at equivalence in acid base titrations. q

Titrating a strong acid (base) with a strong base (acid)pH at equivalencepH at equivalence

Tit ti k id ith t bTitrating a weak acid with a strong base

pH at equivalence

Titrating a weak base with a strong acid

pH at equivalencepH at equivalence

Chem 112 34MJ Bojan

Page 18: Chapter 17Chapter 17 Complex Aqueous Equilibriacourses.chem.psu.edu/chem112/Summer/Lecture Notes/8_Ch17... · Chapter 17Chapter 17 Complex Aqueous Equilibria ... Consider the equilibrium

Titrations of Polyprotic Acidsyp

Chem 112 35MJ Bojan

Find the pH of this solution?Find the pH of this solution?Find the pH of this solution?Find the pH of this solution?

A buffer solution is 0.20M in HPO42− and 0.10M in H2PO4

−. What is the [H+] in this solution?

H3PO4: Ka1 = 7.5 10−3

Ka2 = 6.2 10−8

K = 1 10−12Ka3 = 1 10 12

A. 3.7 10−3 MB. 3.7 10−4 MC. 3.1 10−8 MD. 3.1 10−9 ME. 5.0 10−13 M

Chem 112 36MJ Bojan