Ap chem unit 15 presentation

Acid-Base EquilibriaAP Chem Unit 15

Solutions of Acids or Bases Containing a Common Ion

Buffered Solutions Buffering Capacity Titrations and pH Curves Acid-Base Indicators

Unit 15 Content

Most chemistry in the natural world takes place in an aqueous solution. One of the most significant aqueous reactions is one with acids and bases. Most living systems are very sensitive to pH and yet are subjected to acids and bases. The main idea of this unit is to understand

the chemistry of a buffered system. Buffered systems contain chemical components that enable a solution to be resistant to change in pH.

Introduction

Solutions of Acids or Bases Containing a

Common Ion15.1

Solutions that contain a weak acid and the salt of it conjugate base create a common ion system. A common ion system can also be a weak

base and the salt of its conjugate acid.◦Example: HF and NaF solution or

NH3 and NH4Cl solution.

Common Ions

The soluble salts of conjugate acids and conjugate bases are typically strong electrolytes and therefore dissociate 100%.

This dissociation increases the concentration of the conjugate ion and a shift in equilibrium occurs according to Le Chatelier.◦Example:

Shift Left!

Common Ions

The shift left in the second equation is known as the common ion effect.

This effect makes a solution of NH3 and NH4Cl less basic than NH3 alone

Common Ion Effect

The shift left in the second equation is known as the common ion effect.

This effect makes a solution of NaF and HF less acidic than a solution of HF alone.

Common Ion Effect

Calculations for common ion equilibria is similar to weak acid calculations except that the initial concentrations of the anion are not 0. Initial concentrations for the equilibrium

calculation must include the concentration of ions from the complete dissociation of the salt.

Common Ion Effect

The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF.

Major species?ICE:

Practice Problem #1

I C E

HF

F-

H+

The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF.

ANS: .072%

Practice Problem #1

Buffered Solutions15.2

The most important application of a common ion system is buffering. A buffered solution resists a change in

its pH when either hydroxide ions or protons are added.◦Blood is a practical example of a buffered

solution; it is buffered with carbonic acid and the bicarbonate ion (among others).

Buffered Solutions

A solution can be buffered at any pH by choosing the appropriate components.

Smaller amounts of H+ or OH- can be added with little pH effect.

Buffered Solutions

A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium acetate. Calculate the pH of this solution.

pH = 4.74

Practice Problem #2

When OH- ions are added to a solution of a weak acid, the OH- ions react with the best source of H+ (weak acid) to make water:

The net result is that OH- ions are replaced by A- ions and water until they are all consumed.

How Does Buffering Work?

When H+ ions are added to a solution of a weak acid, the H+ ions react with the strong conjugate base.

The net result is that more weak acid is produced, but free H+ do not accumulate to make large changes in the pH.

How Does Buffering Work?

Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14. When a strong acid or base is added to a

buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilbrium calculations.

Buffered Solution Summary

Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to the buffered solution in #2 (0.50 M acetic and 0.50 M sodium acetate). Compare this pH change with the original (pH=4.74) and the pH that occurs with 0.010 mol of solid NaOH is added to 1.0 L of water.

Practice Problem #3

Major species:

OH- +HC2H3O2 C2H3O2- + H2O

mole table:

Practice Problem #3

HC2H3O2 OH- C2H3O2- H2O

Before

After

ICE table:

pH= 4.76 (+.02 off original)pH of just OH-=12.00

Practice Problem #3

I C E

HC2H3O2

OH-

C2H3O2-

Buffered solutions work well as long as the concentration of the salts and weak acids/bases in solution are much larger than the amount of OH- and H+ being added. The amount of [H+] being in a buffered

solution is often solved using a rearrangement of the equilibrium expression:

Henderson-Hasselbalch Equation

This equation, , can be changed into another

useful form by taking the negative log of both sides:

This log form of the expression is called the Henderson-Hasselbalch Equation.


Henderson-Hasselbalch Equation: When using this equation it is often assumed that the the equilibrium concentrations of A_ and HA are equal to their initial concentrations due to the validity of most approximations.

Since the initial concentrations of HA and A-

are relatively large in a buffered solution, this assumption is generally acceptable.


Calculate the pH of a solution containing 0.75 M lactic acid (Ka=1.4x10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion.

pH = 3.38

Practice Problem #4

Buffered solutions can be formed from a weak base and the corresponding conjugate acid. In these solutions, the weak base B reacts with any H+ added:

The conjugate acid BH+ reacts with any added OH-:

Reminder

A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.

pH=9.05

Practice Problem #5

Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from #5 (.25M NH3 and .40 NH4Cl).

pH= 8.73

Practice Problem #6

1. Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+.

2. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base:

Summary

3. When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present:

4. 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 concentration of the buffering materials (HA and A- or B and BH+) are large compared with the

amounts of H+ or OH- added.

Summary

Buffering Capacity15.3

The buffering capacity of a buffered solution represents the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. The pH of a buffered solution is determined

by the ratio [A-] / [HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].

Buffering Capacity

Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions:

◦ Soln A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2

◦ Soln B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

Ka=1.8x10-5.

Soln A: pH=4.74 Soln B: pH=4.56

Practice Problem #7

The optimal buffering occurs when [HA] is equal to [A-]. This ratio of 1 resists the most amount of

pH change. The pKa of the weak acid to be used in the

buffer should be as close as possible to the desired pH.◦ For example, if a buffered solution is needed with

a pH of 4.0. The most effect buffering will occur when [HA] is equal to [A-] and the pKa of the acid is close to 4.0 or Ka=1.0x10-4.

Optimal Buffering

A chemist needs a solution buffered at pH=4.30 and can choose from the following acids and their sodium salts. Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30. Which system will work best?a. chloroacetic acid (Ka=1.35x10-3)

b. propanoic acid (Ka=1.3x10-5)

c. benzoic acid (Ka=6.4x10-5)

d. hypochlorous acid (Ka=3.5x10-8)

benzoic acid

Practice Problem #8

Titrations and pH Curves

15.4

An acid-base titration is often graphed by plottingthe pH of the solution (y-axis) vs. the amount of titrant added (x-axis). Point (s) is the equivalencepoint/stoichiometric point.This is where [OH-]=[H+].

Titration Curve

The net ionic reaction for a strong acid- strong base titration is:

To compute [H+] at any point in a titration, the moles of H+ that remains at a given point must be divided by the total volume of the solution. Titrations often include small amounts. The

mole is usually very large in comparison. Millimole (mmol) is often used for titration

amounts.

Strong Acid-Strong Base Titrations

Molarity with mmols

50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. a. NaOH has not been added.

Since HNO3 is a strong acid (complete dissociation), pH = 0.699

Strong Acid-Base Titration Example Calculation

50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. b. 10.0 ml of 0.100 M NaOH has been added.

H+ = 10 mmol – 1.0 mmol of OH- = 9.0mmol

9.0 mmol H+/ 60 ml = 0.15 MpH=0.82


50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. c. 20.0 mL of 0.100 M NaOH has been added.

pH=0.942


50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. d. 50.0 mL of 0.100 M NaOH has been added.

pH=1.301


50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. e. 100.0 mL of 0.100 M NaOH has been added.

The stoichiometric point/equivalence point is reached. pH=7


50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. f. 150.0 mL of 0.100 M NaOH has been added.

pH = 12.40


50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. g. 200.0 mL of 0.100 M NaOH has been added.

pH=12.40


The results of these calculations are summarized by this graph:

The pH changes very gradually until the titration is close to the equivalence point, then dramatic change occurs. Near the E.P., small changes produces large changes in the OH-/H+ ratio.


1. Before the equivalence point, [H+] can be calculated by dividing the number of millimoles of H+ remaining by the total volume of the solution in mL.

2. At the equivalence point, pH=7.03. After the equivalence point, [OH-] can be

calculated by dividing the number of millimoles of excess OH- by the total volume of the solution. [H+] is calculated from Kw.

Strong Acid-Base Summary

When acids don’t completely dissociate, calculations used previously need adjusted. To calculate [H+] after a certain amount of

strong base has been added, the weak acid dissociation equilibrium must be used.

Remember: that a strong base reacts to completion with a weak acid.

Titrations of Weak Acids with Strong Bases

Calculating the pH Curve for a Weak Acid-Strong Base Titration is broken down into two steps:1. A stoichiometric problem. The reaction of

hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined.

2. An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.

Calculating the pH Curve for a Weak Acid-Strong Base Titration

50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.

a. NaOH has not been added.

pH = 2.87

Weak Acid-Strong Base Titration Example Calculation


b. 10.0 mL of 0.10 M NaOH has been added.

pH= 4.14



c. 25.0 mL of 0.10 M NaOH has been added.

pH=4.74



d. 40.0 mL of 0.10 M of NaOH has been added.

pH=5.35



e. 50.0 mL of NaOH is added.

pH=8.72



f. 60.0 mL of 0.10 M NaOH is added.

pH = 11.96



g. 75.0 mL of 0.10 M NaOH is added.

pH = 12.30


Some of the differences in the the titration curves: The pH increases more rapidly in the

beginning of the weak acid titration.

Weak Acid-Strong Base Titration

Some of the differences in the the titration curves: The titration curve levels off near the

halfway point due to buffering effects.◦ Buffering happens when [HA]=[A-]. This is exactly

the halfway point of the titration.


Some of the differences in the the titration curves: The value of the pH at the equivalence point

is not 7. The value of the e.p. is greater than 7.


Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka=6.2x10-10) when dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution.a. after 8.00 mL of 0.100 M NaOH has been

added.b. at the halfway point of the titrationc. at the equivalence point of the titration

a) pH=8.49 b) pH=9.21 c)pH=10.96

Practice Problem #9

When comparing the example calculation and the practice problem with weak acids, two major conclusions can be made:1. The same amount of 0.10 M NaOH is

required to reach the equivalence point even though HCN is a much weaker acid.

◦ It is the amount of acid, not its strength that determines the equivalence point.

Important Conclusions

When comparing the example calculation and the practice problem with weak acids, two major conclusions can be made:2. The pH value at the equivalence point is

affected by the acid stregth. The pH at the e.p. for acetic acid is 8.72. The pH at the e.p. for hydrocyanic acid is 10.96.

◦ The CN- ion is a much stronger base than C2H3O2-.

The smaller acid strength and stronger conjugate base produces a higher pH.

Important Conclusions

The strength of a weak acid has a significant effect on the shape of its pH curve. The e.p. occurs at the same point, but the shapes of the curve is dramatically different.

Weak Acid Titration Curves

A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value. To do so, the chemist dissolves 2.00 mmol of the solid acid in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid?

Ka=1.0x10-6

Practice Problem #10

Find the pH of a solution of 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HCl (Kb=1.8x10-

5).a. HCl has not been added.

pH is found with Kb equilibrium. pH = 10.96

Titration of Weak Bases with Strong Acids Example Calculation


5).b. 10.0 mL of HCl is added.

H+ ions react to completion. pH=9.85



5).c. 25.0 mL of HCl is added.

pH = 9.25



5).d. 50.0 mL of HCl is added.

pH = 5.36



5).e. 60.0 mL of HCl is added.

pH= 2.21


The Titration Curve for a Weak Base with a Strong Acid

Acid-Base Indicators15.5

There are two common methods for determining the equivalence point of a titration:1. Use a pH meter to monitor the pH and then

plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point.

2. Use an acid-base indicator, which marks the end point of a titration by changing color.

Acid-Base Indicators

The equivalence point of a titration is defined by the stoichiometry, but it is not necessarily the same as the end point where the indicator changes color). Selection of the right indicator for the

titration is very important.

Acid-Base Indicators

The most common acid-base indicators are complex molecules that are weak acids (HIn). Most exhibit one color when the proton is

attached to the molecule and a different color when the proton is absent.◦ Example: Phenolphthalein is colorless in its HIn form and pink in its In-, or basic form.

Indicators

(RED) (BLUE)

The color will depend on the ratio of [In-] to [HIn]. For most indicators, approximately 1/10th of the initial form must be converted to the final form before a color is perceived by the human eye. the color change will occur at a pH when

Indicator Example

Bromthymol blue, an indicator with Ka=1.0x10-7, is yellow in its HIn form and blue in its In- form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible?

pH=6.0

Practice Problem #11

The H-H equation is very useful in determining the pH at which an indicator changes color.

Example:Bromthymol blue (Ka=1x10-7, or pKa=7), the pH at the color change is pH=7-1=6

Henderson-Hasselbalch and Indicators

When a basic solution is titrated, the indicator will initially exist as In- in solution, but as acid is added more HIn is formed. Color change will occur at:

Substituting this reciprocal into the H-H equation gives us:

◦Bromthymol blue=pH 7+1=8;The useful range of bromthymol blue is pH(6-8)

Indicator Titrations

Indicator Ranges

THE END

Education

Ap chem unit 15 presentation