HOMEWORK Do in this order 51,52,53,55,61,65,67,23,25,27,29,31,33, 37a&e, 39,41,43a,45,47, and...

HOMEWORK• Do in this order • 51,52,53,55,61,65,67,23,25,27,29,31,33, 37a&e, 39,41,43a,45,47, and 49a&b

Acid-Base Equilibria

REMEMBERWhen working problems in this chapter, it will

help if you always……..

Strong Acid/Strong Base Titration

Excess Base

Excess Acid

moles of acid =moles of base

WHAT IS THE DIFFERENCE BETWEEN END POINT AND

EQUIVALENCE POINT?

Strong Acid/Strong Base Titration•Determine the pH if 50 ml of 0.25 M HCl is mixedwith 30 ml of 0.25 M NaOH

Strong Acid/Strong Base TitrationDetermine the pH when 30.0 ml of 0.500 M HClis added to 60 ml of 0.250 M NaOH.

Strong Acid/Strong Base TitrationDetermine the pH when 25.0 ml of 0.400 M HClis added to 85.0 ml of 0.300 M NaOH.

Strong Acid/Strong Base Titrationa. Determine the pH if enough 0.700 M NaOH is

added to 30.00 ml of .500 M HCl to reach the equivalence point.

b. Determine the volume of NaOH required to reach the equivalence point.

Weak Acid/Strong Base Titration

All strong base is converted to CB

Use Kb = X2

moles base – x total volume

Weak Acid/Strong Base Titration

Buffer RegionUse [H+

] = Ka [ moles acid ] [ moles base]You are in in the buffer region when the moles of acid are greater than base.

Excess base regionMoles of base is greater than acid

Determine the moles ofstrong base left over . Divide by total volume gives [OH-

] concentration.

All strong base is converted to CBat the equivalence point

Use Kb = X2

moles base – x total volume

HENDERSON-HASSELBALCH EQUATION (Buffers)

pH pKa + log[Conj. base]

[Acid]

Derivative is

H+ = Ka [HB] [B- ]

If you have a buffer , use the aboveequation. Makes the math easier.

HENDERSON-HASSELBALCH EQUATION (Buffers)

pH pKa + log[Conj. base]

[Acid]Derivative ispKa = -log10Ka

pH = pka when moles of acid = moles of base

H+ = Ka [HB] [B- ]

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH.1. Before the addition of any base (initial pH).2 After the addition of 8.00 mL of 0.500 M NaOH (buffer region).3. After the addition of 10.00 mL of 0.500 M NaOH

(half-neutralization point).4. After the addition of 20.00 mL of 0.500 M NaOH

(equivalence point).5. After the addition of 21.00 mL of 0.500 M NaOH

(beyond the equivalence point).

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH.

1. Before the addition of any base .

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH 2. after the addition of 8.00 mL of 0.500 M NaOH

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH 3. after the addition of 10.00 mL of 0.500 M NaOH

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH 4. after the addition of 20.00 mL of 0.500 M NaOH

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH 5. After the addition of 21.00 mL of 0.500 M NaOH

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Determine the pH of a solution when 25.0 ml of 2.00 M HC2H3O2 is titrated halfway to the equivalence point with NaOH

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Determine the pH of a solution when 25.0 ml of 2.00 M HC2H3O2 is titrated with 1.50 M NaOH to the equivalence point,

Neutralization ReactionWeak Acid - Strong Base

5 Positions on the Titration Curve

Buffer Region

Excess AcidRegion

Use [H+ ] = Ka [ moles acid ]

[ moles base]

You are in in the buffer region when the moles of weak is greater than strong

All strong acid is converted to CA ate the equivalence point Use

Ka = X2

moles base – x total volume

Moles of strong are greater than weak Determine the moles of acid left over then divide by total volume. Gives H+ concentration

WEAK BASE STRONG ACID TITRATION

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia with 0.500 M HCl1. Before the addition of any acid (initial pH).2 After the addition of 8.00 mL of 0.500 M HCl (buffer region).3. After the addition of 10.00 mL of 0.500 M HCl

(half-neutralization point).4. After the addition of 20.00 mL of 0.500 M HCl

(equivalence point).5. After the addition of 21.00 mL of 0.500 M HCl

(beyond the equivalence point).

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 1. Before the addition of any acid.

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 2. After the addition of 8.00 mL of 0.500 M HCl

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 3. After the addition of 10.00 mL of 0.500 M HCl

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

4. After the addition of 20.00 mL of 0.500 M HCl

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia

5. After the addition of 21.00 mL of 0.500 M HCl

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration CurveCalculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia

Determine the pH of a solution when 25.0 ml of 2.00 M NH3 is titrated halfway to the equivalence point with HCl

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

Determine the pH of a solution when 25.0 ml of 2.00 M NH3 is titrated to the equivalence point with 0.125 M HCl.

Neutralization ReactionWeak Base - Strong Acid

5 Positions on the Titration Curve

Titration of a Polyprotic Weak Acid

Titration of a Polyprotic Strong Acid

2nd equivalence point

1st equivalence point

37

pH Review

• pH = - log [H+]• H+ is really a proton• Range is from 0 - 14• If [H+] is high, the solution is acidic; pH < 7• If [H+] is low, the solution is basic or alkaline ;

pH > 7

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The Body and pH

• Homeostasis of pH is tightly controlled• Extracellular fluid = 7.4• Blood = 7.35 – 7.45• < 6.8 or > 8.0 death occurs• Acidosis (acidemia) below 7.35• Alkalosis (alkalemia) above 7.45

39

40

Small changes in pH can produce major disturbances

• Most enzymes function only with narrow pH ranges

• Acid-base balance can also affect electrolytes (Na+, K+, Cl-)

• Can also affect hormones

41

Acidosis• Principal effect of acidosis is depression of the CNS

through ↓ in synaptic transmission.• Generalized weakness• Deranged CNS function the greatest threat• Severe acidosis causes

– Disorientation– coma – death

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Alkalosis• Alkalosis causes over excitability of the central and

peripheral nervous systems.• Numbness• Lightheadedness• It can cause :

– Nervousness– muscle spasms or tetany – Convulsions – Loss of consciousness– Death

43

The function of a buffer is to resist changes in the pH of a solution.

Buffers are just a special case of the common ion effect. Buffer CompositionWeak Acid + Conj. BaseHOAc + OAc-

H2PO4- + HPO4

2-

Weak Base + Conj. AcidNH3 + NH4

+

Buffer Solutions

ACID USES UP ADDED OH-

CONJ. BASE USES UP ADDED H+

OAc- + H2O HOAc + OH-

LeChatelier's Principle

NOTE: THE CONJUGATE ACID/BASE PAIRS ARE ON OPPOSITE SIDES OF THE EQUATION.

Buffer Solutions

Preparing a BufferPreparing a BufferA buffer can be prepared by 1. Using a weak acid and ½ the moles of a strong

base or less2. Using a weak base and ½ the moles of a strong

acid or less3. A weak acid and the salt of a weak base4. A weak base and the salt of a weak acid

Preparing a BufferPreparing a Buffer Using a weak acid and ½ the moles of a strong

base or less

Why does this work? 30.0 ml of 0.200 M acetic acid and 15.0 ml of 0.200 M sodium hydroxide C2H3O2

- + OH- H C2H3O2 + H2O

I C E

Preparing a BufferPreparing a Buffer Using a weak base and ½ the moles of a strong

base or less

Why does this work? 30.0 ml of 0.200 M ammonia and 15.0 ml

of 0.200 M HCl NH3 + H2O NH4

+ + H2O

I C E

Derivative isH+ = Ka [HB] [B- ] If you have a buffer , use the aboveequation. Makes the math easier.

You have a buffer if have a conjugate acid base pair or the moles of weak (acid/base)is greater than the strong (acid/base)

Buffer SolutionspH pKa + log

[Conj. base][Acid]

Buffer SolutionsWhich of the following mixtures would result in a buffered solution when 1.0 L of eachof the two solutions are mixed?

a. 0.2 M HNO3 and 0.4 M NaNO3 b. 0.2 M HNO3 and 0.4 M HFc. 0.2 M HNO3 and 0.4 M NaFd. 0.2 M HNO3 and 0.4 M NaOH

H+ = Ka [HB] [B- ]

Buffer SolutionsHow does adding water to a buffer system affect the pH?

H+ = Ka [HB] [B- ]

Buffer SolutionsHow does adding water to a buffer system affect the pH?

CONCENTRATION of the acid and conjugate base are not important.It is the RATIO OF THE NUMBER OF MOLES of each. The volumes cancel. You can ignore the volume and use

H+ = Ka HB moles B- moles

Buffer SolutionsProblem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? Ka = 1.8 x 10-5

Adding an Acid to a BufferProblem: What is the pH when 1.00 mL of 1.00 M HCl

is added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M

Adding an Base to a Buffer What is the pH when 1.00 mL of 1.00 M NaOH is

added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M .

Buffer SolutionsHow many moles of HCl must be added to 1.0 L of 1.0 M sodium acetate to producea solution buffered ata. pH= pKab. pH =4.20

Buffer SolutionsConsider a solution that contains both ammonia and ammonium chloride. Calculate the ratio of [NH3 ]/[NH4 Cl ] at a pH of 5.23.

Preparing a BufferPreparing a BufferYou want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M

It is best to choose an acid such that [H3O+] is about equal to Ka (or pH pKa).

—then you get the exact [H3O+] by adjusting the ratio of acid to conjugate base.

Preparing a Buffer SolutionBuffer prepared fromHCO3

- weak acid

CO32- conjugate base

HCO3- + H2O H3O+ + CO3

2-

Preparing a BufferYou want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M

POSSIBLE ACIDS Ka

HSO4- / SO4

2- 1.2 x 10-2

HOAc / OAc- 1.8 x 10-5

HCN / CN- 4.0 x 10-10

Best choice is acetic acid / acetate.

Remember pKa = -log10 Ka

pH = pKa when moles acid = moles of base

You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M

Preparing a Buffer

You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M

[H3O] 5.0 x 10-5 = [HOAc]

[OAc- ] (1.8 x 10-5 )

Preparing a Buffer

You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M

Solve for [HOAc]/[OAc-] ratio = 2.78/ 1Therefore, if you use 0.100 mol of NaOAc and 0.278

mol of HOAc, you will have pH = 4.30.

[H3O] 5.0 x 10-5 = [HOAc]

[OAc- ] (1.8 x 10-5 )

Preparing a Buffer

Buffer capacity

The buffer capacity is the measure of the

amount of acid or base that the solution can

absorb without significant change in pH

Depend on how many moles of the conjugate

acid-base pair are present

- for solutions having the same concentration,

the greater the volume greater buffer

capacity

Buffer capacity

To increase the buffering capacity of a buffer

system, increase the moles of acid and moles of base

• CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each.

Buffer Capacity And Buffer Range

• There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed.

• In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize.

• As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal (pH=pKa).

Acid-Base Indicators

• An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless.

HIn + H2O º H3O+ + In-

color 1 color 2• Acid-base indicators are often used for

applications in which a precise pH reading isn’t necessary.

• A common indicator used in introductory chemistry laboratories is litmus.

17-3 Acid-Base Indicators

• Color of some substances depends on the pH.

HIn + H2O In- + H3O+

In the acid form the color appears to be the acid color.In the base form the color appears to be the base color.Intermediate color is seen in between these two states.The complete color change occurs over about 2 pH units.

Remember….LeChatelier’s Principle

Slide 69 of 45

Indicator Colors and Ranges

Slide 27 of 45 General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration Curve ForStrong Acid - Strong Base

Acid-Base Titrations

Indicators should change color close to the equivalence point.