acid-base equilibria. attack method identify what you have. identify where you are on the titration...

Acid-Base Equilibria

ATTACK METHOD• Identify what you have.• Identify where you are on the titration

curve.• When mixing acids and bases, determine

which species wins the mole war.• Make sure your answer makes sense.• Prepare by reviewing notes every night.

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.

Learning Check Strong Acid/Strong Base Titration

• What is the pH at the equivalence point for this titration?

• Label parts A, B, and C• If 30.0 ml of 1.00 M

NaOH is needed to reach the equivalence point, how many moles of acid were used?

A

B

C

Weak Acid/Strong Base Titration

Buffer RegionUse [H+

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

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

new molarity– xHydrolysis region Moles of acid = moles of base at this point

pH =pKa H+ = Ka

Weak Acid/Strong Base Titration

All strong base is converted to CB

Use Kb = X2

new molarity – x

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)

poH = pkb + log [Acid/Conj. Base]

H+ = Ka [B-] [HB]

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 .



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



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



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



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



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



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.



LEARNING CHECKWEAK ACID STRONG BASE

TITRATION• What is the pH at the

equivalence point of a weak acid/strong base titration?

• When does pKa = pH?• How do you know you have a

buffer?• What occurs at the

equivalence point?• How do you know you are at

the equivalence point?

• How do you know you are beyond the equivalence point?

• When does H+ = Ka?• Write reaction for a mixture

of HF and NaOH when enough NaOH is added to reach the equivalence point.

• What equation would you use to solve for pH if you are halfway to the stiochiometric point?

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

New molarity– x

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

Moles of acid = moles of base

pH = pKaH+ = Ka

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


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.



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



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



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



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


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



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.



LEARNING CHECKWEAK ACID STRONG BASE

TITRATION• What is the pH at the

equivalence point of a weak acid/strong base titration?

• When does pKa = pH?• How do you know you have a

buffer?• What occurs at the

equivalence point?• How do you know you are at

the equivalence point?

• How do you know you are beyond the equivalence point?

• When does H+ = Ka?• Write reaction for a mixture

of NH3 and HCl when enough HCl is added to reach the equivalence point.

• What equation would you use to solve for pH if you are halfway to the stiochiometric point?

Titration of a Polyprotic Weak Acid

Titration of a Polyprotic Strong Acid

2nd equivalence point

1st equivalence point

38

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

39

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

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

42

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

Buffers

Preparing a Buffer

Preparing a Buffer

A buffer must contain or produce a conjugate acid/base pair. Example:HA and A-

HF and F-

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

base (ex HF & NaF)4. A weak base and the salt of a weak

acid (ex NH3 & NH4Cl

Remember:

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

Buffer Solutions

ACID USES UP ADDED OH-

BASE USES UP ADDED H+

Equal moles of HF and NaF are used.

HF + H2O F- + H+

LeChatelier's Principle

Buffer SolutionsHow do they resist changes in pH?

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 R C2H3O2

- + OH- H C2H3O2 + H2O

I C E

H+ = 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?

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 solution that has [HOAc] = 0.700 M and [NaOAc-] = 0.600 M? Ka = 1.8 x 10-5

Adding an Acid to a BufferWhat 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 HINT: I would use moles.

Learning Check

30.0 ml of 0.200 M NH3 is mixed with 60.0 ml of 0.100. Determine the pH.

Learning Check

50.0 ml of 1 M HCl is mixed with 100. ml of 1 M NH3. 0.05 moles of NaOH is added to the mixture determine the pH.

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 [NaOAc-] = 0.600 M .

IS IT A BUFFER OR NOT? 50 ml of 2.00 M NaOH is titrated with 60.0 ml of 0.1 M HC2H3O2

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 Buffer SolutionBuffer prepared fromHCO3

- weak acid

CO32- conjugate base

HCO3- + H2O H3O+ + CO3

2-

Preparing a BufferPreparing a BufferYou want to buffer a solution at pH = 4.30. This means [H3O+] = 5.01 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 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 = -log 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


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

[OAc- ] (1.8 x 10-5 )

Preparing a Buffer


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).

Buffer Capacity And Buffer Range

In this course we assume the buffer system is a small system.

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

Titration Curve ForStrong Acid - Strong Base

Acid-Base Titrations/Indicators

Indicators should change color close to the equivalence point.

Acid-Base Titrations/Indicators

Phenolphthalein is used in a strong acid/strong base titration the solution turns pink upon the addition of a certain amount of base. Upon the addition of more base the solution turns colorless.

Explain

2012 AP QUESTION 1

(a) Explain how the data in the table above provide evidence that HA is a weak acid rather than a strong acid.(b) Write the balanced net-ionic equation for the reaction that occurs when the solution of NaOH is added to the solution of HA .

2012 AP QUESTION 1

(c) Calculate the number of moles of HA that were titrated.

2012 AP QUESTION 1

(d) Calculate the molar mass of HA .

2012 AP QUESTION 1

(e) Assume that the initial concentration of the HA solution (before any NaOH solution was added) is 0.200 M. Determine the pH of the initial HA solution.

2012 AP QUESTION 1

(f) Calculate the value of [H3O+] in the solution after 30.0 mL of NaOH solution is added and the total volume of the solution is 80.0 mL.

2006 AP QUESTION 1 FORM B

NET IONIC EQUATIONS ACIDS AND BASES

Frustrated Teacher

This is a great resource it alsocomes with a teachers edition with answers.

The Ultimate Chemical EquationsHandbook by George R Hague, Jr. and Jane D. Smith

Pre- AP & NET IONIC EQUATIONS

Which one(s)?• Anhydrides • Combustion (Hydrocarbons)• Decomposition (simple)• Double replacement• Redox (Single replacement)

NET IONIC EQUATIONS

ANYDRIDES (ACIDIC & BASIC)

• Solid sodium oxide is added to water

• Solid lithium oxide is added to water

ANYDRIDES (ACIDIC)• Carbon dioxide is bubbled through water

• Sulfur dioxide is bubbled through water

ANYDRIDES (SALT)

• Solid sodium oxide is treated with gaseous carbon dioxide

• Solid calcium oxide is treated with gaseous sulfur dioxide

ANYDRIDES (MISC.)

• Dilute nitric acid is added to crystals of pure calcium oxide.

• Carbon dioxide gas is bubbled through a solution of concentrated sodium hydroxide.

• Carbon dioxide gas is bubbled through a dilute solution of sodium hydrioxde.

ANYDRIDES (MISC.)

• Gaseous sulfur dioxide is bubble through a dilute solution of potassium hydroxide.

GROUP I & CBS Group II IN WATER (form basic solutions)

• Solid sodium is added to water.

• Solid lithium is added to water.

• Solid calcium is added to water

Hydroxides • Solid ammonium chloride is added to a

solution of sodium hydroxide.

• Solid barium carbonate is added to concentrated nitric acid

• Solutions of cobalt (II) nitrate and sodium hydroxide are mixed.

Trivial Redox (Single Replacement)

• Solid zinc is placed in hydrochloric acid

• Solid calcium is added to hydrochloric acid

• Concentrated hydrochloric acid is added to solid manganese(II) sulfide

Some essential principles to keep in mind: • elements in their highest positive oxidation state (same as group #, whether A or B) can ONLY be reduced • elements in their lowest oxidation state (0 for metals, negative for non-metals, corresponding to distance from noble gases) can ONLY be oxidized

• intermediate oxidation states can go either way!!!

NON-TRIVIAL REDOX ACIDIC/BASIC

Some essential principles to keep in mind: • if the mixture is acidic, H+ should be included as a reactant; water is one product

• if the mixture is basic, OH- should be included as a reactant; water is one product

• occasionally the acid anion or base cation may precipitate with a product ion

NON-TRIVIAL REDOXACIDIC/BASIC

oxidizers [remember, oxidizers will become reduced] MnO4

- (in acid) → Mn2+

MnO4- (in neutral or basic) → MnO2

MnO2 (in acid) → Mn2+

Cr2O72- (in acid) → Cr3+

HNO3 (conc.) → NO2

HNO3 (dilute) → NO

H2SO4 (hot, concentrated) → SO2

[if not hot and conc., this acts like HCl or other normal acids]

metal cations → lower charge cations or (rarely) free metals free halogens → halide ions

reducers [remember, reducers will become oxidized] halide ions → free halogens free metals → metal cations SO3

2- (or SO2) → SO42-

NO2- → NO3

-

free halogens (dil. basic) → hypohalite ions [like XO-] free halogens (conc. basic) → halate ions [like XO3

-]

metal cations → higher charge cations

Non-Trivial Redox

• A solution of potassium iodide is added to an acidified solution of potassium dichromate.

• A solution of potassium dichromate is aadddede to a solution of sodium hydroxide.

• Potassium dichromate solution is added to a solution of sodium sulfite.

NON-TRIVIAL REDOX

• A solution of potassium iodide is added to an acidified solution of potassium dichromate

• Liquid mercury is added to concentrated nitric acid

• Potassium permanganate solution is added to concentrated hydrochloric acid

NON-TRIVIAL REDOX

• Concentrated ammonia solution is added to a solution of copper (II) hydroxide.

• Excess concentrated ammonia solution is added to a suspension of silver chloride

• Excess concentrated potassium hydroxide solution is added to a precipitate of zinc hydroxide

• Silver chloride is dissolved in excess ammonia solution.

Complex Ions – Hydroxides/Ammonia

• Excess potassium hydroxide solution is added to a solution of sodium hydroxide

• Ammonia gas is bubbled into a solution of potassium hydrogen carbonate.

• Solutions of ammonia and hydrofluoric acid are mixed

• Hydrogen sulfide gas is bubbled through excess potassium hydroxide solution.

Acid/Base Neutralization

acid-base equilibria. attack method identify what you have. identify where you are on the titration...

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