Acids and BasesAcids and Bases

2003 AP 2003 AP

#1 A-E#1 A-E

Write the Write the Equilibrium, Kb, Equilibrium, Kb, for the reaction for the reaction

representedrepresented

Writing an equilibrium Writing an equilibrium expression for 1aexpression for 1a

Using the law of mass action given the Using the law of mass action given the chemical equilibrium equationchemical equilibrium equation

Concentration Products over Reactants Concentration Products over Reactants raised to their stoichiometric coefficients raised to their stoichiometric coefficients excluding pure liquids and solids!excluding pure liquids and solids!

Writing an equilibrium Writing an equilibrium expression for 1aexpression for 1a

In this case, equilibrium expression In this case, equilibrium expression consists of the products of the consists of the products of the concentrations of Conjugate acid of concentrations of Conjugate acid of Aniline and Hydroxide Ion over the Aniline and Hydroxide Ion over the concentration of Aniline. concentration of Aniline.

A Sample of aniline A Sample of aniline is dissolved in is dissolved in

water to produce water to produce 25mL of a 0.10M. 25mL of a 0.10M.

The pH of the The pH of the solution is 8.82. solution is 8.82.

Calculate the Calculate the Equilibrium Equilibrium Constant Constant

Calculate the Kb for the Calculate the Kb for the reactionreaction

Use the equilibrium Use the equilibrium expression above to expression above to determine the Kb. determine the Kb.

The concentration of The concentration of aniline is 0.1M aniline is 0.1M

The Hydroxide and The Hydroxide and Conjugate acid Conjugate acid concentration can both be concentration can both be determined after determined after calculation of the pOH (14-calculation of the pOH (14-8.82).8.82).

Take the pOH and raise it Take the pOH and raise it to the 10^(-pOH) to the 10^(-pOH)

Calculate the Kb for the Calculate the Kb for the reactionreaction

This will yield the This will yield the concentration for concentration for OH- and Aniline OH- and Aniline Conjugate because Conjugate because they are produce in they are produce in the same proportion the same proportion 1:1. 1:1.

Multiply these Multiply these concentrations and concentrations and divide them by the divide them by the initial concentration initial concentration of Aniline.of Aniline.

Calculate Kb Calculate Kb

Assume the initial Assume the initial concentration of concentration of [OH-] is negligible. [OH-] is negligible.

The reason x is not The reason x is not subtracted from the subtracted from the initial concentration initial concentration of aniline is of aniline is because x is so because x is so small that it is small that it is considered considered negligible. negligible.

The solution The solution prepared in part b prepared in part b is titrated with a is titrated with a 0.10M. Calculate 0.10M. Calculate

the pH of the the pH of the solution when solution when

5.0mL of the acid 5.0mL of the acid has been added. has been added.

Calculate pH after 5mL of Calculate pH after 5mL of HCl is added HCl is added

Set up a net chemical equation Set up a net chemical equation for the reaction of Aniline and for the reaction of Aniline and H+ (Strong Acid)H+ (Strong Acid)

Since the initial concentrations Since the initial concentrations of OH- and conjugate acid are of OH- and conjugate acid are small they are not taken into small they are not taken into account. account.

Multiply the Molarity of Aniline Multiply the Molarity of Aniline by the volume in liters (same by the volume in liters (same with the HCl) with the HCl)

From here you have the moles From here you have the moles of both Aniline and HCl of both Aniline and HCl

Calculate pH after 5mL of Calculate pH after 5mL of HCl is addedHCl is added

The H+ will combine The H+ will combine with Aniline to form a with Aniline to form a conjugate acid and goes conjugate acid and goes to completion.to completion.

Subtract the number of Subtract the number of moles of H+ from the moles of H+ from the moles of Aniline. This moles of Aniline. This gives a new number of gives a new number of moles of Aniline. moles of Aniline.

Since all of the H+ Since all of the H+ moles are consumed, it moles are consumed, it is equal to the number is equal to the number of moles of the of moles of the Conjugate acid. Conjugate acid.

Calculate pH after 5mL of Calculate pH after 5mL of HCl is addedHCl is added

To determine the pH, we To determine the pH, we will use a variation of the will use a variation of the Henderson-Hassalbauch. Henderson-Hassalbauch. Below Below

From here we take the From here we take the pKb calculated above and pKb calculated above and the mole ratio of the acid the mole ratio of the acid over the base. over the base.

The pOH can be The pOH can be determined determined

To get the pOH we To get the pOH we subtract the pOH from 14 subtract the pOH from 14 to get the pH. to get the pH.

Calculate the pH Calculate the pH at the at the

equivalence equivalence point. point.

pH at the equivalence pH at the equivalence point point

We already know the We already know the moles of the Aniline. moles of the Aniline.

The moles of Aniline The moles of Aniline must equal the moles must equal the moles of HCl for the of HCl for the equivalence point to equivalence point to be reached. (Ratio 1:1)be reached. (Ratio 1:1)

From here we have From here we have the only Aniline the only Aniline Conjugate Acid moles. Conjugate Acid moles.

We must determine We must determine the concentration of the concentration of conjugate acid and the conjugate acid and the Ka of the conjugate Ka of the conjugate

pH at the equivalence pH at the equivalence pointpoint

25mL initially of aniline25mL initially of aniline To determine the volume of HCl take the moles To determine the volume of HCl take the moles

of HCl and divide it by the concentration which of HCl and divide it by the concentration which yields the volume required. yields the volume required.

Convert all volumes to liters and add the initial Convert all volumes to liters and add the initial volume of aniline with the volume of HCl added. volume of aniline with the volume of HCl added.

Take the moles of conjugate acid and divide it by Take the moles of conjugate acid and divide it by this new volume this new volume

pH at the equivalence pH at the equivalence pointpoint

To determine the Ka dived the Kb into To determine the Ka dived the Kb into (1.0 x 10^-14).(1.0 x 10^-14).

pH at the equivalence pH at the equivalence pointpoint

Write the Chemical Write the Chemical reaction for the behavior reaction for the behavior of Aniline conjugate in of Aniline conjugate in water and make an water and make an equilibrium expression.equilibrium expression.

x is not subtracted from x is not subtracted from the initial concentration the initial concentration because it is considered because it is considered negligible. negligible.

Multiply the Multiply the concentration of concentration of Conjugate aniline Conjugate aniline concentration by the Ka. concentration by the Ka. Then take the square root Then take the square root of the product. of the product.

pH at the equivalence pH at the equivalence pointpoint

Then take –log (x) Then take –log (x) of the answerof the answer

This will give you This will give you the pH at the the pH at the equivalence pointequivalence point

pH = 2.97pH = 2.97

Which of the Which of the following following

indicators listed is indicators listed is most suitable for most suitable for

this titration. this titration.

Selecting an IndicatorSelecting an Indicator

Based on the calculations above the Based on the calculations above the pH at the end point is 2.97pH at the end point is 2.97

Erythrosine is optimal because Erythrosine is optimal because based on the color change in an based on the color change in an acidic pHacidic pH

It is a weak based titration with a It is a weak based titration with a strong acid. strong acid.

20022002

Calculate the value Calculate the value [H+] in an HOBr [H+] in an HOBr

solution that has a solution that has a pH of 4.95pH of 4.95

Calculate [H+] Calculate [H+]

Given the pH is 4.95 Given the pH is 4.95 Use the exponent 10 raised the –pH Use the exponent 10 raised the –pH This will yield the [H+] This will yield the [H+]

Write the Write the equilibrium equilibrium

constant constant expression for the expression for the ionization of HOBr ionization of HOBr

in an HOBr in an HOBr solution with a solution with a

[H+]= 1.8 x 10^(-[H+]= 1.8 x 10^(-5)5)

Equilibrium constant Equilibrium constant expressionexpression

Given the chemical Given the chemical equilibrium reaction and equilibrium reaction and the concentration of the concentration of [H+] [H+]

[H+] = [OBr-]: This is [H+] = [OBr-]: This is true because as HOBr true because as HOBr dissociates the products dissociates the products form in same form in same proportions proportions

Exclude x because it is Exclude x because it is negligible. negligible.

Use the Law of Mass Use the Law of Mass action Products over action Products over Reactants raised to their Reactants raised to their stoichiometric stoichiometric coefficients. coefficients.

Use the Use the equilibrium equilibrium constant. constant.

The product of The product of the concentration the concentration divided by the Ka divided by the Ka yields the yields the concentration of concentration of HOBr.HOBr.

Calculate the Calculate the volume of 0.115M volume of 0.115M

Ba(OH)2 needed to Ba(OH)2 needed to reach equivalence reach equivalence when titrated into when titrated into 65mL sample of 65mL sample of 0.146M of HOBr0.146M of HOBr

Calculate Volume Needed Calculate Volume Needed to Reach Equivalent Point to Reach Equivalent Point

Convert the Volume of Convert the Volume of HOBr to liters HOBr to liters

Multiply the volume in Multiply the volume in Liters by the Molarity of Liters by the Molarity of HOBr in Solution HOBr in Solution

This will give you the This will give you the moles of the HOBr in moles of the HOBr in solution solution

The titrant used is The titrant used is Ba(OH)2 (Strong Base) so Ba(OH)2 (Strong Base) so the concentration of OH- the concentration of OH- must be doubled therefore must be doubled therefore you must multiply the you must multiply the concentration of Ba(OH)2 concentration of Ba(OH)2 to get the concentration of to get the concentration of OH- OH-

Calculate Volume Needed Calculate Volume Needed to Reach Equivalent Pointto Reach Equivalent Point

Take the Moles of Take the Moles of HOBr (HOBr = HOBr (HOBr = OH-) that was OH-) that was calculated and calculated and divide it by the divide it by the molarity of OH- molarity of OH- ([Ba(OH)2] x 2). ([Ba(OH)2] x 2).

This will give you This will give you the volume in liters the volume in liters necessary to added necessary to added to reach eq point to reach eq point

Indicate whether Indicate whether the pH at the pH at

equivalence point equivalence point is less than 7, 7, or is less than 7, 7, or greater than seven. greater than seven.

ExplainExplain

What will the pH at the eq What will the pH at the eq point be? point be?

You can go through the calculations You can go through the calculations to determine the pH specifically at to determine the pH specifically at the eq point. the eq point.

The basic rule of thumb is that if you The basic rule of thumb is that if you are titrating a weak acid with a are titrating a weak acid with a strong base then the pH at the end strong base then the pH at the end point will be greater than 7. point will be greater than 7.

Calculate the Calculate the number of the number of the

moles of NaOBr(s) moles of NaOBr(s) that would have to that would have to be added to 125mL be added to 125mL of 0.160M HOBr to of 0.160M HOBr to produce a buffer produce a buffer solution with a solution with a [H+] of 5.00 x [H+] of 5.00 x

10^(-9)10^(-9)

Number of Moles needed to Number of Moles needed to produce certain produce certain Concentration Concentration

The Henderson-The Henderson-Hasselbauch Hasselbauch equation can be equation can be used for this used for this scenario (See scenario (See equation Below)equation Below)

Given the [H+] Given the [H+] take the –log([H+]) take the –log([H+]) and determine the and determine the pH for the pH for the equation equation

Number of Moles needed to Number of Moles needed to produce certain produce certain ConcentrationConcentration

You have already been given the Ka You have already been given the Ka so plug that in the equation as well so plug that in the equation as well

Separate the ratio of concentrations Separate the ratio of concentrations into log(B) – log (A) (Refer to Log into log(B) – log (A) (Refer to Log Rules)Rules)

Number of Moles needed to Number of Moles needed to produce certain produce certain ConcentrationConcentration

Solve for the Log Solve for the Log (Base) and then (Base) and then eliminate the log eliminate the log function by raising function by raising the solution using the solution using the base of 10. the base of 10.

This will yield your This will yield your concentration of concentration of BaseBase

Number of Moles needed to Number of Moles needed to produce certain produce certain ConcentrationConcentration

Multiply the concentration by the Multiply the concentration by the volume of solution in liters volume of solution in liters

This will yield the number of NaOBr This will yield the number of NaOBr moles moles

HOBr is a weaker HOBr is a weaker acid than HBrO3. acid than HBrO3. Account for this Account for this

fact. fact.

Strong Acid/Weak Acid Strong Acid/Weak Acid The number of oxygen atoms will effect the The number of oxygen atoms will effect the

strength of the acid. HOBr has a single strength of the acid. HOBr has a single oxygen atom while HBrO3 contains three. oxygen atom while HBrO3 contains three. Oxygen is a very electro negative atom and will Oxygen is a very electro negative atom and will

draw electrons away from the H-Br bond thus draw electrons away from the H-Br bond thus weakening the bond making it easier to dissociate weakening the bond making it easier to dissociate when in dissolved in water. The more oxygen when in dissolved in water. The more oxygen atoms the weaker the H-Br bond. atoms the weaker the H-Br bond.

Charge of X Charge of X The charge on the X which in this case is the Br The charge on the X which in this case is the Br

atom has a different charge in HOBr (+1) than in atom has a different charge in HOBr (+1) than in HBrO3 (+5)HBrO3 (+5)

Strength of O-H Bond Strength of O-H Bond Strength of X-O Bond Strength of X-O Bond

Arrhenius Acid Base Arrhenius Acid Base ConceptConcept

Arrhenius Acid/Base Concept Arrhenius Acid/Base Concept Acids produce Hydrogen ions (H+) within an Acids produce Hydrogen ions (H+) within an

aqueous solution aqueous solution Bases produce Hydroxide Ions (OH-) in Bases produce Hydroxide Ions (OH-) in

solution solution Definition is limited because it applies only to Definition is limited because it applies only to

acids and bases that can dissociate OH- and acids and bases that can dissociate OH- and H+ ions H+ ions

Examples: Examples: - NaOH will dissociate into Na+ and OH- NaOH will dissociate into Na+ and OH- - HCl will dissociate into H+ and Cl-HCl will dissociate into H+ and Cl-

Bronsted-Lowry Model Bronsted-Lowry Model

The model definition of Acid/BaseThe model definition of Acid/Base Bronsted Acid – A proton “donor”Bronsted Acid – A proton “donor” Bronsted Base – A proton “acceptor”Bronsted Base – A proton “acceptor” The definition applies to many more molecules The definition applies to many more molecules

that may exhibit Acid/Base qualities but do not that may exhibit Acid/Base qualities but do not directly produce OH- or H+ ions. directly produce OH- or H+ ions.

Every Acid and Base has a conjugate Acid or Every Acid and Base has a conjugate Acid or BaseBase

Water can act as an acid and a base Water can act as an acid and a base HH33O+ (Hydronium ion) (acid) and OH- (Base)O+ (Hydronium ion) (acid) and OH- (Base)

Bronsted-Lowry ModelBronsted-Lowry Model

Example of Bronsted Base Example of Bronsted Base NHNH3(aq)3(aq) + H + H22OO(l)(l) NH NH44

++(aq)(aq) + OH + OH--

(aq)(aq)

Example of Bronsted AcidExample of Bronsted Acid HCHC22HH33OO2(aq)2(aq) + H + H22OO(l)(l) H H33OO++

(aq)(aq) + C + C22HH33OO22--

(aq)(aq)

Base Acid Conjugate Acid Conjugate Base

Weak Acid Base Conjugate Acid Conjugate Base

Lewis Acid/Base Lewis Acid/Base DefinitionDefinition

Lewis Acid/Base DefinitionLewis Acid/Base Definition Lewis Acid – Electron Pair Acceptor Lewis Acid – Electron Pair Acceptor Lewis Base – Electron Pair DonorLewis Base – Electron Pair Donor Encompasses an even wider variety of Encompasses an even wider variety of

molecules (Bronsted and Arrhenius) molecules (Bronsted and Arrhenius) even ones that do not donate protons or even ones that do not donate protons or produce OH- ions. produce OH- ions.

Must be aware of the Lewis structure of Must be aware of the Lewis structure of a particular molecule to determine a particular molecule to determine whether it is a Lewis Acid or Base.whether it is a Lewis Acid or Base.

Lewis Acid/Base Lewis Acid/Base DefinitionDefinition

ExamplesExamples BF3(g) + NH3(g) BF3(g) + NH3(g) F3BNH3(g) F3BNH3(g)

BF3 is the Lewis Acid because it has no free BF3 is the Lewis Acid because it has no free unpaired electrons with only has 6 electrons unpaired electrons with only has 6 electrons around the central atom (Boron will require around the central atom (Boron will require one more pair of electrons to complete the one more pair of electrons to complete the valence shell) valence shell)

NH3 is the Lewis Base because the molecule NH3 is the Lewis Base because the molecule has a completed octet valence shell with free has a completed octet valence shell with free unpaired electrons on the central atom. The unpaired electrons on the central atom. The BF3 will accept these unpaired electrons and BF3 will accept these unpaired electrons and form a covalent bond. form a covalent bond.

Lewis Acid/Base Lewis Acid/Base DefinitionDefinition

ExampleExample Ni2+(aq) + 6NH3(aq) Ni2+(aq) + 6NH3(aq) [Ni(NH3)6]2+ [Ni(NH3)6]2+

(aq) (aq) Ni2+ is the Lewis Acid because it is a cation Ni2+ is the Lewis Acid because it is a cation

which will attract negatively charged which will attract negatively charged electrons to toward itself.electrons to toward itself.

The NH3 is the Lewis Base because it The NH3 is the Lewis Base because it provides the free unpaired electrons for the provides the free unpaired electrons for the Ni2+ Ni2+

Lewis Acid/Base Lewis Acid/Base DefinitionDefinition

Example Example Even earlier definitions are Even earlier definitions are

encompassed in the Lewis Acid Model encompassed in the Lewis Acid Model H+ + H2O H+ + H2O H3O+ H3O+