Unit 17 Acids and BasesChapter 14

Arrhenius ModelAcids produce H+ in aqueous solution while bases produce OH-

Bronsted-Lowry ModelAcid is a proton (H+) donor. Base is proton acceptor

Show how water can act as an acid or a baseHydronium ion- H3O+

Section 14.1: The Nature of Acids and Bases

Conjugate acid/base pair- two substances related to each other by the donating and accepting of a single proton

• Think think of the reaction between an acid and a base as competition for the H+ between the two bases (the original base and the CB)

Acid Dissociation in Water: HA + H2O ↔ H3O+ + A- or

HA + H2O ↔ H+ + A-

Acid dissociation constant (Ka): Ka = [H3O+][A-]/[HA] or Ka = [H+][A-]/[HA]

*Remember that pure solids or pure liquids are not included in equilibrium expressions- this is the reason that water is not present in the equation

*Although water is not included in the equilibrium expression, it does play a role in ionization of the acid

Example: pg. 673 (#28 both CB/CA and Ka)

Strong acid- almost all the original HA is dissociated at equilibrium (→)

• strong acid yields a weak CB• The CB is much weaker than water for the affinity for H+

HA + H2O→H3O+ + A-

• The equilibrium lies far to the right for a strong acid

• Ka is large because the equilibrium lies to the right (product favored)

•All [HA]o = [H+]eq and [HA]eq=0 because we assume 100% dissociation

Section 14.2: Acid Strength

Weak acid- acid for which very little dissociates (↔ or ←)

• weak acid yields a strong CBThe CB is much stronger than water for the affinity for H+

HA + H2O←H3O+ + A-

• The equilibrium lies far to the left for a weak acid

• Ka is small because the equilibrium lies to the left (reactant favored)

• All [HA]eq > [H+]eq because we assume little dissociation

Diprotic acid- acid that has 2 acidic protonsH2SO4 + H2O → HSO4

-1 + H3O+ (strong for 1st H+)

HSO4-1 + H2O ↔ SO4

2- + H3O+ (weak acid-HSO4+)

Monoprotic acids- acids with one proton

Oxyacids- acidic proton is attached to an oxygen atom- most acids (HNO3, H3PO4, etc)

Organic Acids- acids with a carbon atom backbone• Contain carboxyl group (COOH)- this is the acidic proton• usually these acids are weak- acetic acid

Amphoteric- substances that behave either as acids or as bases (H2O and NH3 are the most common)

Autoionization- transfer of a proton from one water molecule to another to produce a hydroxide ion and a hydronium ion

H2O + H2O H3O+ + OH-

NH3 + NH3 ↔ NH4+ + NH2

-

Ion-product constant (dissociation constant or equilibrium constant for water)

Kw = [H+][OH-] = 1.0 x 10-14 @ 25°C(WEAK acid b/c small Ka)

In ANY aqueous solution at 25 degrees Celsius, no matter what it contains, the value of Kw is always 1.0 x 10-14

There are three possibilities:

Neutral solution where [H+] = [OH-]

Acidic Solution where [H+] > [OH-]

Basic Solution where [H+] < [OH-]

1.0 x 10-14 = [H+] [OH-]

Reminder: You can compare the strengths of acids based on their Ka

values!!

Examples: pgs. 673-674 (#32, 36, 38)

HW: 14.1 (pg. 673 # 27, 29) and 14.2 (pg. 674 #33-37 odd)

Sig Figs for logarithms- the number of decimal places in the log is equal to the number of significant figures in the original number

Ex. [H+]= 1.0 x 10-9 (2 sig figs) then the log[H+]=2.00 (2 decimals)

Equations: pH = -log[H+] pOH = -log[OH-] the pH changes by 1 for every power of 10 change Also, the pH decreases as [H+] increases

pK = -log K Kw = [H+] [OH-]

[H+] = 10-pH [OH-] = 10-pOH

pH + pOH = 14 (at 25°C)

Example pg. 674 (#40)HW pg. 674 (#39-45 odd)

Section 14.3: The pH Scale

2 Things to Focus on for Acid-Base Calculations-solution components and their chemistry.

-which components are significant and which can be ignored

Major species- solution components present in relatively large amounts-For strong acids, it is IONS and WATER (because the acid has dissociated)

If the [Strong Acid] > 1.0 x 10-7, then [H+] from autoionization of water can be ignored because the [H+] produced is negligible.

- so 0.10 M HCl = [H+] = 0.10 M, therefore the pH = -log (0.10) = 1.00

If the [Strong Acid] < 1.0 x 10-7, then [H+] from the strong acid can be ignored because the amount dissociated is negligible.

-so [H+] = 1.0 x 10-7 (from water) and pH = 7.00

Example pg. 674 (#48, 50, 52)HW pg. 674-675 (#47-51 odd)

Section 14.4: Calculating the pH of Strong Acid Solutions

What are the steps for calculating the pH of a weak acid solution? • Find the major species in the solution.

• Determine which of the major species supplies the hydrogen ions. • IF Ka> Kw, the H+ comes from the weak acid

• List the initial concentrations. • Initials for the IONS will usually be 0

• Let x be the change in the concentration of the substance required to reach equilibrium.

• “-x” for WEAK ACID and “+x” for the IONS • Assume “-x” is negligible for the weak acid

• Solve• “x” will usually be the amount of H+ dissociated

• Use the 5% rule to verify whether approximations are valid

• x/[HA]o * 100% < 5% (if not, you can’t assume “-x” is negligible for the weak acid

Section 14.5: Calculating the pH of Weak Acid Solutions

Percent dissociation- also percent ionizationamount dissociated initial concentration x 100%

What happens to percent dissociation when a weak acid is diluted?

the percent dissociation increases

When water is added, Q is less than Ka, so the system must adjust to the right to reach the new equilibrium position, so percent dissociation increases when the acid is diluted.

Example: What is the pH of 1.00 M HF? What is the percent dissociation?

HW: pg. 675 (#53, 55, 59-69 odd)

Strong bases- complete dissociation of OH compounds (100% soluble)

- Group 1, calcium, strontium, and barium -KOH → K+ + OH- -Ca(OH)2 → Ca2+ + OH-

Many bases do not contain the hydroxide ion. How do these substances act as bases?

they increase the concentration of hydroxide ion because of their reaction with water.

NH3 +H2O ↔ NH4+ + OH-

What do many of these bases without hydroxide ions have in common?

A lone pair of electrons is located on a nitrogen atom in many cases

Section 14.6: Bases

Equations: Kb = [BH+][OH-] / [B]

Kb x Ka = Kw

HW: pg 675-676 (#71-83 odd, 87-91 odd)

Kb – refers to the reaction of a base with water to form the conjugate acid and the hydroxide ion

Weak bases- bases with small Kb values- where the base competes with the OH- for the H+ ion

- B + HOH ↔ BH+ + OH-

-Calculations are similar to those for acids

Polyprotic acids- acids that furnish more than one proton- dissociates in a stepwise manner- one proton at a time

Triprotic acids- acids with 3 protons- like phosphoric acid.

Describe the relative sizes of dissociation constants for polyprotic acids. (Ka1 > Ka2> Ka3) (see pg. 650-651)

This means that the first proton is easiest to remove and that it is more difficult to remove successive protons- this happens because the negative charge on the acid increases, making it more difficult to remove a positively charged proton

only the first dissociation step makes an important contribution to [H+] this simplifies pH calculations

Section 14.7 Polyprotic Acids

Why is sulfuric acid unique among strong acids?

It is the only diprotic strong acid - its first dissociation step and a weak acid in its second step.

General Rules:

For concentrated solutions, 1M or higher, only use Ka1 for calculations.

For dilute concentrations- the second step does make a significant contribution and the quadratic equation needs to be used

-Use all K values for each successive step

HW: pg. 676 (# 93-97)

Salt- an ionic compound

What type of salts produce neutral solutions when dissolved in water?

-salts that consist of the cations of strong bases and the anions of strong acids

-KCl, NaNO3, NaCl, KNO3

Section 14.8: Acid-Base Properties of Salts

A salt that consist of a cation of a strong base and the anion of a weak acid

A solution of sodium acetate contains sodium and acetate and water.

NaC2H3O2 (aq) + H2O (l) → NaOH (aq) + HC2H3O2 (aq) Weak Acid

Na+ + C2H3O2- + H2O (l) → Na+ + OH- + HC2H3O2 (aq)

C2H3O2- + H2O (l) → OH- + HC2H3O2 (aq)

BASIC

What kind of salt yields a basic solution?

A salt that consist of a cation of a weak base and the anion of a strong acid

NH4Cl (aq) + H2O (l) → NH4OH (aq) + HCl (aq)

NH4+ + Cl- + H2O (l) → NH3(g) + HOH (l) + H+ + Cl-

NH4+ → NH3(g) + H+

Acidic

Salts with highly charged metal ions also produce acidic solutions. (Al+3

, W+6, etc)

The high charge on the metal ion polarizes the O-H bonds in the attached water molecules, making the hydrogens in these water molecules more acidic than those in free water molecules. The higher the charge on the metal ion, the stronger the acidity of the hydrated ion.

What kind of salt yields a acidic solution?

How do you determine whether a solution will be acidic, basic, or neutral when there is a salt made up of 2 ions that can affect the pH of aqueous solutions?

Compare the Ka value for the acidic ion with the Kb value for the basic ion.

If the Ka > Kb, the solution will be acidic.

If the Kb > Ka value, the solution will be basic.

Equal Ka and Kb values mean that the solution is neutral Equations: Ka x Kb =Kw

Example: pg. 676 (# 105)HW: pg. 676 (#99, 101, 109, 111 (ab, de))

What are the two main factors that determine whether a molecule containing a bond between an atom and hydrogen will behave as a Bronsted-Lowry acid?

1. strength of the bond (weak bonds, more acidic)2. polarity of the bond (more polar, more acidic)

C-H bonds are strong and nonpolar, so there is no tendency to donate protons.

What is the trend in strength of oxyacids?

-The strength of the acid grows as the # of oxygen atoms increases.

-This occurs because very electronegative oxygen atoms are used to draw electrons away , which polarizes and weakens the O-H bond

-A proton is easiest to remove from a molecule with the largest number of attached oxygen atoms.

-This is the same reason that hydrated metal ions become acidic.

-The greater the electronegativity of the attached ion, the stronger the acid because the EN strong atom pulls electrons away from others and weakens bonds.

Section 14.9: The effect of Structure on Acid-Base Properties

Substances that contain H-O-X bonds can behave as acids or bases.

What determines which behavior will occur?

-If X has a high electronegativity, the O-X bond will be covalent and strong, meaning that a proton will be released.

-If X has a low electronegativity, the O-X bond will be ionic and can be broken in polar water and OH- is released

Acidic oxides: when a covalent oxide dissolves in water, an acidic solution forms

SO3 + H2O → H2SO4

Basic oxides: when an ionic oxide (metal oxide) dissolves in water and a basic solution results

CaO + H2O → Ca(OH)2

HW: pg.677 (113-118)

Section 14.10: Acid-Base Properties of Oxides

Table 14.9 Comparison of Electronegativity of X and Ka Value for a Series of Oxyacids

Lewis acid- electron-pair acceptor OR it has an empty atomic orbital that it can use to accept or share an electron pair from a molecule that has a lone pair of electrons (usually BORON)

Lewis base- electron-pair donor OR molecule that has a lone pair of electrons (usually oxygen, nitrogen, or fluorine)

Lewis model for acids and bases is beneficial because it covers many reactions that don’t involve Bronsted-Lowry aids

HW: pg. 677 (#119, 123)

Section 14.11: The Lewis Acid-Base Model

Acid-Base Equilibria

Sections 15.1-15.5

(Unit 18A)

15.1 Common Ion Effect

Suppose we have a solution containing the weak acid hydrofluoric acid (Ka=7.2x10-4) and its salt NaF. Remember that when a strong electrolyte dissolves, it completely dissociates.

– What is the common ion?– How does this ion effect pH?

HF (aq)↔ H+ (aq) +F- (aq)

Added F- ions from NaF

Equilibrium shift. Form more HF.

Fewer H+ ions present.Less acidic.

NH3 (aq) + H2O↔ NH4+ (aq) +OH- (aq)

Added NH4+ ions

from NH4ClEquilibrium shift.

Form more NH3 and H2O. Fewer OH- ions present.

Less basic.

Equilibrium Calculations

• Use the same procedures for finding pH of a solution with a weak acid or base as last unit (use ICE).

• The only difference is the initial concentration of the common ion will NOT be zero.

• Example 15.23 on pg. 740

15.2 Buffered Solutions

Buffered solutions resist a change in pH regardless of whether OH- or H+ is added.

– Most common example is our blood, which maintains a constant pH so that cells can survive.

– Contain a weak acid and its common ion salt, such as HF and NaF.

– OR– Contain a weak base and its common ion salt, such

as NH3 and NH4Cl.

How does buffering work?

• Buffer solution of a weak acid (HA) and its conjugate base (A-).

• Ka= [H+][A-] or [H+]= Ka [HA]

[HA] [A-]

– If you add protons,

H+ + A- → HA

– A- converted to HA ; [HA]/[A-] increases; [H+] increases; pH increases.

– ****If [A-] and [HA] are large compared to [H+] added, then there will be little change in the pH.

Addition of OH- to Buffered Solutions

– If amounts of HA and A- originally present are large compared to OH-, change in [HA]/[A-] is small. Therefore, [H+] and pH remain constant.

• Ka= [H+][A-] or [H+]= Ka [HA]

[HA] [A-]

– If you add hydroxide ions,

OH-+ HA → A- + H2O

– HA is converted to A-; [HA]/[A-] decreases; [H+] decreases; pH decreases (but only slightly if [HA] and [A-] are large).

1. Buffered solutions are simply weak acids or bases with a common ion.

2. pH calculations for buffered solutions are exactly the same procedures as last unit.

3. When a strong acid or base is added to a buffered solution, deal with stoichiometry first, then equilibrium calculations.

Buffer Calculations

• Calculating [H+]: [H+]= Ka [HA]

[A-]

• Henderson-Hasselbalch Equation

-log[H+]=-log(Ka) –log ([HA]/[A-])

pH=pKa-log([HA]/[A-])

pH=pKa+log([A-]/[HA])

pH=pKa+log(base/acid)

• See the Summary of the Most Important Characteristics of Buffered Solutions on page 692.

• Examples 15.37• HW: 41 and 43 on pages 740-741

15.3 Buffering Capacity

Buffering capacity is the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.

– pH depends on the [A-]/[HA] ratio– Buffering capacity depends on the

magnitudes of [HA] and [A-]• Larger concentrations allow for more absorbtion of

H+ and OH- with little change in pH

Optimal Buffering– When [HA]=[A-], so that [A-]/[HA]=1.– pKa=pH (or as close as possible)

Examples 15.46 on pg. 741

HW: 45 and 49* on pg. 741

15.4 Titrations and pH Curves

• Titration: used to determine the amount of acid or base in a solution, involving a solution of known concentration (titrant), indicator, and unknown solution.

• pH curve or titration curve: a plot of the pH of an acid-base titration as a function of the amount of titrant added.

Strong Acid-Strong Base Titrations

• Net Ionic Reaction: H+(aq) + OH- (aq)→ H2O (l)

• Computing [H+]:– Determine mmol of H+ at that point and divide by total

volume of solution– 1 mmol= (1 mol/1000) = 10-3 mol

*used because titrations involve small quantities– Molarity = mol solute = (mol solute/1000) = mmol solute

L solution (L soltn/1000) mL soltn

- Mmol =mL x molarity

Case Study

• 50.0 mL of 0.200 M HNO3 is titrated with 0.100 M NaOH. We will calculate the pH of the solution at selected points during the course of the titration.

– A. No NaOH has been added.

[H+]=0.200 M and pH=0.699

– B. 10.0 mL of 0.100 M NaOH has been added.

• [H+]=9.00 mmol/(50.0+10.0)= 0.15 M and pH=0.82

– C. 20.0 mL (total) of 0.100 M NaOH has been added.

• [H+]=8.00 mmol/(50.0+20.0)= 0.11 M and pH=0.942

H+ + OH- → H2O

Before 50.0 mL x 0.200 M =10.0 mmol

10.0 mL x 0.100 M=1.00 mmol

Reaction -1.00 mmol -1.00 mmol 1.00 mmol

After 9.00 mmol left 0 mmol 1.00 mmol

H+ + OH- → H2O

Before 50.0 mL x 0.200 M =10.0 mmol

20.0 mL x 0.100 M=2.00 mmol

Reaction -2.00 mmol -2.00 mmol 2.00 mmol

After 8.00 mmol left 0 mmol 2.00 mmol

– D. 50.0 mL (total) of 0.100 M NaOH has been added.• Proceed exactly as for points B and C.• pH=1.301

– E. 100.0 mL (total) of 0.100 M NaOH has been added.

• Stoichiometric point or equilvalence point (pH=7.00)

– F. 150.0 mL (total) of 0.100 M NaOH has been added.

• [OH-]=5.00 mmol/(50.0+150.0)= 0.025 M, pOH=1.60, pH=12.40

H+ + OH- → H2O

Before 50.0 mL x 0.200 M =10.0 mmol

100.0 mL x 0.100 M=10.0 mmol

Reaction -10.0 mmol -10.0 mmol 10.0 mmol

After 0 mmol 0 mmol 10.0 mmol

H+ + OH- → H2O

Before 50.0 mL x 0.200 M =10.0 mmol

150.0 mL x 0.100 M=15.0 mmol

Reaction -10.0 mmol -10.0 mmol 10.0 mmol

After 0 mmol 5.0 mmol left 10.0 mmol

– G. 200.0 mL (total) of 0.100 M NaOH has been added• Proceed the same as point F.• pH = 12.60

Characteristics of Strong-Strong pH Titration Curves• pH changes are gradual until the titration is close to

the equivalence point, due to the large amount of original H+ or OH- ions depending on the titration.• At the equivalence point, pH=7.00.• Pg. 742 (#53)

Titrations of Weak Acids with Strong Bases

Calculating the pH curve:1. Stoichiometry problem (BRA): The rxn of OH- with

the weak acid runs to completion, and [remaining acid] and [formed conjugate base] are determined.

2. Equilibrium problem (ICE): The position of the weak acid equilibrium is determined; pH is calculated.

Case Study50.0 mL of 0.10 M acetic acid (Ka=1.8x10-5) is titrated with 0.10 M

NaOH. We will calculate pH at various points.

• A. No NaOH has been added.– Complete this the same as last unit.– pH=2.87 (check it yourself)

• B. 10.0 mL of 0.10 M NaOH has been added.

OH- + CH3COOH → CH3COO- + H2O

Before 10.0 mL x 0.10 M =1.0 mmol

50.0 mL x 0.10 M=5.0 mmol

0 mmol

Reaction -1.0 mmol -1.0 mmol +1.0 mmol

After 0 mmol 4.0 mmol left 1.0 mmol

• Since only CH3COOH and CH3COO- are the only species left in solution:

– Ka=1.8x10-5 = [H+][CH3COO-] = (0.017)x

[CH3COOH] 0.067

– X=7.2 x 10-5 M = [H+]

– pH=4.14

CH3COOH → CH3COO- + H+

Initial 4.0 mmol(50.0+10.0) mL

1.0 mmol(50.0 +10.0) mL

0

Change -x +x +x

Equilibrium 0.067-x ≈ 0.067 0.017+x ≈ 0.017 x

– C. 25.0 mL (total) of 0.10 M NaOH has been added• This is the halfway point.• [CH3COOH]=[CH3COO-]

• [H+]=Ka

• pH=pKa

– D. 50.0 mL (total) of 0.10 M NaOH has been added.• This is the equivalence point because 5.0 mmol of OH- will

react with the 5.0 mmol of CH3COOH originally present.

• Species left is CH3COO-.

– CH3COO- + H2O (l) ↔ CH3COOH (aq) + OH-

– Find Kb.

– Determine pH.

– E. 60.0 mL (total) of 0.10 M NaOH has been added• At this point, excess OH- has been added.

• pH is determined by the excess OH- because it is a strong base• [OH-]=1.0 mmol / (50.0+60.0)mL= 9.1x10-3 M• pOH=2.04• pH=11.96

– HW: pg 742 (#55)

OH- + CH3COOH → CH3COO- + H2O

Before 60.0 mL x 0.10 M =6.0 mmol

50.0 mL x 0.10 M=5.0 mmol

0 mmol

Reaction -5.0 mmol -5.0 mmol +5.0 mmol

After 1.0 mmol in excess 4.0 mmol left 5.0 mmol

Comparisons:• Near the beginning, pH

increases more rapidly than strong/strong.

• Weak curve levels off near the halfway point due to buffering.

• Weak curve, the pH at the equivalence point > 7 due to basicity of the conjugate base.

• Shapes of both curves are the same after the equivalence point.

*Remember that equivalence point is determined by stoichiometry (enough titrant has been added to react exactly with acid or base), not the pH.

• The amount of acid present determines the equivalence point, not its strength.

• pH value at the equivalence point is affected by the acid strength. – Weak acid=high pH

• Notice the difference in the shapes of the curves.

Titrations of Weak Bases with Strong Acids

1st: Think about the major species in solution.

2nd: Decide whether a reaction occurs that runs to completion and do stoichiometric calculations.

3rd: Choose the dominant equilibrium and calculate pH.

-See Case Study on page 709

Note that the pH at the equivalence point < 7 since the solution contains the weak acid NH4

+.

15.5 Acid-Base Indicators

2 methods for determining equivalence points:– Use a pH meter to plot the titration curve. The center

of the vertical region indicates the equivalence point.

– Use an acid-base indicator, which marks the end point by changing colors. Although equivalence point ≠ end point, careful selection of the indicator will ensure error is negligible.

Common Acid-Base Indicators are Complex Molecules

• HIn: represents the acid form of a complex molecule that is a particular color or colorless.

• In-: represents the conjugate base of the complex molecule that shows as a different color.

• Assume that the ratio of [In-]/[HIn] must be 1/10 for the human eye to detect a color change.

a.) Yellow acid form of bromthymol blue; b.) a greenish tint is seen when the solution contains 1 part blue and 10 parts yellow; c.) blue basic form

• Henderson-Hasselbalch equation useful to determine pH at the indicator color change.– Acid Solution being titrated.

• pH=pKa + log ([In-]/[HIn])

• pH=pKa + log (1/10)= pKa-1

• FOR ACID– pH=pKa - 1

– Basic solution being titrated (indicator initially in the form of In-)• pH=pKa + log ([In-]/[HIn])

• pH=pKa + log(10/1)=pKa+1

• FOR BASE– pH=pKa + 1

• The useful pH range for an indicator is pKa(indicator)± 1.

Choosing an Indicator

• The indicator end point (when it changes colors) should be as close as possible to the titration equivalence point.– What indicator would you use for the titration of

100.00 mL of 0.100 M HCl with 0.100 M NaOH?• Equivalence point occurs at pH of 7.00

(strong/strong)• Titrating an acid; pH=pKa-1

– pKa=pH+1=8

– Ka=1x10-8

• Indicators for strong acid-strong base titrations– Color changes will

be sharp, occurring with the addition of a single drop of titrant.

– There is a wide choice of suitable indicators.

• Titration of Weak Acids– Smaller vertical

area around the equivalence point.

– Less flexibility in choices of indicators.

– Choose an indicator whose useful pH range has a midpoint as close as possible to the equivalence point.

• Examples 15.65, 15.69 on pages 742-743

•THE END!!