Acids and Bases Chapter 14 AP Chemistry Seneca Valley.

Acids and BasesAcids and Bases

Chapter 14Chapter 14

AP ChemistryAP Chemistry

Seneca ValleySeneca Valley

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What do you know about acids & bases?What do you know about acids & bases?

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Some DefinitionsSome Definitions

Arrhenius– An acid is a substance that, when

dissolved in water, increases the concentration of hydrogen ions.

– A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.

Problem with this definition is that it limits us to aqueous solns.

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Some DefinitionsSome Definitions

Brønsted-Lowry– An acid is a proton (H+ or H3O+) donor.

– A base is a proton acceptor. It does not need to contain OH-.

HCl + HHCl + H22O O Cl Cl + H + H33OO++

acid baseacid base

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A Brønsted-Lowry acid…

…must have a removable (acidic) proton.

A Brønsted-Lowry base…

…must have a pair of nonbonding electrons.

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If it can be either…If it can be either…

…it is amphoteric.

H2O

HCO3-

HSO4-

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What Happens When an Acid Dissolves What Happens When an Acid Dissolves in Water?in Water?

• Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid.

• As a result, the conjugate base of the acid and a hydronium ion are formed.

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Conjugate Acids and BasesConjugate Acids and Bases• The term conjugate comes from the Latin

word “conjugare,” meaning “to join together.”

• Reactions between acids and bases always yield their conjugate bases and acids.

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

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Acid and Base StrengthAcid and Base Strength• Strong acids are completely

dissociated in water.– Their conjugate bases are

quite weak.

• Weak acids only dissociate partially in water.– Their conjugate bases are

weak bases.

• OH- is the strongest base that can exist in equilibrium in aqueous solution.

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Acid and Base StrengthAcid and Base Strength

• Substances with negligible acidity do not dissociate in water.– Their conjugate bases

are exceedingly strong.

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Strong Acids and BasesStrong Acids and BasesConjugate basesConjugate bases• Using the idea of conjugate bases previously covered

arrange the following according to their strength as bases.

H2O F- Cl- NO2- CN-

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• In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

• H2O is a much stronger base than Cl-, so the equilibrium lies so far to the right that Ka is not measured (Ka>>1).

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Again, in any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

• Acetate is a stronger base than H2O, so the equilibrium favors the left side (Ka<1).

CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)

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Relative Strengths of AcidsRelative Strengths of Acids

Consider two separate aqueous solutions: one of a weak acid HA and one of HCl. Assume 10 molecules of each acid. Draw a picture of what each solution looks like at equilibrium. What are the major species in each beaker? From your pictures, calculate the Ka values of each acid. Order the following from strongest to weakest base and explain: H2O, A-(aq), Cl-(aq).

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Relative Strengths of AcidsRelative Strengths of Acids

Draw molecular-level pictures of the following:• concentrated weak acid• dilute weak acid• concentrated weak base• dilute weak base• concentrated strong acid• dilute strong acid• concentrated strong base• dilute strong base

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Weak AcidsWeak Acids

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Autoionization of WaterAutoionization of Water

• As we have seen, water is amphoteric.

• In pure water, a few molecules act as bases and a few act as acids.

• This is referred to as autoionization.

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

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Ion-Product ConstantIon-Product Constant

• The equilibrium expression for this process is

Kc = [H3O+] [OH-]

• This special equilibrium constant is referred to as the ion-product constant for water, Kw.

• At 25C, Kw = 1.0 10-14

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The pH ScaleThe pH Scale

• In most solutions [H+(aq)] is quite small.

• We define pH = -log[H+] = -log[H3O+].

• Most pH and pOH values fall between 0 and 14.• There are no theoretical limits on the values of pH or

pOH. (e.g. pH of 2.0 M HCl is -0.301)

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pHpH

• In pure water,

Kw = [H3O+] [OH-] = 1.0 10-14

• Since in pure water [H3O+] = [OH-],

[H3O+] = 1.0 10-14 = 1.0 10-7

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pHpH

• Therefore, in pure water,pH = -log (1.0 10-7) = 7.00

• An acid has a higher [H+] than pure water, so its pH is <7.• A base has a lower [H+] than pure water, so its pH is >7.

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pHpH

These are the pH values for several common substances.

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Other “p” ScalesOther “p” Scales

• The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions).

• Some similar examples are– pOH: -log [OH-]

– pKw: -log Kw

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Watch This!Watch This!

Because

[H3O+] [OH-] = Kw = 1.0 10-14,

we know that

-log [H3O+] + -log [OH-] = -log Kw = 14.00

or, in other words,

pH + pOH = pKw = 14.00

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How Do We Measure pH?How Do We Measure pH?

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How Do We Measure pH?How Do We Measure pH?

For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

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The pH ScaleThe pH Scale

The number of decimal places in the log (pH) is equal to the number of significant figures in the original number.

For example: [H+] = 1.0 x 10-9 2 SF

pH = 9.00 2 decimal places

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The pH ScaleThe pH ScaleSolving for [H+], [OH-] and pHSolving for [H+], [OH-] and pH• 1.0x10-5 M OH-

• 1.0x10-7 M OH-

• 10.0 M H+

See sample problems 1-3.

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The pH ScaleThe pH ScaleMeasuring pHMeasuring pH

•Calculate the pH of 0.10 M HNO3.

•Step 1: List the materials used to prepare solution & the major species in the solution.

•Step 2: Indicate the major & minor sources of H+. (Write the reactions that produce H+.)

•Step 3: Indicate the approximations to be made.

•Step 4: Calculate the pH.

See sample problem 4.

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Strong AcidsStrong Acids

• You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

• These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.

• For the monoprotic strong acids,

[H3O+] = [acid].

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Calculating the pH for solutions containing Calculating the pH for solutions containing weak acidsweak acids

• This situation is more complex than for a strong acid.

• Since a weak acid does not completely dissociate, an equilibrium calculation must be performed to find [H+].

• A weak acid can be recognized by its small Ka value.

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Solving Acid-Base ProblemsSolving Acid-Base Problems

1. List the major species. Strong acids are written as completely dissociated. Weak acids are written as HA.

2. Write the balance equations for major species that produce H+ and determine the dominant source of H+.

3. Write the equilibrium expression for the dominant acid.

4. List the initial concentrations of all species involved in the dominant equilibrium.

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Solving Acid-Base ProblemsSolving Acid-Base Problems5. Define x, the change required to reach

equilibrium.6. Write the equilibrium concentrations in terms of

the initial concentrations and x.7. Substitute the equilibrium concentrations in the

expression for Ka.8. Simplify the expression by neglecting x where

possible. That is, assume that [HA] = [HA]0 – x ≈ [HA]0. Then solve for x.

9. Check the validity of the assumption made in step 8. (Use 5% rule).

10.Calculate [H+] and pH.

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Calculating pH from Calculating pH from KKaa

Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C.

HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)

Ka for acetic acid at 25C is 1.8 10-5.

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The equilibrium constant expression is

[H3O+] [C2H3O2-]

[HC2H3O2]Ka =

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We next set up a table…

[C2H3O2], M [H3O+], M [C2H3O2-], M

Initially 0.30 0 0

Change -x +x +x

At Equilibrium 0.30 - x 0.30 x x

We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

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Now,

(x)2

(0.30)1.8 10-5 =

(1.8 10-5) (0.30) = x2

5.4 10-6 = x2

2.3 10-3 = x

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pH = -log [H3O+]pH = -log (2.3 10-3)pH = 2.64

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• Calculate the pH of a 0.50 M aqueous solution of the weak acid HF (Ka = 7.2 x 10-4).

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Solving for pHSolving for pH

HF(aq) H+(aq) + F-(aq)

Initial 0.50 M 0 0

Change -x +x +x

Equilibrium 0.50-x x x

Ka = 7.2 x 10-4

See Sample Problem #5.

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Weak AcidsWeak AcidsUsing Using KKaa to Calculate pH to Calculate pH

• Using Ka, the concentration of H+ (and hence the pH) can be calculated.– Write the balanced equation showing the equilibrium.

– Write the equilibrium expression. Find the value for Ka.

– Write down the initial and equilibrium concentrations for everything except pure water. We usually assume that the change in concentration of H+ is x.

• Substitute into the equilibrium constant expression and solve. Remember to turn x into pH if necessary.

Ka3H O A

HA

H A

HA

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Weak AcidsWeak AcidsUsing Using KKaa to Calculate pH to Calculate pH

• Percent ionization is another method to assess acid strength.

• For the reaction

• Percent ionization relates the equilibrium H+ concentration, [H+]eqm, to the initial HA concentration, [HA]0.

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

100]HA[

][Hionization %

0

eqm

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Calculating Calculating KKaa from the pH from the pH

The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature.

We know that

[H3O+] [COO-][HCOOH]

Ka =

To calculate Ka, we need the equilibrium concentrations of all three things.We can find [H3O+], which is the same as [HCOO-], from the pH.

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Calculating Calculating KKaa from the pH from the pH

pH = -log [H3O+]

2.38 = -log [H3O+]

-2.38 = log [H3O+]

10-2.38 = 10log [H3O+] = [H3O+]

4.2 10-3 = [H3O+] = [HCOO-]

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Calculating Calculating KKaa from pH from pH

Now we can set up a table…

[HCOOH], M [H3O+], M [HCOO-], M

Initially 0.10 0 0

Change - 4.2 10-3 + 4.2 10-3 + 4.2 10-3

At Equilibrium 0.10 - 4.2 10-3

= 0.0958 = 0.10

4.2 10-3 4.2 10-3

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Calculating Calculating KKaa from pH from pH

[4.2 10-3] [4.2 10-3][0.10]

Ka =

= 1.8 10-4

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Calculating Percent IonizationCalculating Percent Ionization

• Percent Ionization = 100• In this example

[H3O+]eq = 4.2 10-3 M

[HCOOH]initial = 0.10 M

[H3O+]eq[HA]initial

Percent Ionization = 1004.2 10-3

0.10= 4.2%

See sample problem #6.

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Weak BasesWeak Bases

Bases react with water to produce hydroxide ion.

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The equilibrium constant expression for this reaction is

[HB+] [OH-][B-]

Kb =

where Kb is the base-dissociation constant.

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Kb can be used to find [OH-] and, through it, pH.

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pH of Basic SolutionspH of Basic Solutions

What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH-]

[NH3]Kb = = 1.8 10-5

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

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Tabulate the data.

[NH3], M [NH4+], M [OH-], M

Initially 0.15 0 0

At Equilibrium 0.15 - x 0.15 x x

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(1.8 10-5) (0.15) = x2

2.7 10-6 = x2

1.6 10-3 = x2

(x)2

(0.15)1.8 10-5 =

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Therefore,

[OH-] = 1.6 10-3 M

pOH = -log (1.6 10-3)

pOH = 2.80

pH = 14.00 - 2.80

pH = 11.20

See sample problems 8 & 9.

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KKaa andand K Kbb

Ka and Kb are related in this way:

Ka Kb = Kw

Therefore, if you know one of them, you can calculate the other.

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Weak AcidsWeak AcidsPolyprotic AcidsPolyprotic Acids• Polyprotic acids have more than one ionizable proton.• The protons are removed in steps not all at once:

• It is always easier to remove the first proton in a polyprotic acid than the second.

• Therefore, Ka1 > Ka2 > Ka3 etc.

• Most H+(aq) at equilibrium usually comes from the first ionization (i.e. the Ka1 equilibrium).

H2SO3(aq) H+(aq) + HSO3-(aq)

HSO3-(aq) H+(aq) + SO3

2-(aq)

Ka1 = 1.7 x 10-2

Ka2 = 6.4 x 10-8

Let’s review using sample problems 10, 11 & 12.

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Weak AcidsWeak AcidsPolyprotic AcidsPolyprotic Acids

See sample problem # 13.

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Sulfuric Acid…Sulfuric Acid…• … is unique among the common acids.• It is a strong acid in the first dissociation step.

– Ka1 = LARGE

• It is a weak acid in the second dissociation step.– Ka2 = 1.2 x 10-2

• This means calculating the pH for sulfuric acid is a little different.

• Only in dilute sulfuric acid solutions does the second dissociation step contribute significantly to [H+].

See sample problem # 14.

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• Calculate the pH of a 1.00 M solution of H3PO4.

• Ka1 = 7.5 x 10-3

• Ka2 = 6.2 x 10-8

• Ka3 = 4.8 x 10-13

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Acid-Base Properties of Salt SolutionsAcid-Base Properties of Salt Solutions• Nearly all salts are strong electrolytes.• Therefore, salts exist entirely of ions in solution.• Acid-base properties of salts are a consequence of the

reaction of their ions in solution.• The reaction in which ions produce H+ or OH- in

water is called hydrolysis.• Anions from weak acids are basic.• Anions from strong acids are neutral.

• Anions with ionizable protons (e.g. HSO4-) are

amphoteric.

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Acid-Base Properties of Salt SolutionsAcid-Base Properties of Salt Solutions• To determine whether a salt has acid-base properties

we use:– Salts derived from a strong acid and strong base are neutral

(e.g. NaCl, Ca(NO3)2).

– Salts derived from a strong base and weak acid are basic (e.g. NaOCl, Ba(C2H3O2)2).

– Salts derived from a weak base and strong acid are acidic (e.g. NH4Cl, Al(NO3)3).

– Salts derived from a weak acid and weak base can be either acidic or basic. Equilibrium rules apply!

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• Arrange the following 1.0 M solutions from lowest to highest pH. Justify your answer.

HBr NaOH NH4ClNaCN NH3 HCN NaCl

HF

React

See sample problems 15 & 16.

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Lewis Acids and BasesLewis Acids and Bases• Brønsted-Lowry acid is a proton donor.• Focusing on electrons: a Brønsted-Lowry acid can be

considered as an electron pair acceptor.• Lewis acid: electron pair acceptor.• Lewis base: electron pair donor.• Note: Lewis acids and bases do not need to contain

protons.• Therefore, the Lewis definition is the most general

definition of acids and bases.

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Three Models for Acids & BasesThree Models for Acids & Bases

Model Definition of Acid

Definition of Base

Arrhenius H+ producer OH- producer

Bronsted-Lowry H+ donor H+ acceptor

Lewis Electron pair acceptor

Electron pair donor

Acids and Bases Chapter 14 AP Chemistry Seneca Valley.

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