05/09/23 Chapter 17 1

CH160 General Chemistry IILecture Presentation

Applications of Acid-Base Equilibria

Chapter 17

Sections 1-4

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Common Ion Effect• Consider the ionization of weak acid HA:

HA + H2O <=> H3O+ + A-

• What affect will adding salt NaA to the solution have on the acid ionization and solution pH?

NaA Na+ + A-

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Common Ion Effect• Consider the ionization of weak acid HA:

HA + H2O <=> H3O+ + A-

A- has 2 sources: HA and NaA. Adding

NaA increases [A-].

05/09/23 Chapter 17 4

Common Ion Effect• Consider the ionization of weak acid HA:

HA + H2O <=> H3O+ + A-

A- is a “common ion”

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Example 1(1a on the Example Problem Handout)

• Calculate the percent ionization of the acid and the pH of the solution that contains 0.500M HC2H3O2 (Ka = 1.8 x 10-5) and 0.200 M NaC2H3O2.

• (ans.: pH = 4.34, 0.0090%)

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Calculations Using Ka • Basic Steps for Weak Acid Calculations Using KBasic Steps for Weak Acid Calculations Using Kaa

– Write balanced chemical equation and the expression for Ka

• Look up value for Ka

– For each chemical species involved in the equilibrium (except H2O), write:

• Initial concentration• Equilibrium concentration

– Let the change in the [H3O+] be the variable “x”

– Substitute the equilibrium concentrations into Ka and solve for x using either

• quadratic approach• simplified approach

– Calculate pH, equilibrium concentrations, % ionization, etc., as specified in the problem.

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0.5 M CH3COOH

pH = 2.5

0.5 M CH3COOH +

0.200 M CH3COONa

pH = 4.3

0.6%

0.009%

+ CH3COONa

(problem 8a)

(problem 1a)

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Common Ion Effect• What affect does NaA have on weak acid

HA ionization?:HA + H2O <=> H3O+ + A-

Which way does equilibrium shift? What do the results of problems

8a and 1a tell us? Does this agree with LeChatelier’s principle?

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Common Ion Effect• What affect does NaA have on weak acid

HA ionization?:HA + H2O <=> H3O+ + A-

In presence of NaA, HA ionization shifts left.

“common-ion effect”

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Buffer Solutions• What is a buffer solution?

– solution with ability to resist pH changes upon addition of small amounts of either acid or base

• Requirements– must contain an acid to neutralize added OH-

ions – must contain a base to neutralize added H3O+

ions – acidic and basic species in buffer must not

neutralize each other.

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Buffer Action

• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of acid

A- + H+ HA

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Buffer Action

• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of acid (small amount)

A- + H+ HA

[A-] decreases slightly

[HA] increases slightly

Added acid is neutralized.

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Buffer Action

• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of base

HA + OH- H2O + A-

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Buffer Action

• How do buffers work? Consider buffer with weak acid HA and salt NaA:– Addition of base

HA + OH- H2O + A-

[HA] decreases slightly

[A-] increases slightly

Added base is neutralized.

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Buffer Action

• Two important properties of buffer solutions:– Buffer capacity

• Amount of acid or base the buffer can react with before giving a significant pH change (1 pH unit)

• Determined by how much buffer acid and base are used to make buffer

– pH• Determined by Ka and relative amounts of buffer

acid and base present

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Calculation of Buffer pH

• Calculating pH for buffer containing both weak acid HA and salt NaA. The major equilibrium is:– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]

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Calculation of Buffer pH

• Calculating pH for buffer containing both weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]

A- has 2 sources: HA and NaA

(This seems familiar! Didn’t we

just do this?)

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Calculation of Buffer pH

• Calculating pH for buffer containing both weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]

Neither CHA nor CNaA change much since only a very small amount of HA ionizes.

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Calculation of Buffer pH

• Calculating pH for buffer containing both weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]

• Since % ionization is small: CA- [A-] and CHA [HA]

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Calculation of Buffer pH• Calculating pH for buffer containing both

weak acid HA and salt NaA.– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]

• Since % ionization is small: CA- [A-] and CHA [HA]

Solving for [H3O+] gives: [H3O+] = KaCHA/CA-

pH = -log[H3O+](This is just a common ion effect problem.)

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Calculation of Buffer pH

• We can also take the –log of our equation:

[H3O+] = KaCHA/CA-

-log[H3O+] = -log(KaCHA/CA-)

-log[H3O+] = -logKa - log CHA/CA-

pH = pKa - log CHA/CA-

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Example 2(2a on Example Problem Handout)

• Calculate the pH of a buffer solution that contains 0.25 M sodium acetate, NaC2H3O2, and 0.35 M acetic acid, HC2H3O2 (Ka = 1.8 x 10-5).

• (ans.: pH = 4.60)

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Calculating pH Changes in Buffers

• How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?

H3O+ + A- HA + H2O

or

OH- + HA A- + H2O

HA/NaA+ H3O+ or OH-

Buffer

Neutralization Rxn

Calculate new [HA] & [A-]

Calculate new [H3O+] from [H3O+]

= KaCHA/CA- pH

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Calculating pH Changes in Buffers

• How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?

H3O+ + A- HA + H2O

or

OH- + HA A- + H2O

HA/NaA+ H3O+ or OH-

Buffer

Neutralization Rxn

Calculate new [HA] & [A-]

This much is a stoichiometry problem. (Oh, oh! General

Chemistry I stuff here.).

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Calculating pH Changes in Buffers

• How do we calculate buffer pH after adding acid (H3O+) or base (OH-)?

Calculate new [H3O+] from [H3O+]

= KaCHA/CA- pH

This part is an equilibrium calculation.

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Example 3(3a on Example Problem Handout)

• Calculate the pH of the solution formed and the change in pH observed when (a) 0.050 moles of HCl and (b) 0.025 moles of NaOH are added to 500 mL of the buffer in example (2a). (c) Calculate the change in pH that occurs when 0.050 moles HCl are added to 500 mL H2O.

(ans.: (a) pH = 4.27, pH = -0.33, (b) pH = 4.74, pH = +0.14, (c) pH = -6)

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Buffer Preparation

• What if I need to make a buffer solution of known pH? Select:– Buffer system.

• Often pKa of buffer acid is close to desired pH.

– Relative amounts of buffer acid and base.• Buffer capacity increases with concentrations.• Buffer effectiveness best with concentrations on

same order of magnitude.

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Example 4 (4 on Example Problem Handout)

• Starting with 1.0L of 0.100 M CH3COOH (Ka = 1.8 x 10-5), how many grams of sodium acetate, CH3COONa (FW = 82.034 g/mol), to give a buffer with a pH of 4.40? (Assume no volume change.)

• (ans.: 3.7 g)

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Quantitative Acid-Base Chemistry• How do we calculate the pH of a solution

formed by mixing an acid solution with a base solution?

• Consider addition of 0.1 M strong base, MOH, to 0.1 M strong acid, HX– pH changes can be observed from titration curve

• pH vs. volume standard

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Titration of 25 mL of 0.1 M HCl with 0.1 M NaOH

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50 55volume 0.1 M NaOH

pH

Rxn is complete = equivalence point. pH = 7.0 since only neutral NaX present.

Strong Acid-Strong Base Titration

HX + NaOH NaX + H2O

pH

mL NaOH

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Calculating pH in Acid-Base Reactions

• How do we calculate pH after adding a strong base to a strong acid?

HX + MOH MX + H2OHX+ MOH

Strong acid

Neutralization Rxn

Calculate [HX] or [MOH] left (ignore

neutral MX)

Calculate new [H3O+] from [HX] or [MOH]

pH

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Calculating pH in Acid-Base Reactions

• How do we calculate pH after adding a strong base to a strong acid? Considerations:

• Stoichiometry– Limiting reagent

• At what point in rxn does calculation take place?– Initial– Pre-equivalence– Equivalence– Post-equivalence

• Does dilution occur?– If mixing 2 solutions: Vtotal = V1 + V2

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Titration of 25 mL of 0.1 M HCl with 0.1 M NaOH

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50 55volume 0.1 M NaOH

pH

-All HX & MOH consumed

-NaX left (neutral)

-pH = 7.0

Strong Acid-Strong Base TitrationHX + NaOH NaX + H2O

-HX in excess

-HX/NaX left

-[H3O+] = CHX

-NaOH in excess

-NaOH/NaX left

-[OH-] = CMOH

Only HX

[H3O+] = CHX

pH

mL NaOH

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Example 5(5 of Example Problem Handout)

• Calculate the pH for a solution prepared by mixing 25.00 mL of 0.100 M HCl(aq) with a) 10.00 mL b) 25.00 mL and c) 35.00 mL of 0.100 M NaOH(aq). (ans.: a) 1.37, b) 7.0, c) 12.22)

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Titration of 25 mL of 0.1 M HOAc with 0.1 M NaOH

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1011121314

0 5 10 15 20 25 30 35 40 45 50volume 0.1 M NaOH

pH

Weak Acid-Strong Base TitrationHA + NaOH NaA + H2O

Rxn is complete = equivalence point.

pH > 7.0 since only basic NaA present.

pH

mL NaOH

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Titration of 25 mL of 0.1 M HOAc with 0.1 M NaOH

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0 5 10 15 20 25 30 35 40 45 50volume 0.1 M NaOH

pHWeak Acid-Strong Base Titration

HA + NaOH NaA + H2O

-HA in excess

-HA/NaA left = buffer

-[H3O+] = Ka(CHA/CNaA)-All HA & MOH consumed

-NaA left (weak base)

-[OH-] (KbCNaA)1/2

-pH > 7.0

-NaOH in excess

-NaOH/NaA left

-[OH-] CMOH

-only HA

[H3O+] (KaCHA)1/2

pH

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Titration of 25 mL of 0.1 M ammonia with 0.1 M HCl

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0 5 10 15 20 25 30 35 40 45 50volume 0.1 M HCl

pH

Rxn is complete = equivalence point. pH < 7.0 since only acidic HB+ present.

Strong Acid-Weak Base TitrationHX + B HB+ + X-

pH