unit 17: acid-base equilibria

AcidsandBases

Unit 17:Acid-Base Equilibria

Dr. Jorge L. AlonsoMiami-Dade College –

Kendall CampusMiami, FL

CHM 1046: General Chemistry and Qualitative Analysis

Textbook Reference:

•Chapter 18

•Module # 6

AcidsandBases

Acid-Base Theories

Arrhenius (1883)

Brønsted–Lowry (1923)

Lewis (1923-38)

HCl NaOH

NH3

BF3

H2O

H2SO4

AlI3

CO32-

AcidsandBases

Acid-Base Definitions• Svante Arrhenius (1883)

Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions.

HX H+ + X-

{acid}

Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

MOH M+ + OH-

Neutralization:

HX + MOH MX + HOH

acid base salt water

AcidsandBases

• Brønsted–Lowry (1923)Acid: Proton (H+) donor.

… have a removable (acidic) proton which are donated to a bases.

{base}

Acid-Base Definitions

HX + H2O X- + H3O+

acid base conjugate b conjugate a.

Neutralization:

HX + B X- + BH+

acid base conj. b conj. a

B + H2O BH+ + OH-

base acid conj. a conj. b

Base: Proton (H+) acceptor.

…must have a pair of nonbonding electrons which accepts protons.

AcidsandBases

Strong AcidsHCl, HBr, HI, HNO3, H2SO4, HClO3, & HClO4

SOLUBILITY RULES: for Ionic Compounds (Salts)• All salts containing the anions: NO3

-, ClO3-, ClO4

-, (C2H3O2-) are soluble.

• All Cl-, Br-, and I- are soluble except for Hg22+, Ag+, and Pb2+ salts.

• All SO42- are soluble except for Pb2+, Ba2+, and Sr2+.

Are strong electrolytes and exist totally as ions in aqueous solution.

For the monoprotic strong acids, [acid]i = [H3O+] or [H+]

HX H+(aq) + X-

(aq)

HX + H2O X- + H3O+

acid base conj. b conj. a

Strong acids have very weak conjugate bases

Arrhenius:

Brønsted–Lowry:

100%

100%

For strong acid: [1M HX]i = [1M H3O+] or [1M H+] For weak acid: [1M HA]i > [0.1M H3O+] or [0.1M H+]

AcidsandBases

these substances dissociate completely (100%) in aqueous solution.

SOLUBILITY RULES: for Ionic Compounds (Salts)All OH- are insoluble except for IA metals, (NH4

+), Ca2+, Ba2+ , & Sr2+ (heavy IIA).

MOH M+ + OH-

Na+OH- + H2O Na+ + HOH + OH-

base acid conjugate acid conjugate base

Arrhenius:

Brønsted–Lowry:

these substances accept protons from water to form very weak conjugate acids.

For the monobasic strong bases, [base]i = [OH-] [1M NaOH] = [1M OH-]

Strong Bases

AcidsandBases

Review of Strong Acid & Bases

Strong acids have very weak conjugate bases

HX + H2O H3O+ + X-

acid base conj. a conj. b

Strong bases have very weak conjugate acids

MOH M+ + OH-

base conj. a conj. b

Salts of Strong acids & bases have very weak conjugate acids & bases

MX + H2O M+(aq)

+ X- (aq)

salt conj. a conj. b

Has no affinity for H+

Has no affinity for OH-

Have no affinity for H+ or OH-

Example: HCl

Example: NaOH

Example: NaCl

Later on this course: Salts that are neutral

(according Brønsted–Lowry)

AcidsandBases

Weak Acid Dissociation

HX

HA

Strong Acid Dissociation

HXH3O

+ X-

Concentrated 10M Strong Acid

Concentrated 10M Very Weak Acid

AcidsandBases

Weak Acids

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

HA + H2O ↔ H3O+ + A-

acid base conj. a conj. b

H+

H+

Weak acids (HA) have very strong conjugate bases (A-)

•The weaker the acid, the stronger their conjugate base.

AcidsandBases

H+

H+

Weak Bases

Bases react with water to produce hydroxide ion.

B + HOH BH+ + OH-

• Weak bases have strong conjugate acids

base acid conj. a conj. b

•The weaker the base, the stronger their conjugate acid.

AcidsandBases

Acid / Base Strength

The stronger an acid, the weaker its conjugate base.

The stronger an base, the weaker its conjugate acid.

conjugate

HA + H2O ↔ H3O+ + A-

acid base conj. a conj. b

B + HOH ↔ BH+ + OH-

base acid conj. a conj. b

AcidsandBases

The Leveling Effect of Water

Strong acids react with water to produce H3O+, the strongest acid that can exist in an aqueous solution.

Strong bases react with water to produce OH-, the strongest base that can exist in an aqueous solution.

+

+

AcidsandBases

Salts: (1) of strong base & weak acid:

MA M+(aq) + A-

(aq)

A- + HOH HA + OH- (hydrolysis)

(2) of a weak base & strong acid:

BHX BH+(aq) + X-

(aq)

BH+ + HOH B + H3O+ (hydrolysis)

(3) of a weak base & weak acid:

BHA ↔ BH+(aq) + A-

(aq)

Weak acids (HA) have very strong conjugate bases (A-)

HA + H2O ↔ H3O+ + A-

acid base conj. a conj. b

Review of Weak Acid & Bases

Weak bases (B) have very strong conjugate acids (BH+)

B + HOH ↔ BH+ + OH-

base acids conj. a conj. b

Has affinity for H+

Has affinity for OH-

Example: HF

Example: NH3

Example: NaFExample: NH4ClExample: NH4F

Later on this course: Salts that are acidic or basic

AcidsandBases

• Lewis acids are electron-pair acceptors (have empty valance orbitals).

• Lewis bases are electron-pair donors.

{Lewis A-B}

Lewis Acids and Bases (1923-38)

+

H+ + O H- H O H..

: : : : ......

+

+

+

• Lewis acid-base neutralization reactions

+

AcidsandBases

Is Water an Acid or a Base? The Autoionization of Water

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

• This is referred to as autoionization.

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

H2O(l) H+(aq) + OH−(aq)

{Movie}

1 H3O+ and 1 OH- for every 10 million (107) H2O molecules

AcidsandBases

Autoionization of Water

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

• The equilibrium expression for this process is

[H3O+] [OH−]

• Kw is referred to as the ion-product constant for water (@ 25°C).

[H2O]2Kw = = 1.0 10−14Kw = = 1.0 10−14

AcidsandBases

What proportion of H2O molecules

dissociate into H3O+ & OH- ?

• In pure water,

Kw = [H3O+] [OH−] = 1.0 10−14

• Because in pure water [H3O+] = [OH−],

Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14

1 H3O+ and 1 OH- for every 10 million (107) H2O molecules

AcidsandBases

pH…..defined as the negative base-10 logarithm of the hydronium ion concentration.

pH = −log [H3O+]

Problem: Calculate pH when [H3O+] = 2.3 x 10-3 M

Problem: Calculate [H3O+] when pH = 2.3 ?

10x

log

antilog10^ -2.3

10x

logbase 10 log log - 2.3 * 10^- 3

For pure H2O: [1.0 10−7 ] = 7.0

= 2.64

= 5.0 * 10-3

AcidsandBases

Other “p” Scales

• The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).

• Some examples of other “p” scales are: pH = −log [H3O+]

pOH = −log [OH−]

pKw = −log Kw

pKa = −log Ka

pKb = −log Kb

[H3O+] = 1 x 10-7

[OH-] = 1 x 10-7

Kw = 1 x 10-14

Ka = 6.8 x 10-4

Kb = 1.8 x 10-5

pH = 7

pOH = 7

pKw = 14

pKa = 3.2

pKb = 4.7

AcidsandBases

pH and pOH equilibrium in pure Water

−log [H3O+] + −log [OH−] = −log Kw

• In pure water,

[H3O+] [OH−] = Kw

• Because in pure water [H3O+] = [OH−],

Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14

pH + pOH = pKw

7 + 7 = 14

[1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14

AcidsandBases

pH and pOH equilibrium in Water to which

Acids & Bases are Added

Kw = [1.0 10−6 ] [1.0 10−8 ] = 1.0 10−14

pH + pOH = pKw

6 + 8 = 14

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

H3O+

Add acid

[H3O+] [OH−]H2O

Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14

AcidsandBases

pH and pOH equilibrium in Water to which

Acids & Bases are Added

Kw = [1.0 10−8 ] [1.0 10−6 ] = 1.0 10−14

pH + pOH = pKw

8 + 6 = 14

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

OH-Add base

[H3O+] [OH−]H2O

Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14

AcidsandBases

pH

0 1.0 (100) 0.00000000000001

1 1.0 (10-1) 0.00000000000012 0.01 0.000000000001

3 0.001 0.00000000001 4 0.0001 0.0000000001

5 0.00001 0.000000001

6 0.000001 0.00000001

7 0.0000001 (10-7) 0.0000001 (10-7)8 0.00000001 0.0000001

9 0.000000001 0.00001

10 0.0000000001 0.0001

11 0.00000000001 0.001 12 0.000000000001 0.01

13 0.0000000000001 1.0 (10-1)

14 0.00000000000001 1.0 (100)

[H3O+] [OH−]

1

[OH−][H3O+]∝

Notice the relationship

between

[H3O+] and [OH−]

It is an inverse relationship!

pH + pOH = 14

AcidsandBases

pH

These are the pH values for several common substances.

pH + pOH = 14

AcidsandBases

Dissociation Constants for Weak Acids

• For a generalized acid dissociation,

the equilibrium expression would be

• This equilibrium constant is called the acid-dissociation (ionization) constant, Ka.

[H3O+] [A−][HA]

Ka =

HA(aq) + H2O(l) A−(aq) + H3O+(aq)

AcidsandBases

Dissociation Constants

The greater the value of Ka, the stronger the acid.

pKa

3.2

3.3

4.2

4.7

7.5

9.3

9.9

pKa

Salts of weak acids are strong bases

AcidsandBases

Calculating Ka from the pH

A 0.10 M solution of formic acid, HCOOH, at 25°C has a pH = 2.38. Calculate Ka for formic acid at this temperature. Also calculate % ionization.

Problem: (to determine Ka, simply dissolve a known [HA]i then measure pH)

HCOOH (aq) + H2O(l) HCOO−(aq) + H3O+

(aq)

pH = −log [H3O+]

2.38 = −log [H3O+]

−2.38 = log [H3O+]

10−2.38 = 10log [H3O+] = [H3O+]

4.2 10−3 = [H3O+] = [HCOO−]

[H3O+] [A−][HA]

Ka =

What is the Ka expression?

AcidsandBases

Calculating Ka from pH

[HCOOH], M [H3O+], M [HCOO−], M

Initially 0.10 0 0

Change −4.2 10-3 +4.2 10-

3

+4.2 10−3

Equilibrium 0.0958 = 0.10 4.2 10−3 4.2 10−3

[H3O+] [A−][HA]

Ka = [4.2 10−3] [4.2 10−3][0.10]

= = 1.8 10−4

[H3O+] = [HCOO−] = 4.2 10−3

• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Also calculate % ionization.

Problem:

HCOOH (aq) + H2O(l) HCOO−(aq) + H3O+

(aq)

AcidsandBases

Calculating Percent Ionization

• Percent Ionization = 100

• In this example

[H3O+]eq

[HA]initial

Percent Ionization = 1004.2 10−3

0.10

= 4.2%

HCOOH(aq) + H2O(l) HCOO−(aq) + H3O

+(aq)

Whole (Initial)

Part (@ Equi)

[HCOOH], M [H3O+], M [HCOO−], M

I 0.10 0 0

C −4.2 10-3 +4.2 10-3 +4.2 10−3

E 0.0958 = 0.10 4.2 10−3 4.2 10−3

AcidsandBases

Calculating pH from Ka

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

Problem: For acetic acid at 25°C is Ka =1.8 10−5

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

[H3O+] [C2H3O2−]

[HC2H3O2]Ka =

[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.

[x] [x][0.3]

=

AcidsandBases

Calculating pH from Ka

(x)2

(0.30)1.8 10−5 =

(1.8 10−5) (0.30) = x2

5.4 10−6 = x2

[H3O+] = 2.3 10−3 = x

[H3O+] [C2H3O2−]

[HC2H3O2]Ka =

• pH = −log [H3O+] = − log x

• pH = −log (2.3 10−3)

• pH = 2.64

What is the pH?

AcidsandBases

(b) Calculate the pH of 0.50 M Lactic Acid.

2002B Q1

AcidsandBases

Polyprotic Acids• Have more than one acidic proton.

• If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

AcidsandBases

Bases react with water to produce hydroxide ion.

Dissociation Constants for Weak Bases

The generalized equilibrium constant expression for these reactions is

[HB+] [OH−][B]

Kb =

where Kb is the base-dissociation constant.

AcidsandBases

Weak Bases

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

pKb

4.7

8.8

8.0

3.4

6.7

3.7

6.5

pKb

Salts of weak bases are strong acids

AcidsandBases

NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

pH & pOH for Weak Acids & Bases

Kb = 1.8 x 10-5

[NH3], M [NH4+], M [OH−], M

Initially 0.15 0 0

Change - x + x + x

[NH4+] [OH−]

[NH3]Kb = = 1.8 10−5 =

At Equilibrium 0.15 - x 0.15

x x

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

(x)2

(0.15)

AcidsandBases

pH of Basic Solutions

(1.8 10−5) (0.15) = x2

2.7 10−6 = x2

1.6 10−3 = x2

(x)2

(0.15) =

[NH4+] [OH−]

[NH3]Kb = = 1.8 10−5

[OH−] = 1.6 10−3 M

pOH = −log (1.6 10−3)

pOH = 2.80

pH = 14.00 − 2.80

pH = 11.20

What is pH of soln?What is x?

AcidsandBases

2002A Q1(a-b)

AcidsandBases

2005 A acid base equilibria

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2005 A acid base equilibria ANSWERS

AcidsandBases

c) ii

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2005 B acid base equilibria

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2005 B acid base equilibria ANSWERS

a)

b)

c)

AcidsandBases

d)i

d)ii

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d)iii

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2001 a/b equilibria titration

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2001 answers

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2002

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2002 B

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2003 A

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2003 B

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2005 A

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AcidsandBases

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

Hydroxide is a much stronger base than H2O, so the equilibrium favors the left side (K<<<1).