Acid-Base Acid-Base

EquilibriumEquilibrium

VCE Chemistry VCE Chemistry

Unit 4: Chemistry at WorkUnit 4: Chemistry at Work

Area of Study 1 – Industrial ChemistryArea of Study 1 – Industrial Chemistry

BRØNSTED-LOWRY THEORY

ACID proton donor HCl H+ (aq) + Cl- (aq)

BASE proton acceptor NH3 (aq) + H+ (aq) NH4+ (aq)

Conjugate systems

Acids are related to bases ACID PROTON + CONJUGATE BASE

Bases are related to acids BASE + PROTON CONJUGATE ACID

For an acid to behave as an acid, it must have a base present to accept a proton...

HA + B BH+ + A-

acid base conjugate conjugate acid base

ACIDS AND BASESACIDS AND BASES

STRONGACIDS completely dissociate (split up) into ions in aqueous solution

e.g. HCl H+ (aq) + Cl- (aq) MONOPROTIC 1 replaceable H

HNO3 H+ (aq) + NO3- (aq)

H2SO4 H+ (aq) + HSO4- (aq)

HSO4- H+ (aq) + SO4

2- (aq) DIPROTIC 2 replaceable H’s

STRONGBASES completely dissociate into ions in aqueous solution

e.g. NaOH Na+(aq) + OH-(aq)

STRONG ACIDS AND BASESSTRONG ACIDS AND BASES

Weak acids partially dissociate into ions in aqueous solution

When a weak acid dissolves in water an equilibrium is set up

e.g. ethanoic acid

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

The water stabilises the ions

The weaker the acid: the less it dissociates the more the equilibrium lies to the left.

The relative strengths of acids can be expressed as Ka or pKa values

For ethanoic acid: [CH3COO-] [H3O+]

Ka =

[CH3COOH]

WEAK ACIDSWEAK ACIDS

Partially react with water to give ions in aqueous solution

When a weak base dissolves in water an equilibrium is set up

e.g. ammonia

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

The weaker the base the less it dissociatesthe more the equilibrium lies to the left

The relative strengths of bases can be expressed as Kb or pKb values.

For ammonia: [NH4+] [OH-]

Kb =

[NH3]

WEAK BASESWEAK BASES

HYDROGEN ION CONCENTRATION [HHYDROGEN ION CONCENTRATION [H++]]

pH Hydrogen ion concentration can be converted to pH pH = - log10 [H+]

To convert pH into hydrogen ion concentration [H+] = 10-pH

pOH An equivalent calculation for bases convertsthe hydroxide ion concentration to pOH pOH = - log10 [OH-]

In both the above, [ ] represents the concentration in mol L-1

STRONGLY ACIDIC

100 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14

10-14 10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 10-0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14pH

OH¯

[H+]

WEAKLY ACIDIC

NEUTRAL STRONGLY ALKALINE

WEAKLY ALKALINE

IONISATION CONSTANT OF WATER - KIONISATION CONSTANT OF WATER - Kww

Despite being covalent, water conducts electricity to a very small extent.

This is due to its self ionisation ... H2O (l) H+ (aq) + OH- (aq)

Applying the equilibrium law Kc = [H+] [OH-][ ] is the equilibrium concentration in mol L-1 [H2O]

As the dissociation is small, the water concentration is large (~ 56 M) compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as constant.

This “constant” is combined with(Kc) to get a new constant (Kw). Kw = [H+] [OH-]

= 10-7 x 10-7

= 1 x 10-14 M2 (at 25°C)

Because Kw is based on an equilibrium, it VARIES WITH TEMPERATURE

The value of Kw varies with temperature because it is based on an equilibrium.

Temperature / °C 0 20 25 30 60

Kw / M2 0.11 0.68 1.0 1.47 5.6

H+ / M 0.33 0.82 1.0 1.27 2.37pH 7.48 7.08 7 6.92 6.63

What does this tell you about the equation H2O (l) H+ (aq) + OH- (aq) ?

• Kw gets larger as the temperature increases

• This means the concentration of H+ and OH- ions gets greater

• This means the equilibrium has moved to the right

• If the concentration of H+ increases then the pH decreases

• pH decreases as the temperature increases

Because the equation moves to the right as the temperature goes up, it must be an ENDOTHERMIC process

IONISATION CONSTANT OF WATER - KIONISATION CONSTANT OF WATER - Kww

RELATIONSHIP BETWEEN pH AND pOHRELATIONSHIP BETWEEN pH AND pOH

Because H+ and OH- ions are producedin equal amounts when water dissociates [H+] = [OH-] = 1 x 10-7 Mtheir concentrations will be the same.

Kw = [H+] [OH-] = 1 x 10-14 M2

take logs of both sides log [H+] + log [OH-] = -14

multiply by minus - log [H+] - log [OH-] = 14

change to pH and pOH pH + pOH = 14 (at 25°C)

N.B. As they are based on the position of equilibrium and that varies withtemperature, the above values are only true if the temperature is 25°C

Neutral solutions may be regarded as those where [H+] = [OH-].

Therefore a neutral solution is pH 7 only at a temperature of 25°C

Kw is constant for any aqueous solution at the stated temperature

BUFFER SOLUTIONS – A BUFFER SOLUTIONS – A BRIEFBRIEF INTRODUCTIONINTRODUCTION

Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.”

Acidic Buffer (pH < 7) made from a weak acid + its sodium salt ethanoic acid sodium ethanoate

Alkaline Buffer (pH > 7) made from a weak base + its chloride ammonia ammonium chloride

Uses Standardising pH metersBuffering biological systems (e.g. in blood)Maintaining the pH of shampoos

pH IN THE BODYpH IN THE BODY

• A number of biochemical reactions involve acid-base chemistry. Without a means of regulating pH, the pH in the body can fluctuate to dangerous levels.

• For example the blood supply must be maintained within the narrow pH range of 7.35 to 7.45.

• Illnesses and diseases such as pneumonia, diabetes and emphysema can cause blood pH to fall leading to acidosis while hyperventilation can lead to an increase in blood pH (alkalosis).

• Natural buffers in the body help maintain the pH within this narrow range.

• They are made from a mixture of weak acids and their conjugate base.

• One important buffer is made up of carbonic acid (H2CO3) and the hydrogen carbonate ion (HCO3

-):

H2CO3 (aq) HCO3- (aq) + H+ (aq)

• If H+ ions were added, a net back reaction occurs removing most of these ions.

• If OH- ions were added, they react with H+ ions. A net forward reaction results producing more H+ ions. The OH- ions have effectively reacted with carbonic acid

pH Calculations

VCE Chemistry VCE Chemistry

Unit 4: Chemistry at WorkUnit 4: Chemistry at Work

Area of Study 1 – Industrial ChemistryArea of Study 1 – Industrial Chemistry

Calculating pH - Calculating pH - Strong Acids and BasesStrong Acids and Bases

Strong acids and alkalis completely dissociate in aqueous solution

It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.

Calculate the pH of 0.02 M HCl

HCl completely dissociates in aqueous solution HCl H+ + Cl-

One H+ is produced for each HCl dissociating so [H+] = 0.02 M

pH = - log [H+]pH = - log [0.02] = 1.7

Calculate the pH of 0.1 M NaOH

NaOH completely dissociates in aqueous solution NaOH Na+ + OH-

One OH- is produced for each NaOH dissociating [OH-] = 0.1 M

The ionic product of water (at 25°C) Kw = [H+][OH-]

Therefore [H+] = Kw / [OH-] pH = - log [H+] = 13

Calculating pH - Calculating pH - Weak AcidsWeak Acids

A weak acid, HA, dissociates as follows HA H+ + A- (1)

Applying the Equilibrium Law Ka = [H+ ] [A-] (2)

[HA]

The ions are formed in equal amounts, so [H+] = [A- ]

therefore Ka = [H+]2 (3)

[HA]

Rearranging (3) gives [H+]2 = [HA] Ka

therefore [H+] = [HA] Ka

pH = -log [H+]

ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration.

A weak acid is one which only partially dissociates in aqueous solution

Calculating pH - Calculating pH - Weak AcidsWeak Acids

Calculate the pH of a weak acid HA of concentration 0.1 M ( Ka = 4 x 10-5 M ) HA dissociates as follows HA H+ + A-

Dissociation constant for a weak acid Ka = [H+] [A- ]

[HA] Substitute for A- as ions are formed in [H+] = [HA] Ka

equal amounts and the rearrange equation

ASSUMPTIONHA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration

[H+] = 0.1 x 4 x 10-5 M

= 4.00 x 10-6 M

= 2.00 x 10-3 M

ANSWER pH = - log [H+] = 2.70

Calculate the pH of a mixture of 25 mL of 0.1 M NaOH is added to 20 mL of 0.1 M HCl

1. Original moles of H+ = 0.1 x 20/1000 = 2 x 10-3 mol2. Original moles of OH- = 0.1 x 25/1000 = 2.5 x 10-3 mol

Moles of excess OH- = 5 x 10-4 mol3. Final volume = 45 mL = 0.045 L4. [OH-] = 5 x 10-4 / 0.045 = 0.0111 M

pOH = 1.955. pH = 14 – 1.95 = 12.05

pH of MixturespH of Mixtures

Strong Acids and Strong Alkalis (Either in EXCESS)Strong Acids and Strong Alkalis (Either in EXCESS)

Weak Acid and EXCESS Strong AlkaliWeak Acid and EXCESS Strong Alkali

Calculate the pH of a mixture of 25 mL of 0.1 M NaOH and 22 mL of 0.1 M CH3COOH

1. Original moles of H+ = 0.1 x 22/1000 = 2.2 x 10-3 mol2. Original moles of OH- = 0.1 x 25/1000 = 2.5 x 10-3 mol

Moles of excess OH- = 3 x 10-4 mol3. Final volume = 47 mL = 0.047 L4. [OH-] = 3 x 10-4 / 0.047 = 6.38 x 10-3 M

pOH = 2.205. pH = 14 – 2.20 = 11.80

Calculations Involving Acidity ConstantsCalculations Involving Acidity Constants

Chloroacetic acid is a weak monoprotic acid with a Ka of 1.3 x 10-3 M. For a 1.0 M solution of chloroacetic acid, calculate:

a. The pH

b. The percentage hydrolysis

CH2ClCOOH (aq) CH2ClCOO- (aq) + H+ (aq)

Ka = [CH2ClCOO-] [H+] / [CH3COOH] = 1.3 x 10-3 M

• From the equation, for every mole of chloroacetic acid that ionises one mole of [CH2ClCOO-] and one mole H+ will be formed. Thus [CH2ClCOO-] = [H+].

[H3O+]2 / [CH3COOH] = 1.3 x 10-3 M

• Because chloroacetic acid is a weak acid, it only ionises to a small extent.

• You can assume that its equilibrium concentration is approximately that of its original concentration. [CH3COOH] = 1.0 M.

• Hence you get:

• [H3O+]2 / 1.0 M = 1.3 x 10-3 M

• [H3O+]2 = 1.3 x 10-3

• [H3O+] = 0.036 M

• pH = -log [0.036] = 1.4

• The percentage hydrolysis (percentage ionisation) measures the extent of the reaction.

• It is the fraction of the acid that is ionised (a ratio of the conjugate base and the acid).

• Percentage hydrolysis = [CH2ClCOO-] / [CH3COOH] x 100

• Percentage hydrolysis = 0.036 / 1.0 x 100

• Percentage hydrolysis = 3.6%

Calculations Involving Acidity ConstantsCalculations Involving Acidity Constants