Applying Acid-Base Dissociation Constants

A. Strong Acids and Bases- Equilibrium expressions are not part of the calculations involved with the ionization of strong acids and

strong bases because the equilibrium constant are either very large or undefined- There is no significant difference between the analytical concentration and the equilibrium concentration of

the ionization products- Common strong acids and bases

ACIDS BASESHClHBrHIHClO4

HNO3

H2SO4

Hydrochloric AcidHydrobromic AcidHydriodic AcidPerchloric AcidNitric AcidSulfuric Acid

LiOHNaOHKOHBa(OH)2

Lithium hydroxideSodium HydroxidePotassium HydroxideBarium Hydroxide

B. Weak Acids and Bases- When a weak acid or weak base is dissolved in water, partial dissociation occurs

HA + H2O ↔ H3O+ + A-

- HA = weak acid; Ka = acid dissociationKa = [H3O+] [A-]/ [HA]

- For weak base, Kb = base dissociation constantB + H2O ↔ BH+ + OH-

Kb = [BH+] [OH-]/ [B]Polyprotic Acids = acids that can donate more than one proton

Ex. H3PO4 + H2O ↔ H3O+ + H2PO4- Ka1 = [H3O + ] [H 2PO4

- ] [H3PO4]

H2PO4- + H2O ↔ H3O+ + HPO4

2- Ka2 = [H3O + ] [ HPO 42- ]

[H2PO4- ]

HPO42- + H2O ↔ H3O+ + PO4

3- ka3 = [H3O + ] [ PO 43- ]

[HPO42- ]

Polyequivalent Base =bases that can accept more than one proton

Ex. CO32- + H2O ↔ HCO3

- + OH- Kb1 = [HCO3- ] [ OH - ]

[CO32-]

HCO3- + H2O ↔ H2CO3 + OH- Kb2 = [H2CO3] [ OH - ]

[HCO3-]

- The subscripts are used to identify the individual ionization steps and K1>K2>K3

Relation Between Ka and Kb for Conjugate Pairs- The Ka and Kb values are used to measure acid and base strength

KaKb = Kw

C. Calculation of pHpH = - log [H3O+]

- When an acid HA dissolves in water, both acid and water are suppliers of H3O+:HA + H2O ↔ H3O+ + A- (a)H2O + H2O ↔ H3O+ + OH- (b)

- When a base B dissolves in water, both the base and water are suppliers of OH-: B + H2O ↔ BH+ + OH- (c)H2O + H2O ↔ H3O+ + OH- (d)

- Calculation of pH depends on three cases

For AcidsCase 1. The acid is the major supplier of H3O+

o This means that the determination of hydronium ion concentration is based on the equilibrium expression for reaction (a)

Case 2. Water is the major supplier of H3O+

o The hydronium ion concentration can be determined by solving the equilibrium expression for reaction (b)

Case 3. Both the acid and the water are major suppliers of H3O+

o Both reactions mustbe considered

For BasesCase 1. The base is the major supplier of OH-

o This means that the determination of hydroxide ion concentration is based on the equilibrium expression for reaction (c)

Case 2. Water is the major supplier of OH-

o The hydroxide ion concentration can be determined by solving the equilibrium expression for reaction (d)

Case 3. Both the base and the water are major suppliers of OH-

o Both reactions (c) and (d) must be considered

Calculation of pH of strong acids and bases- Identify which of the above cases applies.- For acids

Case Major Supplier of H3O+ Necessary condition123

Strong acidWaterBoth

CHX >> 10-7

CHX << 10-7

CHX = 10-7

Case 1. The acid is the major supplier of H3O+.[H3O+] = CHX

Case 2. The water is the major supplier of H3O+.[H3O+] = √ Kw

Case 3. Both acid and water are the major supplier of H3O+.[H3O+] - CHX [H3O+] - Kw = 0

- For basesCase Major Supplier of OH- Necessary condition

123

Strong baseWaterBoth

CB >> 10-7

CB << 10-7

CB = 10-7

Case 1. The base is the major supplier of OH-

[OH-] = CB

Case 2. The water is the major supplier of OH-.[OH-] = √ Kw

Case 3. Both acid and water is the major supplier of OH-.[OH-]- CB [OH-] - Kw = 0

To solve for pH, use the equation belowpH + pOH = pKw

where pOH = - log [OH-]

Calculation of pH of weak acids and bases- For weak acids:

Case Major Supplier of H3O+ Necessary condition123

Weak acidWaterBoth

Ka x CHA >> 10-7

Ka x CHA << 10-7

Ka x CHA = 10-7

Case 1. The weak acid is the major supplier of the H3O+.[H3O+]2 + Ka[H3O+] - KaCHA = 0

If CHA/Ka ≥ 10-3,[H3O+] = √ Ka CHA

- For weak baseso techniques used on weak acids are the same ones that are used to calculate hydroxide ion

concentration in solutions of weak bases.Case Major Supplier of OH- Necessary condition

123

Weak baseWaterBoth

Ka x CHA >> 10-7

Ka x CHA << 10-7

Ka x CHA = 10-7

o Case 1. The weak base is the major supplier of the OH-.[OH-]2 + Kb[OH-] – KbCB = 0

If CB/Kb ≥ 10-3,[OH-] = √ Kb CB

Calculation of pH of conjugate acid-base pairso There are two equilibria involved:

HA + H2O ↔ H3O+ + A- (e)A- + H2O ↔ HA + OH- (f )

o The ionization of water is neglected, and [H3O+] is taken from reaction (e)Ka = [H3O + ][A - ]

[HA] Where:

[HA] = CHA - [H3O+] + [OH-][A-] = CNaA + [H3O+] - [OH-]

Ka =[H3O + ]( C NaA + [H3O + ] - [OH - ]) CHA - [H3O+] + [OH-]

Solving for [H3O+],[H3O+]3 + (CNaA + Ka) [H3O+]2 - (KaCHA + Kw) [H3O+] - KaKw = 0

o If CHA & CNaA ≥ 10-3 and Ka & Kb ≤ 10-3, then CHA and CNaA are larger compared to the difference between [H3O+] and [OH-], therefore

[HA] = CHA

[A-] = CNaA

Ka becomes,Ka =[H3O + ]C NaA

CHA o Note also that conjugate acid-base do not react with each other

pH buffers- a mixture of weak acid or base and its conjugate- resists changes in pH upon dilution or addition of acids or base

o effect of dilution:pH = pKa - log [HA]/[A-] pH depends on the ratio of the concentrations of acids and conjugate bases rather than

their absolute value

o effect of addition of acids and bases

Buffer Capacity, β- depends on both the concentration of components and the concentration ratio- Defined as the quantity of strong acid or strong base needed to cause 1.0 L of buffer to undergo a pH change

of 1.0 unit- To determine the useful pH range of a buffer use the previous equation

[HA] / [A-] pH from the equation 1 / 10 pHmax = pKa - log 1/10 = pKa + 1 10 / 1 pH min = pKa - log 10/1 = pKa - 1Therefore the useful pH range is pKa ± 1.

- In selecting a buffer for a given application there are two considerations to be considered:1. The desired pH and2. The chemical compatibility of the buffer components with the sample

- Preparing a buffer:o By combining the calculated quantities of an acid-base conjugate pairo In cases one of the conjugate pair is unavailable, combining of excess of the available weak acid or

baser with an appropriate amount of strong base or acid can be done.o Preparation of buffers uses the aid of pH meter

Prepared by:

Raissa E. DycocoInstructor