pH = - log [H3O+]

[H3O+] = 10- pH mol/L

For pure water at 25oC

pH = - log (1.0 x 10-7) = 7.00

For a change in pH by 1, H3O+ concentration changes by 10

Higher pH, lower H3O+ concentration

pH of pure water is 7

pH of an acidic solution is less than 7

pH of a basic solution is greater than 7

pOH = - log [OH-]

pKw = - log Kw

pKw = 14.00 at 298 K

[H3O+ (aq)] [OH- (aq)] = Kw

- log[H3O+ (aq)] - log[OH- (aq)] = - log Kw

pH + pOH = pKw

At 298 K pH + pOH = 14.00

Strengths of Acids and Bases

The pH of 0.10 M HCl(aq) will be recorded as close to 1

The pH of a 0.10 M solution of CH3COOH(aq) solution is recorded as ~ 3.

H3O+(aq) concentration in 0.10 M HCl(aq) is greater than that in 0.10 M CH3COOH(aq)

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

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

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

Ka = [CH3COOH(aq)]

[H3O+(aq)][CH3COO- (aq)]

At 298 K, Ka for CH3COOH(aq) = 1.8 x 10-5

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

Kb = [NH3(aq)]

[NH4+(aq)][OH- (aq)]

At 298 K, Kb for NH3(aq) = 1.8 x 10-5

Ka: acidity constant or acid dissociation constant

Kb: basicity constant or base dissociation constant

The proton donor strength of an acid is measured by the value of Ka; higher Ka, stronger the acid

The proton acceptor strength of a base is measured by Kb; higher Kb, stronger the base

pKa = - log Ka

pKb = - log Kb

The larger the pK values, weaker the acid or base

Ka Kb = [H3O+(aq)] [OH- (aq)] = Kw

Or pKa + pKb = pKw

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

Kb = [NH3(aq)]

[NH4+(aq)][OH- (aq)]

NH4+(aq) + H2O(l) H3O+(aq) + NH3 (aq)

Ka = [NH4+( aq)]

[NH3 (aq)][H3O+(aq)]

Relationship between conjugate acid/base pairs

The stronger the acid/base, the weaker its conjugate base/acid

pKa - pink

pKb - blue

HClO2(aq)/ ClO2-(aq)

HOCl(aq)/ OCl-(aq)

CH3COOH(aq)/ CH3COO-(aq)

NH4+(aq) / NH3(aq)

CH3NH3+(aq) / CH3NH2(aq)

Using tabulated Ka and Kb values determine which species is stronger as an acid or base

1) as acid HF(aq) or HIO(aq)

2)as base C6H5NH2(aq) or (CH3)3N(aq)

3) as acid C6H5NH3+(aq) or (CH3)3NH+(aq)

Molecular Structure and Acid Strength

The more polar or weaker the H-A bond, the stronger the acid

Effect of bond strength

HF < HCl < HBr < HI

H - I bond is weakest

H2O < H2S < H2Se < H2Te

H-Te bond weakest

For an acid HA, greater the electronegativity of A, stronger the acid

electronegativity difference

N-H 0.8

F-H 1.8

HF is an acid in water, NH3 is a base

Solutions of Weak Acids/Bases

For a strong acid and base; assume that deprotonation/protonation reactions go to completion

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

pH = - log [H3O+ (aq)]

Knowing the concentration of HCl, can determine pH

For weak acids/bases, set up equilibrium table to determine the H3O+ (aq) / OH- (aq) concentration at equilibrium, knowing the value of Ka/Kb.

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

Calculate the pH and percentage deprotonation of 0.10 M CH3COOH(aq) given that Ka is 1.8 x 10-5.

Ka = [CH3COOH(aq)]

[H3O+(aq)][CH3COO- (aq)]

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

Initial 0.10 0 0

Change - x x x

Equilibrium 0.10 - x x x

1.8 x 10-5 = x2/(0.10 - x)

1.8 x 10-5 = x2/(0.10 - x)

Since Ka is so small, assume that x << 0.10

1.8 x 10-5 ≈ x2/(0.10)

x = 1.3 x 10-3 M

[H3O+(aq)] = 1.3 x 10-3 M

pH = 2.89

% deprotonation = 100% x ([CH3COO-(aq)]/[CH3COOH]initial)

= 100% x (1.3 x 10-3 M)/(0.10 M) = 1.3 %

Note: x < 5% of 0.10 , OK to make this approximation

For a weak base

B(aq) + H2O(l) HB+ (aq) + OH- (aq)

Use a similar approach to determine pOH knowing Kb, and then determine pH

Determine the pH and percentage protonation of a 0.20 M aqueous solution of methylamine, CH3NH2. The Kb for CH3NH2 is 3.6 x 10-4.

pH = 11.9

% protonation = 4.2%

pH of Salt Solutions

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

“neutralization” reaction

If a 0.3M solution of CH3COOH(aq) is added to a 0.3M solution of NaOH, pH of resulting solution is not 7.0 but ~ 9.0

Solution of a salt is a solution of an acid (usually the cation) and a base (usually the anion), and the pH depends on their relative strength.

CH3COO-(aq) determines the pH of the solution

Ni(H2O)62+(aq) + H2O(l) H3O+(aq) + Ni(H2O)5(OH)+(aq)

Estimate the pH of 0.15 M NH4Cl(aq).

Kb (NH3(aq)) = 1.8 x 10-5

NH4+ (aq) is an acid and Cl- (aq) is neutral; expect pH < 7

NH4+ (aq) + H2O(l) H3O+ (aq) + NH3(aq)

Ka = [NH4+ (aq) ]

[H3O+(aq)][NH3(aq)]

NH4+ (aq) NH3 (aq) H3O+(aq)

Initial 0.15 0 0

Change -x x x

Equilibrium 0.15-x x x

Ka (NH4+ (aq)) =

Kw

Kb (NH3 (aq))

5.6 x 10-10 =x2

0.15 - x

Assume x << 0.15

x ≈ 9.2 x 10-6 (agrees with the assumption)

pH = - log(9.2 x 10-6 ) = 5.04

Polyprotic Acids & Bases

A polyprotic acid can donate more than one H+

Carbonic acid: H2CO3(aq); dissolved CO2 in water

Sulfuric acid: H2SO4(aq)

Phosphoric acid: H3PO4(aq)

A polyprotic base: can accept more than one proton

Carbonate ion: CO32-(aq)

Sulfate ion: SO42-(aq)

Phophate ion: PO43-(aq)

Treat each step of protonation or deprotonation sequentially

H2CO3 (aq) + H2O(l) H3O+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7

HCO3-(aq) + H2O(l) H3O+(aq) + CO3

2-(aq) Ka2 = 4.8 x 10-11

Typically:

Ka1 >> Ka2 >> Ka3 >>…

Harder to loose a positively charged proton from a negatively charged ion, because of attraction between opposite charges.