CM4106 Chemical Equilibria & Thermodynamics

Lesson 2Acid-Base EquilibriaA Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/

Fundamentals:

Acid Base

1. proton H+ donor2. electron acceptor

(vacant orbital)

1. proton H+ acceptor2. electron donor

(lone pair)

1. Identify acids/ bases

Acid Conjugate Base– H+

+ H+

Base Conjugate Acid+ H+

– H+

2. Identify conjugate acids/ bases

Fundamentals: 3. Identify stronger acids/ bases

(A) In any acid-base reaction, the equilibrium will favor the reaction where the stronger acid reacts with the stronger base. i.e. (In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.)

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

CH3CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + CH3CO2- (aq)

(B) The stronger an acid, the weaker its conjugate base.

The stronger a base, the weaker its conjugate acid.

p = -log10

At 25ºC:

pH + pOH = pKw = 14

pKa + pKb = 14

[H+][OH–] = 10-14

Ka x Kb = Kw = 10-14

Important relations

Fundamentals:

(I) Calculations

pH, pKa, [H+], Ka pOH, pKb, [OH–], Kb

Step 1: Determine what is present in the solution.

(A)Acid : Strong Acid vs Weak Acid Monoprotic Acid / Diprotic Acid / Triprotic AcidConcentration of Acid

(B) Base: Strong Base vs Weak BaseMonoprotic base / Diprotic base / Triprotic baseConcentration of Bas

Step 2: Use the appropriate equations for the respective species.

Take note of concentration of acid / bases (For dilute solutions, we need to take into consideration of [H+] / [OH-] from auto-ionization from water)

pH of Acid/ Base

Strong acids dissociate completely into ions in aqueous solution.

[H+] = [HA]

Strong bases dissociate completely into ions in aqueous solution.

[OH-] = [B]

HA(aq) ⇌ H+(aq) + A-

(aq)

I [HA] 0 0

C - x + x + x

E [HA] - x + x + x

B + H2O ⇌ BH+ + OH-

I [B] - 0 0

C - x - + x + x

E [B] - x - x x

Strong Acid

Strong Base

Weak Acid

Weak Base

Ka =x2

([HA] – x)

Kb =x2

([B] – x)

pH of Acid/ Base

1. Determine if acid/ base is strong or weak (more common)

2. For weak acids, if asked to determine Ka, pH or [H+],

you can save time by using the formula:

[H+] = Ka × c [OH–] = Kb × cExample:Calculate the pH of 0.50M HF solution at 25ºC where the Ka = 7.1 x 10-4

Assumption: x is negligible

[H+][F-][HF]

Ka =x2

(0.50 – x)= = 7.1 x 10-4

Assumption: For weak acids, x must be very small 0.50 – x ≈ 0.5

x2

0.50= 7.1 x 10-4 x = [H+] = 0.0188 M

pH = 1.73 (to 2 d.p.)

Assumption is valid; x < 5% of [HF]initial

Calculate the pH of the following solutions at 298K

0.10 mol dm−3 CH3COOH (pKa = 4.75) Weak acid solution (monoprotic acid)

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

(aq)

Initial (M) 0.10 - 0.00 0.00

Change (M) - x - + x + x

Eqm (M) 0.10 - x - + x + x

Assumption: For weak acids, x must be very small 0.10 – x ≈ 0.10

x = [H3O+] = 1.338 x 10-3 M

pH = 2.87 ( 2 d.p.)

[CH3COO][H3O+]

[CH3COOH]Ka =

x2

(0.10 – x)= 10 4.75= = 1.79 x 10-5

x2

0.10 = 1.79 x 10-5 Assumption is justified, x < 5% of [CH3COOH]initial

Concentration: 2 s.f.pH: 2 d.p.

Calculate the pH of the following solutions at 298K

0.30 mol dm3 ethylamine, CH3CH2NH2 (pKb= 3.27)

pH = 14.00 – 1.896 pH = 12.10 (2 d.p.)pOH = - log10[OH]

= - log10[0.01269 ]= 1.896

CH3CH2NH2 (aq) + H2O (l) ⇌ CH3CH2NH3+(aq) +

OH-(aq)

Initial (M) 0.30 - 0.00 0.00

Change (M) -x - + x + x

Eqm (M) 0.30 - x - + x + x

[CH3CH2NH3+][OH-]

[CH3CH2NH2]Kb =

x2

(0.30 – x)= 10 3.27= = 5.37 x 104

Assumption: For weak bases, x must be very small 0.30 – x ≈ 0.30

x2

0.30 = 5.37 x 10-4

x = [OH] = 0.01269 M

Assumption is justified, x < 5% of [CH3CH2NH2]initial

Concentration: 2 s.f.pH: 2 d.p.

Weak base solution (monoprotic base)

pH of salt solutions

Basic Salt

Acidic Salt

CH3COONa (aq) → CH3COO- (aq) + Na+ (aq)

conjugate base

CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)

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

NH4Cl (aq) → NH4+ (aq) + Cl- (aq)

conjugate acid

Salt solutions can be (i) neutral (ii) weak acids or (iii) weak bases

pH > 7

pH < 7