CM4106 Chemical Equilibria & Thermodynamics

Lesson 3 (Part 1)Additional Aspects of Acid-Base Equilibria(Topic 3.1 – 3.5)

A Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/

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 for acids OR bases

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)

(I) pH calculations of Acid OR 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)

(I) pH calculations of Acid OR Base1. 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 × c

Example: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 298K0.10 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 10-5Assumption is justified, x < 5% of [CH3COOH]initial

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

Remember to validate Assumption

(II) Calculations for mixture of acids AND bases

pH, pKa, [H+], α pOH, pKb, [OH–]

Step 1: Determine what is present in the solution.

(A) acidbase

(B) buffer

(C) salt

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

Stoichiometric amounts of acid and base – salt solution

Non-stoichiometric amounts of acid and base – likely to be a buffer

(II) (a) or (c) pH calculation 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

COVERED IN GREATER DETAILS IN TOPIC 3.8

(II) (b) pH of Buffer SolutionsAcid Buffer

Basic Buffer

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

CH3COONa(aq) → CH3COO-(aq) + Na+(aq)

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

NH4Cl(aq) → NH4+(aq) + Cl-(aq)

pH = pKa + lg[salt][acid]

pOH = pKb + lg [salt][base]

weak acid + conjugate base

weak base + conjugate acid

weak acid

conjugate base

Assume negligible dissociation of acid due to common ion effect

Assume full dissociation of salt (strong electrolyte)

Assume negligible dissociation of base due to common ion effectweak base

conjugate acidAssume full dissociation of salt (strong electrolyte)

(II) (b) pH of Buffer Solutions1. Determine if buffer is acidic/ basic

2. When using H-H equation, you can save time by just calculating no. of moles of salt and acid/base since total volume is the same and cancels out.

3. Need to be sensitive to the condition (maximum buffering capacity)

where [salt] = [acid] which simply means pH = pKa

Similarly for basic buffer,

[salt] = [base] pOH = pKb

nsalt/V

nacid/VpH = pKa + lg

Useful for MCQs!

nsalt/V

nbase/VpOH = pKb + lg

Sample Calculation • Find the pH of a buffer made from adding 3.28 g of sodium

ethanoate, CH3COO-Na+ to 1.00 dm3 of 0.0100 M ethanoic acid, CH3COOH . (Ka = 1.80 x 10-5)

[acid]

[salt]lgKH a pp

pH = 5.3466 = 5.347 (3 d.p.)

Useful InformationMolar Mass of CH3COO-Na+ = 82.03 g/mol

Page 46

Amount of CH3COO-Na+ present in 3.28 g = (3.28 g / 82.03) = 0.0400 mol

[CH3COO-Na+] = 0.0400 M[CH3COOH] = 0.0100 M

0.0100

0.0400lg)1080.1lg( 5

[acid]

[salt]lgKH a pp

Sample Calculation• Find the pH of a buffer made from adding 3.28 g of sodium ethanoate,

CH3COO-Na+ to 1.00 dm3 of 0.0100 M ethanoic acid, CH3COOH. (Ka = 1.80 x 10-5)

= 5.347 (3 d.p.)

Useful InformationMolar Mass of CH3COO-Na+ = 82.03 g/mol

n(CH3COO-Na+) present in 3.28 g = (3.28 g / 82.03)

= 0.0400 mol[CH3COO-Na+] = 0.0400 M[CH3COOH] = 0.0100 M

An alternative approach – ICE Table

CH3COOH(aq) + H2O (l) H⇌ 3O+ (aq) + CH3COO (aq)

Initial [ ] (M) 0.0100 M - 0 0.0400 M

Change [ ] (M) - x - + x + x

Equilibrium [ ] (M)

0.0100 - x - x 0.0400 + x

[H3O+][CH3COO-][CH3COOH]

Ka =x (0.0400 + x)(0.0100 – x)

= = 1.80 x 10-5

Solve for x, x = [H3O+] = 4.497 x 10-6 M

pH = - lg [H3O+] = - lg (4.497 x 10-6) = 5.3470

Calculating pH change after small amounts of acid / base is added to buffer solutions

1) Determine new [salt]new and [acid]new / [base]new

How?a) Determine n(acid / base) added to buffer solutionb) Determine the change to n(salt)buffer and n(acid)buffer /

n(base)buffer by considering the neutralization action of a buffer

c) Determine new total volume of the buffer solution

new

newanew [acid]

[salt]lgKH pp

CM4106 Chemical Equilibria &Thermodynamics

Review of Pre-Quiz 3Topic 3 (3.1 – 3.5)

Additional Aspects of Acid-Base Equilibria

Buffer Solutions, pH calculation, Maximum Buffer Capacity

Question 1 (a)• Hypochlorous acid, HOCl, is a weak acid

commonly used as a bleaching agent. It dissociates in water as represented by the equation below.

HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq)

Ka = 3.2 x10-8

a) Write the equilibrium-constant expression for the dissociation of HOCl in water.

][HOC

]O][H[OCK 3

a l

l

Question 1(b)b) Calculate the molar concentration of H3O+

in a 0.14 M solution of HOCl.

y)(0.14

yK

2

a

)14.0(102.3

28

y

y

Solve for y,

y = 6.7 x 10-5

[H3O+] = 6.7 x 10-5 M (2 s.f.)

Note: units must be present

HOCl (aq) + H2O (l) ⇌ H3O+ (aq) + OCl- (aq)

Initial [ ] (M) 0.14 - 0 0

Change [ ] (M) - y - + y + y

Equilibrium [ ] (M)

0.14 – x - x y

HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq) Ka = 3.2 x10-8

Question 1 (c)c) A mixture of HOCl and sodium hypochlorite

(NaOCl) can be used as a buffer. Write two equations to show how this buffer solution can control pH.

• Addition of H3O+:

OCl(aq) + H3O+(aq) HOCl(aq) + H2O(l)

• Addition of OH : HOCl(aq) + OH(aq) OCl (aq) + H2O(l)

Note: state symbols must be present

Question 1(d)HOCl reacts with NaOH according to the reaction represented below:

HOCl(aq) + OH-(aq) OCl-(aq) + H2O(l)

Bob, a budding young chemist, decides to make a buffer by adding a volume of 10.0 mL of 0.56 M NaOH to 50.0 mL of 0.14 M HOCl solution. Assume that the volumes are additive.

HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq) Ka = 3.2 x10-8

Preliminary Considerations:

Reaction of NaOH with HOCl will form NaOCl (salt containing conjugate base OCl-).

Hence, if HOCl is in excess, the final mixture will contain some excess HOCl that is unreacted and some NaOCl formed as a result of the neutralization reaction. BUFFER SOLUTION (Mixture of acid and conjugagte base (salt)

Question 1(d)(i)• Calculate the pH of the buffer solution

n HOCl = (50.0/1000)(0.14) = 7.0 x 10-3 mol (excess reagent)n NaOH = (10.0/1000)(0.56) = 5.6 x 10-3 mol

n HOCl remaining

= 7.0 x 10-3 - 5.6 x 10-3 = 1.4 x 10-3 mol

HOCl (aq) + NaOH (aq) NaOCl (aq) + H2O(l)

n NaOCl formed = 5.6 x10-3 mol

[OCl] = 5.6 x 10-3 mol / 0.0600 L= 9.33 x 10-2 M

Total Volume = 50.0 + 10.0 = 60.0 cm3 = 0.0600 L

[HOCl] = 1.4 x 10-3 mol / 0.0600 L= 2.33 x 10-2 M

[acid]

[salt]logKH a pp

Henderson-Hasselbalch Equation to calculate pH of buffer equation

Need to determine [salt] / [acid]

Question 1(d)(i)• Calculate the pH of the buffer solution.

[OCl] = 5.6 x 10-3 mol / 0.0600 L= 9.33 x 10-2 M

[HOCl] = 1.4 x 10-3 mol / 0.0600 L= 2.33 x 10-2 M

[acid]

[salt]logKH a pp

)10 (2.33

)10 (9.33lg)102.3lg(H

2-

-28

p

097.8H p

Question 1(d)(ii)• Bob would like to prepare a new buffer solution. How many

grams of solid NaOH must he add to a 50.0 mL of 0.20 M HOCl to obtain a buffer that has a pH of 6.0?

• Assume that the addition of the solid NaOH results in a negligible change in volume. (Mr NaOH = 40)

Let the no of mole of NaOH required be y mol: n HOCl = (50/1000)(0.20) = 0.01 mol

n NaOH = y mol

n HOCl remaining = (0.01 - y) mol

n NaOCl formed = y mol

NaOH (s) + HOCl (aq) NaOCl (aq) + H2O (l)

Question 1(d)(ii)• Bob would like to prepare a new buffer solution. How many

grams of solid NaOH must he add to a 50.0 mL of 0.20 M HOCl to obtain a buffer that has a pH of 6.0?

• Assume that the addition of the solid NaOH results in a negligible change in volume. (Mr NaOH = 40)

n HOCl remaining = (0.01 - y) moln NaOCl formed = y mol

[acid]

[salt]logpK pH a

Total Volume = 50.0 mL = 0.0500 L

0500.0

01.00500.0

lg)10 x (3.2 log- 6.0 8-

y

y

y

y

01.0lg4949.1

y

y

01.010 4949.1

Solve for y,

y = n (NaOCl) formed = n (NaOH) required = 3.101x 10-4 mol

Mass of NaOH (s) required= 3.101 x 10-4 x 40 = 0.1240 g