Easier using pH scale than Conc [H+] • Conc H+ increase 10x from 0.0001(10-4) to 0.001(10-3) - pH change by 1 unit from pH 4 to 3 • pH 3 is (10x) more acidic than pH 4 • 1 unit change in pH is 10 fold change in Conc [H+]

Conc OH- increase ↑ by 10x – pH increase ↑ by 1 unit

pOH with Conc OH-

pOH = -log [OH-] [OH-] = 0.0000001M pOH = -log [0.0000001] pOH = -log1010-7 pOH = 7 pH + pOH = 14 pH + 7 = 14 pH = 7 (Neutral)

pH with Conc H+

pH = -log [H+] [H+] = 0.0000001M pH = -log [0.0000001] pH = -log1010-7

pH = 7 (Neutral)

Conc H+ increase ↑ by 10x – pH decrease ↓ by 1 unit

pH measurement of Acidity of solution

• pH is the measure of acidity of a solution in logarithmic scale •pH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration

← Acidic – pH < 7 Alkaline – pH > 7 →

pOH with Conc OH-

pOH = -log [OH-] [OH-] = 0.1M pOH = -log[0.1] pOH = 1 pH + pOH = 14 pH + 1 = 14

pH = 13 (Alkaline)

pH with Conc H+

pH = -log [H+] [H+] = 0.01M pH = -log [0.01] pH = -log1010-2

pH = 2 (Acidic)

Easier scale

pH measurement of Acidity of solution

H2O dissociate forming H3O+ and OH- (equilibrium exist)

• H2O + H2O ↔ H3O+ + OH− • Kc = [H3O+][OH−]/[H2O]2

• Kc [H2O]2 = [H3O+][OH−] Dissociation H2O small, conc [H2O] is constant Kc [H2O]2 is constant called Kw = [H3O+][OH−] Kw = 1.0 x 10-14 mol2 dm-6 - dissociation constant water at -25C Kw = [H3O+][OH−] 1.0 x 10-14 = [H3O+][OH−] 1.0 x 10-14 = [1.0 x 10-7][1.0 x 10-7] [H3O+]= 1.0 x 10-7, [OH-] = 1.0 x 10-7

Conc [H+] = 1 x 10-12 pH = -lg[H+] pH = -lg [10-12] pH = 12

Conc [OH-] = 1 x 10-2 pOH = -log10[OH-]

= -log1010-2 pOH = 2 pH + pOH = 14 pH + 2 = 14 pH = 12

Conc [H+] = 1 x 10-2 pH = -lg[H+] pH = -lg [10-2] pH = 2

Alkaline Alkaline

Acidic Acidic

Kw - dissociation constant water - ionic product water

Using conc [H+] pH = -log10[H+] Using conc [OH-]

pOH = -log10[OH-]

Conc [OH-] = 1 x 10-12 pOH = -log10[OH-]

= -log1010-12 pOH = 12 pH + pOH = 14 pH + 12 = 14 pH = 2

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Formula for acid/base calculation

pH = -log10[H+] Kw = [H+][OH-] pKa = - lg10Ka pOH = -log10[OH-] Ka x Kb = Kw pKb = - lg10Kb

pH + pOH = pKw Ka x Kb = 1 x 10-14 pKa + pKb = pKw

pH + pOH = 14 pKa + pKb = 14

Weak acid HA ↔ H+ + A-

Ka = (H+) (A-) (HA)

Weak acid CH3COOH + H2O ↔ CH3COO- + H3O+

Ka = (CH3COO-) (H3O+) = (H+)2 (CH3COOH) (CH3COOH)

Weak base NH3 + H2O ↔ NH4

+ + OH-

Kb = (NH4+) (OH-) = (OH-)2

(NH3) (NH3)

Weak base B + H2O ↔ BH+ + OH-

Kb = (BH+) (OH-) (B)

Dissociation constant Weak acid CH3COOH + H2O ↔ CH3COO- + H3O+

Ka = (CH3COO-) (H3O+) (CH3COOH)

Dissociation constant Conjugate base CH3COO- + H2O ↔ CH3COOH + OH-

Kb = (CH3COOH) (OH-) (CH3COO-)

Ka x Kb = Kw

Ka x Kb = Kw

(CH3COO-) (H3O+) x (CH3COOH) (OH-) = (H3O+)(OH- ) = Kw

(CH3COOH) (CH3COO-)

H2O (base) - H3O+ (conjugate acid)

gain H

CH3COOH (acid) - CH3COO- (conjugate base)

lose H

Weak Acid Weak Base

CH3COOH + H2O ↔ CH3COO- + H3O+

Video on how Ka x Kb = Kw derived

Formula for acid/base calculation

Ka /Kb measures the equilibrium position Ka /Kb large ↑ – ↑ dissociation – product favour – shift to right Ka /Kb large ↑ – pKa /pKb small ↓ – strong acid/base

Strong acid Large ↑ Ka

Weak acid Small ↓ Ka

Strong base Large ↑ Kb

Weak base Small ↓Kb

↑ Ka → ↓ pKa , pKa = - lg10Ka

Ka /Kb measures the equilibrium position Ka /Kb small ↓ – ↓ dissociation – reactant favour – shift to left Ka /Kb small ↓ – pKa /pKb high ↑– weak acid/base

↑ Kb → ↓ pKb , pKb = - lg10Kb ↓ Ka → ↑ pKa , pKa = - lg10Ka

↓ Kb →↑ pKb , pKb = - lg10Kb

Click here on weak acid dissociation

Click here on CH3COOH dissociation animation Click here on weak acid dissociation animation

Weak acid Animation •Dissociate partially, ↔ used • Ka /Kb value used

Animation on weak acid dissociation

H2O dissociate forming H3O+ and OH- (equilibrium exist)

• H2O + H2O ↔ H3O+ + OH− • Kw = [H3O+][OH−]/[H2O]2 Dissociation H2O is small, conc [H2O] is constant - Kw = [H3O+][OH−] Kw = 1.0 x 10-14 mol2 dm-6 - dissociation constant water at -25C Kw = [H3O+][OH−] 1.0 x 10-14 = [H3O+][OH−] 1.0 x 10-14 = [1.0 x 10-7][1.0 x 10-7] [H3O+]= 1.0 x 10-7, [OH-] = 1.0 x 10-7

Animation on weak acid

Strong acid • 100% dissociation (complete) • HCI → H+ + CI-

Strong base • 100% dissociation (complete) • NaOH → Na+ + OH-

Find pH of 0.100M H2SO4 H2SO4 → 2H+ + SO4

2- 0.100 0.200 pH = -lg(H+) = -lg(0.200) pH = 0.700

Strong Acid/Base calculation

http://www.clker.com/clipart-lab-beaker.html

Find pH of 0.10M HCI HCI → H+ + CI- 0.10 0.10 pH = -lg(H+) = -lg(0.100) pH = 1.00

HCI → H+ + CI-

H2SO4 → 2H+ +SO42-

Find pH of 0.001M NaOH NaOH → Na+ + OH- 0.001 0.001 (H+)(OH-) = 1 x 10-14

H+ = (1 x 10-14)/1 x 10-3 = 1 x 10-11

pH = -lg(H+) = -lg(1 x 10-11) pH = 11.0

Find pH of 0.001M NaOH NaOH → Na+ + OH- 0.001 0.001 pOH = -lg(OH-) pOH = -lg(0.001) = 3 pH + pOH = 14 pH + 3 = 14 pH = 11.0

NaOH → Na+ + OH-

NaOH → Na+ + OH-

2nd Method

1st Method

Click here strong acid ionization

Animation on Acid Dissociation

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Find pH of 0.100M HA, Ka = 1.80 x 10-5

HA ↔ H+ + A-

Ka = (H+)(A-) or (H+)2

(HA) (HA) Ka = (H+)2 /HA (H+)2 = Ka x HA

(H+)2 = Ka x HA (H+)2 = 1.80 x 10-5 x 0.100 H+ = 1.34 x 10-3

pH = -lg(H+) pH = -lg (1.34 x 10-3) pH = 2.872

Find Ka of 0.02M HA, pH= 3.9

HA ↔ H+ + A-

Ka = (H+)(A-) or (H+)2

(HA) (HA) Ka = (H+)2 /HA

Ka = (H+)2 /HA Ka = (10-3.9 x 10-3.9)/0.02 Ka = 7.92 x 10-7

pH = -lg(H+) 3.9 = -lg(H+) H+ = 10-3.9

Find Conc of HA, pH = 4.5, Ka = 4.1 x 10-6

HA ↔ H+ + A-

Ka = (H+)(A-) or (H+)2

(HA) (HA) Ka = (H+)2 /HA HA = (H+)2/Ka

HA = (H+)2/Ka HA = (10-4.5)2/(4.1 x 10-6) HA = 2.4 x 10-4

pH = -lg(H+) 4.5 = -lg(H+) H+ = 10-4.5

Click here weak acid ionization

Weak Acid calculation

HA ↔ H+ + A-

HA ↔ H+ + A-

Animation on Acid Dissociation

Ka (HF) x Kb (F-) = Kw Kb F- = Kw = 1.0 x 10-14

Ka HF 6.8 x 10-4

Kb F- = 3.98 x 10-4

Determine Kb for F-, Ka HF = 6.8 x 10-4

HF + H2O ↔ F- + H3O+ Ka HF = 6.8 x 10-4

H2O (base) - H3O+ (conjugate acid)

HF (acid) - F- (conjugate base)

HF + H2O ↔ F- + H3O+

gain H

lose H

Ka (HF) x Kb (F-) = Kw HF + H2O ↔ F- + H3O+

HA ↔ H+ + A-

Find Kb of 0.03 B , pH= 10.0

B + H2O ↔ BH+ + OH-

Kb = (OH-)2

(B)

pOH = -lg(OH-) 4 = -lg(OH-) OH- = 10-4

Find Conc of B, pH = 10.8, Kb = 4.36 x 10-4

B + H2O ↔ BH+ + OH-

Kb = (OH-)2

(B) B = (OH-)2/ Kb

pH + pOH = 14 pOH = 14 – 10.8 = 3.2 pOH = -lg[OH-] 3.2 = -lg[OH-] OH- = 10-3.2

Find pH of 0.01 B, Kb = 1.80 x 10-5

B + H2O ↔ BH+ + OH- Kb = (BH+)(OH-) or (OH-)2

(B) (B) (OH-)2 = Kb x B

(OH-)2 = 1.80 x 10-5 x 0.01 OH- = 4.24 x 10-4

pOH = -lg(OH-) pOH = -lg(4.24 x 10-4 ) pOH = 3.37 pH + pOH = 14 pH = 14 – 3.37 pH = 10.6

Kb = (OH-)2 /B Kb = (10-4 x 10-4)/0.03 Kb = 3.33 x 10-7

B = (OH-)2/ Kb B = (10-3.2)2/ 4.36 x 10-4

B = 9.13 x 10-4 M

Weak Base calculation

pH + pOH = 14 pOH = 4

Find pH of 0.05 B, pKb = 3.40 B + H2O ↔ BH+ + OH- Kb = (OH-)2

(B) Kb = (OH-)2 / B (OH-)2 = Kb x B

pKb = -lg(Kb) 3.40 = -lg(Kb) Kb = 10-3.40

Kb = 3.98 x 10-4

(OH-)2 = 3.98 x 10-4 x 5.00 x 10-2

OH- = 4.46 x 10-3

pOH = -lg(OH-) pOH = -lg(4.46 x 10-3) = 2.40 pH = 14 – 2.40 pH = 11.6

Click here on equilibria animation

Animation on Base Dissociation

B + H2O ↔ BH+ + OH- B + H2O ↔ BH+ + OH-

B + H2O ↔ BH+ + OH- B + H2O ↔ BH+ + OH-

Weak Acid /Base Calculation

Find Ka of 0.01 HA, pH= 5.00

HA ↔ H+ + A-

Ka = (H+)(A-) or (H+)2

(HA) (HA) Ka = (H+)2 /HA

pH = -lg(H+) 5.0 = -lg(H+) H+ = 1 x 10-5

Ka = (10-5 x 10-5)/0.01 Ka = 1.00 x 10-8

Find pH of 0.1 HA, pKa = 4.20

HA ↔ H+ + A-

Ka = (H+)(A-) or (H+)2

(HA) (HA) Ka = (H+)2 /HA (H+)2 = Ka x HA

(H+)2 = Ka x HA (H+)2 = 6.31 x 10-5 x 0.1 H+ = 2.51 x 10-3

pH = -lg(H+) pH = -lg (2.51 x 10-3) pH = 2.60

pKa = -lg(Ka) 4.2 = -lg(Ka) Ka = 6.31 x 10-5

Find Conc of B, pH = 10.8, pKa = 10.64

B + H2O ↔ BH+ + OH-

Kb = (OH-)2

(B)

B = (OH-)2 / Kb B = (OH- )2/4.36 x 10-4

B = (10-3.2)2/ 4.36 x 10-4

B = 9.13 x 10-4

pKa + pKb = 14 pKb = 14 – pKa

pKb = 14 – 10.64

pKb = 3.36 pKb= -lg(Kb) 3.36 = -lg(Kb) Kb = 10-3.36

Kb = 4.36 x 10-4

HA ↔ H+ + A-

HA ↔ H+ + A-

pH + pOH = 14 pOH = 14 – 10.8 pOH = 3.2 pOH = -lg[OH-] 3.2 = -lg[OH-] OH- = 10-3.2

B + H2O ↔ BH+ + OH- Click here on weak base simulation

Click here on weak base simulation

Simulation on Base Dissociation

Calculate conc of OH- and pH for 0.001 HCI. HCI → H+ + CI- (100% dissociate) 0.001 0.001 Kw = [H+][OH−]= 10-14 (assume all H+ from HCI and H+ from water is negligible) [0.001][OH-]= 10-14 [OH-] = 10-14/0.001 = 10 -11

pH = -log1o[H]+ =-log10o.oo1

pH = 3

Calculate conc of OH- when 3.o x 10-4 [H+] was added to pure water HCI → H+ + CI- (100% dissociate) 3.o x 10-4 3.o x 10-4 Kw = [H+][OH−] = 10-14 (assume all H+ from HCI and H+ from water is negligible) [OH] = 10-14 = 10 -14 = 3.3 x 10 -11 M

[H+] 3.o x 10-4

What is the pH for [H+] = 10-12 M pH = -lg[H+] pH = -lg [10-12] pH = 12

What is the conc of H+ in solution with pH 3? pH = -lg[H+] 3 = -lg[H+] [H+] = 10 –pH

[H+] = 10 -3

What is the pH for [OH-] = 0.1 M pOH = -lg[OH-] pOH = -lg [0.1] pOH = 1 pH + pOH = 14 pH = 14 – 1 = 13

What is the pH of 1.0M NaOH ? NaOH → Na+ + OH- (100% dissociate) 1M 1M 1M Kw = [H3O+][OH−] = 10-14 (assume all OH- from NaOH and OH- from water is negligible) [H+] = 10-14 = 10 -14 = 1.0 x 10 -14

[OH] 1.0 pH = -log [H+] pH = -log [1.0 x 10 -14] pH = 14

Acid/base Calculation

pH = -log10[H+] pOH = -log10[OH-] pH + pOH = 14 Kw = [H+][OH-]

Formula acid/base calculation

Questions on Acids and Base

Which list contains only strong acids ?

A. CH3COOH, H2CO3, H3PO4

B. HCI, HNO3, H2CO3

C. CH3COOH, HNO3, H2SO4

D. HCI, HNO3, H2SO4

When equal volume of four 1M solutions are arranged in order of increasing pH (lowest pH first), what is the correct order?

A. CH3COOH < HNO3 < CH3CH2NH2 < KOH

B. HNO3 < CH3COOH < CH3CH2NH2 < KOH

C. CH3CH2NH2 < HNO3 < CH3COOH < KOH

D. KOH < CH3CH2NH2 < CH3COOH < HNO3

pH of a solution changes from pH =2 to pH =5. What happens to the concentration of H+ ions during this pH change?

A. Decrease by factor of 1000

B. Increase by factor of 1000

C. Decrease by factor of 100

D. Increase by a factor of 100

Solution of acid A has a pH of 1 and a solution of acid B has a pH of 2. Which statement is correct ?

A. Acid A is stronger than acid B

B. [A] > [B]

C. Concentration of H+ ions in A is higher than B

D. Concentration of H+ ions in B is twice the concentration of H+ in A

100ml of NaOH solution of pH 12 is mixed with 900ml of water. What is the pH of resulting solution?

A. 1

B. 3

C. 11

D. 13

1

2

3

4

5

О

О

О

О

О

List two ways to distinguish between strong and weak acid/base 6

By Conductivity measurement

1M Strong Acid – Ionise completely – More H+ ion – pH lower ↓ 1M Weak Acid – Ionise partially – Less H+ ion – pH higher ↑

1M Strong Acid – Ionise completely – More H+ ion – Conductivity higher ↑ 1M Weak Acid – Ionise partially – Less H+ ion – Conductivity lower ↓

By pH measurement

Click here on pH calculation

Video on Acid/ Base

Click here on pKa /pKb calculation How pH = pOH = 14 derived How Ka x Kb = Kw derived

Simulation on Acid/ Base

Click here on pH animation Click here to acid/base simulation

Click here on weak base simulation Click here strong acid ionization Click here on weak acid dissociation