ib chemistry on titration curves, acid base titration and buffer regions

Titration between strong acid(flask) with strong base(burette) HCI + NaOH → NaCI + H2O

Titration curves Strong Acid with Strong Base

Click here titration simulation

NaOH M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

7

HCI + NaOH → NaCI + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

2.7

11.3

• Rapid jump in pH (2.7 – 11.3) • Rapid change at equivalence point • Equivalence point → amt acid = amt base • pH at equivalence point = 7 • Neutral salt, NaCI - neutral

1

HCI M = 0.1M V = 25ml

NaOH M = 0.1M V = 25ml

HCI left → 25.0ml, 0.1M Conc H+ = 0.1M pH = -lg[0.1] pH = 1

HCI M = 0.1M V = 1ml left

NaOH M = 0.1M V = 24 ml added

HCI left → 1ml, 0.1M Moles H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Moles/Vol = 0.0001/0.049 = 0.002M pH = -lgH+ = -lg(0.002) pH = 2.7

NaOH M = 0.1M V = 25ml added


Neutral Salt, NaCI Conc H+ = 1 x 10-7M (dissociation of water) pH = -lg[1 x 10-7] pH = 7

NaOH M = 0.1M V = 26ml added

NaOH left → 1ml left, 0.1M Moles OH-= (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.051 = 0.002M pOH = -lgOH- = -lg(0.002) pOH = 2.7 pH = 14 - 2.7 pH = 11.3

NaOH V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml

11.3

2.7

Neutralization

Mole ratio – 1: 1

http://chem-ilp.net/labTechniques/AcidBaseIdicatorSimulation.htm


Titration between strong acid(flask) with weak base(burette) HCI + NH4OH → NH4CI + H2O

Titration curves Strong Acid with Weak Base

NH4OH M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

5.3

HCI + NH4OH → NH4CI + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

2.7

7.8

• Rapid jump in pH (2.7 – 7.8) • Rapid change at equivalence point • Equivalence point → amt acid = amt base • pH at equivalence point = 5.3 • Acidic salt, NH4CI – pH = 5.3

1

HCI M = 0.1M V = 25ml

NH4OH M = 0.1M V = 25ml

HCI left → 25.0ml, 0.1M Conc H+ = 0.1M pH = -lg[0.1] pH = 1


NH4OH M = 0.1M V = 24 ml added

HCI left → 1ml, 0.1M Moles H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Moles/Vol = 0.0001/0.049 = 0.002M pH = -lgH+ = -lg(0.002) pH = 2.7

NH4OH M = 0.1M V = 25ml added


Acidic Salt, NH4CI NH4

+ hydrolysis to produce H+

pH = 5.3

NH4OH M = 0.1M V = 26ml added

NH4OH left → 1ml left, 0.1M Moles OH-= (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.051 = 0.002M Conc NH4CI = Moles/Vol = 2.5 x 10-3/0.051 = 0.05M

NH4OH V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml

7.8

2.7

Neutralization


pOH = pKb -lg (base) (salt) pOH = 4.74 – lg (0.002) (0.05) pOH = 6.13 pH + pOH = 14 pH = 7.8

pH buffer – salt and weak base

pH buffer region

salt and weak base

Mole ratio – 1: 1



Titration between weak acid (flask) with strong base(burette) CH3COOH + NaOH → CH3COONa + H2O

Titration curves Weak Acid with Strong Base

NaOH M = 0.1M V = 0 ml


9

CH3COOH + NaOH → CH3COONa + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

11.3

6.11

• Rapid jump in pH (6.11 – 11.3 ) • Rapid change at equivalence point • Equivalence point → amt acid = amt base • pH at equivalence point = 9 • Basic salt, CH3COONa = pH 9

2.87

CH3COOH M = 0.1M V = 25ml


CH3COOH left → 25.0ml, 0.1M CH3COOH ↔ (CH3COO- )(H+) Ka = (CH3COO-) (H+) CH3COOH

(H+) = √Ka x CH3COOH (H+) = 1.34 x 10-3

pH = -lg(1.34 x 10-3) pH = 2.87

CH3COOH M = 0.1M V = 1ml left


NaOH M = 0.1M V = 25ml add


Basic Salt (CH3COONa) CH3COO- hydrolysis produce OH-

pH = 9

NaOH M = 0.1M V = 26ml add

NaOH V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


CH3COOH left → 1ml, 0.1M Mole CH3COOH = (MV)/1000 = (1 x 0.1)/1000 = 0.0001mol Conc CH3COOH = Moles/Vol = 0.0001/0.049 = 2.04 x 10-3

Conc CH3COONa = Moles/Vol = 2.4 x 10-3/0.049 = 0.048M Ka = 1.8 x 10-5

NaOH left → 1ml left, 0.1M Moles OH-= (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.051 = 0.002M pOH = -lgOH- = -lg(0.002) pOH = 2.7 pH = 14 - 2.7 pH = 11.3

11.3

6.11

Neutralization

pH buffer region

salt and weak acid

pH buffer – salt and weak acid

pH = pKa -lg [acid] [salt] pH = 4.74 – lg [2.04 x 10-3] [0.048] pH = 4.74 + 1,37 pH = 6.11

Mole ratio – 1: 1


Titration curves Weak Acid with Weak Base

NH4OH M = 0.1M V = 0 ml


7

CH3COOH + NH4OH → CH3COONH4 + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

7.8

• No sharp rise in pH • pH changes gradually over a range • no inflection point

2.87


NH4OH M = 0.1M V = 25ml

NH4OH M = 0.1M V = 25ml add


Neutral Salt CH3COONH4 pH = 7


NH4OH V = 1ml left

Total = 25 + 26

Vol = 51ml

7.8

6.11

Neutralization

Titration between weak base(burette) with weak acid(flask) CH3COOH + NH4OH → CH3COONH4 + H2O

6.11




CH3COOH left → 25.0ml, 0.1M CH3COOH ↔ (CH3COO- )(H+) Ka = (CH3COO-) (H+) CH3COOH

(H+) = √Ka x CH3COOH (H+) = 1.34 x 10-3

pH = -lg(1.34 x 10-3) pH = 2.87


Conc CH3COONH4 = Moles/Vol = 2.4 x 10-3/0.049 = 0.048M



pH buffer region

salt and weak acid

NH4OH left → 1ml, 0.1M pH = 7.8

Mole ratio – 1: 1



Titration between strong acid(flask) with strong base(burette) HCI + NaOH → NaCI + H2O


Titration curves Acid with Base

Titration between strong acid(flask) with weak base(burette) HCI + NH4OH → NH4CI + H2O

11.3

2.7

Titration between weak acid(flask) with weak base(burette) CH3COOH + NH4OH → CH3COONH4 + H2O


HCI M = 0.1M V = 25ml

6.11

11.3 NaOH M = 0.1M V = 25ml


2.87

1

Vs

•Start at pH = 1 → End at 11.3 • Rapid change at equivalence point • Rapid jump in pH (2.7 – 11.3) • Equivalence point → amt acid = amt base • pH at equivalence point = 7 • Neutral salt, NaCI - neutral

•Start at pH = 2.87 → End at 11.3 • Rapid change at equivalence point • Rapid jump in pH (6.11 – 11.3) • Equivalence point → amt acid = amt base • pH at equivalence point = 9 • Basic salt, CH3COONa - basic

9

7

Vs

•Start at pH = 1 → End at 7.8 • Rapid change at equivalence point • Rapid jump in pH (2.7 – 7.8) • Equivalence point → amt acid = amt base • pH at equivalence point = 5.3 • Acidic salt, NH4CI - acidic

1

2.7

5.3

7.8

2.87

NH4OH M = 0.1M V = 25 ml

HCI M = 0.1M V = 25ml

NH4OH M = 0.1M V = 25 ml


•Start at pH = 2.87 → End at 7.8 • pH changes gradually over a range • No sharp rise in pH • Equivalence point → amt acid = amt base • pH at equivalence point = 7 • Neutral salt, CH3COONH4 - neutral

6.11

7

7.8

pH buffer region

salt and weak base

pH buffer region

salt and weak acid

Buffer region form • Slow gradual increase pH due to buffering effect

Buffer region form • Slow gradual increase pH due to buffering effect

Titration curves Strong Acid with Strong Base

HCI M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

7

HCI + NaOH → NaCI + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

2.7

11.3

• Rapid drop in pH (11.3– 2.7 ) • Rapid change at equivalence point • Equivalence point → amt acid = amt base • pH at equivalence point = 7 • Neutral salt, NaCI - neutral



NaOH left → 25.0ml, 0.1M Conc OH- = 0.1M pOH = -lg[0.1] pOH = -1 pH + pOH = 14 pH = 13

NaOH M = 0.1M V = 1ml left

HCI M = 0.1M V = 24 ml added

NaOH left → 1ml, 0.1M Moles OH- = (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.049 = 0.002M pOH = -lgOH- = -lg(0.002) pOH = 2.7 pH + pOH = 14 pH = 11.3

HCI M = 0.1M V = 25ml added


Neutral Salt ,NaCI Conc H+ = 1 x 10-7M (dissociation of water) pH = -lg[1 x 10-7] pH = 7

HCI M = 0.1M V = 26ml add

HCI left → 1ml left, 0.1M Moles H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Moles/Vol = 0.0001/0.051 = 0.002M pH = -lg(0.002) pH = 2.7

HCI V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


13

11.3

2.7

Neutralization

Titration between strong base(flask) with strong acid(burette) HCI + NaOH → NaCI + H2O

Mole ratio – 1: 1



Titration curves Weak Acid with Strong Base

CH3COOH M = 0.1M V = 0 ml


9


M = 0.1M M = 0.1M

V = 25ml V = 25ml

11.3

6.13

• Rapid drop in pH ( 11.3 - 6.13) • Rapid change at equivalence point • Equivalence point → amt acid = amt base • pH at equivalence point = 9 • Basic salt, CH3COONa = pH 9

13

NaOH M = 0.1M V = 25.0ml



CH3COOH M = 0.1M V = 24 ml added

CH3COOH M = 0.1M V = 25ml add


Basic Salt (CH3COONa) CH3COO- hydrolysis to produce OH-

pH = 9


CH3COOH V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


NaOH left → 25.0ml, 0.1M Conc OH- = 0.1M pOH = -lg[0.1] pOH = -1 pH + pOH = 14 pH = 13

NaOH left → 1ml, 0.1M Moles OH- = (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.049 = 0.002M pOH = -lgOH- = -lg(0.002) pOH = 2.7 pH + pOH = 14 pH = 11.3


Conc CH3COONa = Moles/Vol = 2.5 x 10-3/0.051 = 0.049M

11.3

6.13

Neutralization

pH buffer region

salt and weak acid

Titration between strong base (flask) with weak acid(burette) CH3COOH + NaOH → CH3COONa + H2O



Mole ratio – 1: 1



Titration curves Strong Acid with Weak Base

HCI M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

5.3

HCI + NH4OH → NH4CI + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

7.8

2.7

• Rapid drop in pH (7.8 – 2.7) • Rapid change at equivalence point • Equivalence point → amt acid = amt base • pH at equivalence point = 5.3 • Acidic salt, NH4CI – pH = 5.3

11.1

NH4OH M = 0.1M V = 25ml

NH4OH M = 0.1M V = 25ml

NH4OH M = 0.1M V = 1ml left




Acidic Salt, NH4CI NH4

+ hydrolysis to produce H+

pH = 5.3


NH4OH left → 1ml left, 0.1M Moles NH4OH = (0.1 x 1)/1000 = 0.0001mol Conc NH4OH = Moles/Vol = 0.0001/0.049 = 0.002M Conc NH4CI = Moles/Vol = 2.4 x 10-3/0.049 = 0.048M

HCI V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml

7.8

2.7

Neutralization




NH4OH left → 25.0ml, 0.1M Conc NH4OH = 0.1M NH4OH ↔ NH4

+ + OH-

Kb = (NH4+) (OH-)

(NH4OH) 1.8 x 10-5 = (OH-)2

0.1 OH- = √0.1 x 1.8 x 10-5 OH- = 1.34 x 10-3

pOH = -lg 1.34 x 10-3 pOH = 2.87 pH = 14 – 2.87 pH = 11.1

pH buffer region

salt and weak base

HCI left → 1ml left, 0.1M Moles H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Moles/Vol = 0.0001/0.051 = 0.002M pH = -lg(0.002) pH = 2.7

Titration between weak base(flask) with strong acid(burette) HCI + NH4OH → NH4CI + H2O

Mole ratio – 1: 1



Titration curves Weak Acid with Weak Base

CH3COOH M = 0.1M V = 0 ml


7

CH3COOH + NH4OH → CH3COONH4 + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

6.11

• No sharp drop in pH • pH changes gradually over a range • no inflection point

11.1

NH4OH M = 0.1M V = 25ml

NH4OH M = 0.1M V = 25ml



Neutral Salt CH3COONH4 pH = 7


CH3COOH V = 1ml left

Total = 25 + 26

Vol = 51ml

7.8

6.11

Neutralization


NH4OH left → 25.0ml, 0.1M Conc NH4OH = 0.1M NH4OH ↔ NH4

+ + OH-

Kb = (NH4+) (OH-)

(NH4OH) 1.8 x 10-5 = (OH-)2

0.1 OH- = √0.1 x 1.8 x 10-5 OH- = 1.34 x 10-3

pOH = -lg 1.34 x 10-3 pOH = 2.87 pH = 14 – 2.87 pH = 11.1

Titration between weak base(flask) with weak acid(burette) CH3COOH + NH4OH → CH3COONH4 + H2O

CH3COOH left → 1ml, 0.1M pH = 6.11

7.8



NH4OH left → 1ml left, 0.1M Moles NH4OH = (0.1 x 1)/1000 = 0.0001mol Conc NH4OH = Moles/Vol = 0.0001/0.049 = 0.002M Conc NH4CI = Moles/Vol = 2.4 x 10-3/0.049 = 0.048M



pH buffer region

salt and weak base

Mole ratio – 1: 1



Titration between strong base(flask) with strong acid(burette) HCI + NaOH → NaCI + H2O

Titration between strong base(flask) with weak acid(burette) CH3COOH + NaOH → CH3COONa + H2O

Titration curves Acid with Base

Titration between weak base(flask) with strong acid(burette) HCI + NH4OH → NH4CI + H2O

11.3

2.7

Titration between weak base(flask) with weak acid(burette) CH3COOH + NH4OH → CH3COONH4 + H2O


HCI M = 0.1M V = 25ml

6.13

11.3



2.87

13

Vs

•Start at pH = 1 3 → End at 2.7 • Rapid change at equivalence point • Rapid drop in pH (11.3 – 2.7) • Equivalence point → amt acid = amt base • pH at equivalence point = 7 • Neutral salt, NaCI - neutral

•Start at pH = 13 → End at 6.13 • Rapid change at equivalence point • Rapid drop in pH (11.3 – 6.13) • Equivalence point → amt acid = amt base • pH at equivalence point = 9 • Basic salt, CH3COONa - basic

9

7

Vs

•Start at pH = 11.1 → End at 2.7 • Rapid change at equivalence point • Rapid drop in pH (7.8 – 2.7) • Equivalence point → amt acid = amt base • pH at equivalence point = 5.3 • Acidic salt, NH4CI - acidic

11.1

2.7

5.3

7.8 11.1

NH4OH M = 0.1M V = 25 ml

HCI M = 0.1M V = 25ml


•Start at pH = 11.1 → End at 6.11 • pH changes gradually over a range • No sharp drop in pH • Equivalence point → amt acid = amt base • pH at equivalence point = 7 • Neutral salt, CH3COONH4 - neutral

6.11

7

7.8

13

NH4OH M = 0.1M V = 25 ml

Buffer region form • Slow gradual drop pH due to buffering effect

Buffer region form • Slow gradual drop pH due to buffering effect

pH buffer region

salt and weak acid

pH buffer region

salt and weak base

Acidic Buffer Region CH3COOH (acid)/ CH3COO-(salt)

Titration bet weak acid + strong base CH3COOH + NaOH → CH3COONa + H2O

Click here buffer simulation


Initial 0.0025 mol 0.00125mol added 0

Change (0.0025-0.00125)mol 0 mol 0.00125mol form

Final 0.00125mol left 0 mol 0.00125mol form

At half equivalence point : • Amt acid = Amt salt : ( 0.00125 = 0.00125) • pH = pKa -lg [acid] → pH = pKa → 4.74 = 4.74

[salt] Buffer at pH = 4.74 form when half amt of acid neutralise by base or at half equivalence point when amt acid = amt salt

Prepare Acidic Buffer at pH = 4.74 • Choose pKa acid closest to pH 4.74 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 4.74 = 4.74 – lg [acid] [salt] • [acid] = 1.00 (amt acid = amt salt) [salt]

Buffer region at half equivalence point

Amt acid = Amt salt

Weak acid 25ml, 0.1M

(0.0025 mol)



Amt acid = Amt salt

At equivalence point (Neutralization)

Amt acid = Amt base


M = 0.1M M = 0.1M

V = 25ml V = 25ml

At half equivalence point : • Vol base = 12.5ml • pH = pKa • Most effective buffering capacity

At equivalence point: • vol base = 25ml • Amt acid = amt base •(Neutralization) • Salt and water result

Mole ratio – 1: 1


NaOH M = 0.1M V = 12.5 ml added

CH3COOH M = 0.1M V = 12.5ml left


Strong base

12.5ml , 0.1M added

(0.00125mol)

pH buffer calculation

pH buffer region

salt and weak acid

Titration curve can be use to find pKa or Ka for weak acid

pH = pKa



NH4OH + HCI → NH4CI + H2O

Initial 0.0025 mol 0.00125mol added 0.

Change (0.0025-0.00125)mol 0 mol 0.00125mol form

Final 0.00125mol left 0 mol 0.00125mol form

At half equivalence point : • Amt base = Amt salt : (0.00125 = 0.00125) • pOH = pKb - lg [base] → pOH = pKb → 4.74 = 4.74 [salt] Buffer at pOH = 4.74 form when half amt of base neutralise by acid or at half equivalence point when amt base = amt salt

Prepare Buffer pH = 9.26 /pOH = 4.74 • Choose pKb base closest to pOH = 4.74 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.74 = 4.74 – lg [base] [salt] • [base] = 1.00 (amt base = amt salt) [salt]

NH4OH + HCI → NH4CI + H2O

M = 0.1M M = 0.1M

V = 25ml V = 25ml

Titration bet weak base + strong acid NH4OH + HCI → NH4CI + H2O



Amt acid = Amt salt

Weak base 25ml, 0.1M

(0.0025 mol)

Titration between weak base (flask) with strong acid(burette) NH4OH + HCI → NH4CI + H2O


Amt acid = Amt salt

At equivalence point (Neutralization)

Amt acid = Amt base

At half equivalence point : • Vol base = 12.5ml • pH = pKb • Most effective buffering capacity

At equivalence point: • vol base = 25ml • Amt acid = amt base •(Neutralization) • Salt and water result

Mole ratio – 1: 1


HCI M = 0.1M V = 12.5 ml added

NH4OH M = 0.1M V = 12.5ml left


Strong acid

12.5ml , 0.1M added

(0.00125mol)

pH buffer calculation

pH buffer region

salt and weak acid

Titration curve can be use to find pKb or Kb for weak base

pH = pKb

Basic Buffer Region NH3(base)/ NH4CI(salt)



Sample Titration Calculation

Strong base with Strong acid

HCI + NaOH → NaCI + H2O M = 0.1M M = 0.1M

V = 25ml V = ?ml Mole ratio – 1: 1

Strong base with Weak acid

CH3COOH + NH4OH → CH3COONH4 + H2O M = 0.1M M = 0.1M

V = 25ml V = ? ml

CH3COOH + NaOH → CH3COONa + H2O M = 0.1M M = 0.1M

V = 25ml V = ? ml

Weak base with Strong acid Weak base with Weak acid

NaOH M = 0.1M V = ? ml

HCI M = 0.1M V = 25ml

NH4OH M = 0.1M V = ? ml

Mole ratio – 1: 1

Mole ratio – 1: 1

HCI + NH4OH → NH4CI + H2O M = 0.1M M = 0.1M

V = 25ml V = ? ml Mole ratio – 1: 1


HCI M = 0.1M V = 25ml


Will volume used the same ?


Yes

25ml

Yes

25ml

Regardless whether strong or weak acid/base • stoichiometric mole ratio is followed for neutralization • 1 mole of weak/strong acid will neutralize 1 mole of weak/strong base

HNO3

M = 0.1M V = 25ml

NaOH M = 0.1M V = ? ml

Student perform a titration between • Strong base NaOH with Strong acid HNO3. • Determine the volume of base needed to neutralize acid.

Student perform a titration between • Strong base NaOH with Weak acid CH3COOH. • Determine the volume of base needed to neutralize acid.

HNO3 + NaOH → NaNO3 + H2O M = 0.1M M = 0.1M


Moles of Acid = MV = (0.1 x 0.025) = 2.5 x 10-3 Mole ratio ( 1 : 1) • 1 mole acid neutralize 1 mole base • 2.5 x 10-3 acid neutralize 2.5 x 10-3 base Moles of Base = MV =0.1 x V o,.1 x V = 2.5 x 10-3

V = 25ml

M aVa = 1 Mb Vb 1 0.1 x 25 =1 0.1 x Vb 1

V b = 25ml

CH3COOH M = 0.1M V = 25ml Acid of same volume and Conc

CH3COOH + NaOH → CH3COONa + H2O M = 0.1M M = 0.1M


Using mole ratio Using formula


Yes

Moles of Acid = MV = (0.1 x 0.025) = 2.5 x 10-3 Mole ratio ( 1 : 1) • 1 mole acid neutralize 1 mole base • 2.5 x 10-3 acid neutralize 2.5 x 10-3 base Moles of Base = MV =0.1 x V o,.1 x V = 2.5 x 10-3

V = 25ml

M aVa = 1 Mb Vb 1 0.1 x 25 =1 0.1 x Vb 1

V b = 25ml

Using formula Using mole ratio

Sample Titration Calculation

Regardless whether strong or weak acid/base • stoichiometric mole ratio is followed for neutralization • 1 mole of weak/strong acid will neutralize 1 mole of weak/strong base

Calculation Calculation

25ml weak acid, HA is titrated with 0.155M NaOH and graph is shown below a) Determine the pH at equivalence point b) Explain using eqn, why the equivalence point is not at pH = 7 c) Calculate the conc of weak acid before the addition of any NaOH d) Estimate, using data from the graph, the dissociation constant Ka of weak acid

Sample IB Question on Acid Base Titration

a) pH is = 9 At equivalence point Amt acid = Amt base

HA M = ? M V = 25.0ml

NaOH M = 0.155M V = 22 ml

c) HA + NaOH → NaA + H2O Moles of Base = MV = (0.155 x 0.022) = 3.41 x 10-3 Mole ratio ( 1 : 1) • 1 mole base neutralize 1 mole acid • 3.41 x 10-3 base neutralize 3.41 x 10-3 acid Moles of Acid = MV = M x 0.025 M x 0.025 = 3.41 x 10-3

M = 0.136M

d) pH = 5.3 At half equivalence point: • vol base 11ml • amt acid = amt salt pH = pKa - lg [acid] [salt] pH = pKa = 5.3 pKa = -lg Ka

5.3 = -lgKa

Ka = 5 x 10-6

b) Neutralization bet strong base with weak acid HA + NaOH → A- + H2O A- is a strong conjugate base A- + H2O → HA + OH- (basic salt)

3.41 x 10-3 base added

Example Buffer Calculation

Find pH buffer prepared by titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3 pKb = 4.75

1 Titration bet strong acid with weak base NH3 + HCI → NH4CI + H2O

Strong acid HCI 1.8 x 10-3 mol

Weak base NH3 3.2 x 10-3 mol

Buffer region Weak base + salt

NH3 + HCI → NH4CI + H2O

Initial 3.2 x 10-3 mol 1.8 x 10-3 mol added 0. 0

Change (3.2 – 1.8) x 10-3mol 0 mol 1.8 x 10-3 mol form

Final 1.4 x 10-3 mol 0 mol 1.8 x 10-3 mol form

Change moles to Conc → Moles ÷ total volume

Conc (1.4 x 10-3)/ 0.05 (1.8 x 10-3)/ 0.05

Conc 2.8 x 10-2 M 3.6 x 10-2 M

(base) (salt) • pOH = pKb - lg [base] [salt] • pOH = 4.75 – lg [2.8 x 10-2]/[3.6 x 10-2] • pOH = 4.86 pH + pOH = 14 pH = 9.14


Strong acid 18ml, 0.1M HCI added

Weak base

32ml, 0.1M NH3

Number = (M x V) = 18 x 0.1 moles 1000 1000


Total vol = 50ml or 0.05dm3

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3) 1.77 x 10-5 = 3.6 x 10-2 x OH- 2.8 x 10-2 OH- = 2.8 x 10-2 x 1.77 x 10-5 3.6 x 10-2 OH- = 1.37 x 10-5

pOH = -lgOH-

pOH = -lg 1.37 x 10-5 pOH = 4.86 pH + pOH = 14 pH = 9.14

1st method (formula) 2nd method (Kb)

pH calculation using



2 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M CH3COOH Ka CH3COOH = 1.8 x 10-5M, pKa = 4.74

Strong base 50ml, 0.1M NaOH added

Weak acid 100ml, 0.1M CH3COOH



Strong base NaOH

5 x 10-3 mol

Weak acid CH3COOH

10 x 10-3 mol

Titration bet strong base with weak acid NaOH + CH3COOH → CH3COONa + H2O



Amt base = Amt salt

Buffer Calculation

pH calculation using

1st method (formula) 2nd method (Ka)

NaOH + CH3COOH → CH3COONa H2O

Initial 5 x 10-3 mol added 10 x 10-3 mol 0. 0

Change 0 mol (10-5) x 10-3 mol 5 x 10-3 mol form

Final 0 mol 5 x 10-3 mol 5 x 10-3 mol form


Conc (5 x 10-3)/0.15 (5 x 10-3)/0.15

Conc 3.3 x 10-2 M 3.3 x 10-2 M

(acid) (salt) • pH = pKa - lg [acid] [salt] • pH = 4.74 – lg [3.3 x 10-2]/[3.3 x 10-2] • pH = 4.74


CH3COOH ↔ CH3COO- + H+

Ka = (CH3COO-)(H+) (CH3COOH) 1.8 x 10-5 = 3.3 x 10-2 x (H+) 3.3 x 10-2 H+ = 1.8 x 10-5

pH = -lg H+ pH = -lg(1.8 x 10-5 ) pH = 4.74



Example pH Calculation

3 Find pH when 50ml 0.1M NaOH added to 100ml, 0.1M HCI

Strong base 50ml, 0.1M NaOH added

Strong acid

100ml, 0.1M HCI



Strong base NaOH

5 x 10-3 mol

Strong acid HCI 10 x 10-3 mol


Titration bet strong base with strong acid NaOH + HCI → NaCI + H2O

pH region

pH calculation

NaOH + HCI → NaCI + H2O

Initial 5 x 10-3 mol added 10 x 10-3 mol 0 0

Change 0 mol (1 0 - 5) x 10-3 mol

Final 0 mol 5 x 10-3 mol


Conc (5 x 10-3)/ 0.15

Conc 3.3 x 10-2 M • pH = -lg[H+] • pH = -lg 3.3 x 10-2

pH = 1.48


NH4+ + H2O ↔ NH3 + H3O

+

Ka = (NH3)(H3O

+) (NH4

+) (H3O

+)2 = Ka x NH4+

H+ = √5.56 x 10-10 x 0.10 H+ = 7.45 x 10-6

pH = -lg 7.45 x 10-6 pH = 5.13

Find pH of 0.10M NH4CI in water. Kb NH3 = 1.8 x 10-5 M

4

Acid dissociation constant

0.10M NH4CI

Ka (NH4) x Kb(NH3) = Kw

Ka = Kw /Kb

Ka = 10-14/ 1.8 x 10-5 Ka = 5.56 x 10-10

Using Ka

Find pH of 0.50M NH3 in water. Kb NH3 = 1.8 x 10-5 M

4

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3) 1.8 x 10-5 = (OH-)2

0.50 OH- = √0.50 x 1.8 x 10-5 OH- = 3.0 x 10-3

pOH = -lg 3.0 x 10-3 pOH = 2.52 pH = 14 – 2.52 pH = 11.48

0.50M NH3

Base Dissociation constant

Using Kb



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Simulation and Animation on Buffer and Titrations

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http://www.youtube.com/watch?v=BBIGR0RAMtY&list=PL39E45DB611F4C7FE

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