ib chemistry on acidic, basic buffer and ph calculation

Acidic Buffer Solution

Acidic Buffer • Resist a change in pH when acid/base is added. • Mixing weak acid + salt/conjugate base • CH3COOH ↔ CH3COO- + H+ (dissociate partially)

• CH3COONa → CH3COO- + Na+ (dissociate fully)

Acidic buffer

CH3COOH ↔ CH3COO- +H+

Click here on acidic buffer simulation

Click here on acidic buffer simulation

Base OH- added Acid H+ added

Acid part

Neutralize each other

CH3COONa → CH3COO- + Na+

Neutralize each other

Acidic Buffer - contain weak acid and its salt • CH3COOH (weak acid) + CH3COONa (salt) • CH3COOH (weak acid) + CH3COO- (base) to neutralise added H+ or OH− • CH3COOH ↔ CH3COO- + H+ → H+ to neutralise added OH− • CH3COONa → CH3COO- + Na+ → CH3COO- to neutralise added H+ • Effective buffer have equal amt of weak acid CH3COOH and base CH3COO-

Salt part

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/acidbasepH/pHbuffer20.html


http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf


Basic Buffer • Resist a change in pH when acid/base is added. • Mixing weak base + salt/conjugate acid • NH3 + H2O ↔ NH4

+ + OH_ (dissociate partially)

• NH4CI → NH4+ + CI_ (dissociate fully)

Basic Buffer Solution

Click here on basic buffer simulation

Base part Salt part

Base OH- added Acid H+ added

Contain weak base and its salt • NH3(base) + NH4CI (salt) • NH3(base) + NH4

+ (acid) will neutralise added H+ or OH− • NH3 + H2O ↔ NH4

+ + OH− → NH3 molecule to neutralise added H+ • NH4CI → NH4

+ + CI− → NH4+ to neutralise added OH−

• Effective buffer have equal amt of weak base NH3 and conjugate acid NH4+

Neutralize each other Neutralize each other

Base part Salt part

Basic buffer

NH3 + H2O ↔ NH4+ + OH_ NH4CI → NH4

+ + CI_



How to prepare acidic/ basic buffer

Acid Dissociation constant CH3COOH + H2O ↔ CH3COO- + H3O+

Ka = (CH3COO-) (H3O+) (CH3COOH) -lgKa = -lgH+ -lg (CH3COO-) (CH3COOH) -lgH+ = -lg Ka + lg (CH3COO-) (CH3COOH) pH = pKa + lg (CH3COO-) (CH3COOH)

Acidic Buffer Formula • Mixture Weak acid + Salt/Conjugate base • CH3COOH ↔ CH3COO- + H+ (dissociate partially)

• CH3COONa → CH3COO- + Na+ (dissociate fully)

Basic Buffer Formula • Mixture Weak base + Salt/Conjugate acid • NH3 + H2O ↔ NH4

+ + OH_ (dissociate partially)

• NH4CI → NH4+ + CI_ (dissociate fully)

pH = pKa - lg (acid) (salt)

pH = pKa + lg (salt) (acid)

Base Dissociation constant NH3 + H2O ↔ NH4

+ + OH-

Kb = (NH4+) (OH-)

(NH3) -lgKb = -lgOH- -lg (NH4

+) (NH3) -lgOH- = -lgKb + lg (NH4

+) (NH3) pOH = pKb + lg (NH4

+) (NH3)

pOH = pKb + lg (salt) (base)

pOH = pKb - lg (base) (salt)

Basic Buffer Acidic Buffer

salt salt

acid base

Henderson Hasselbalch Equation Henderson Hasselbalch Equation

multiply -lg both sides

Acidic Buffer Calculation

Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid) Ka = 1.8 x 10-5M

CH3COOH ↔ CH3COO- + H+

Ka = (CH3COO-)(H+) (CH3COOH) 1.8 x 10-5 = 0.40 x (H+) (0.10) H+ = 4.5 x 10-6

pH = -lg H+ pH = -lg(4.5 x 10-6) pH = 5.34

pH = pKa -lg(acid) (salt) pH = 4.74 – lg (0.10) (0.40) pH = 5.34

Conc CH3COO- = Moles/volume = 0.20/0.5 = 0.40M

Ka = (1.8 x 10-5) pKa = -lgKa

pKa = -lg(1.8 x 10-5 ) pKa = 4.74

Click here for videos from Khan Academy

Find conc of CH3CH2COONa(salt) added to 1.0dm3 of 1.0M CH3CH2COOH(acid) Ka = 1.8 x 10-5M, pKa = 4.74 , pH 4.5

pH = pKa -lg (acid) (salt) 4.5 = 4.74 – lg (1.0) (salt) lg (1.0) = 0.24 (salt) Conc (salt) = 0.0578M

CH3CH2COOH ↔ CH3CH2COO- + H+

Ka = (CH3CH2COO-)(H+) (CH3CH2COOH) 1.8 x 10-5 = CH3CH2COO- x (3.16 x 10-5) (1.0) CH3CH2COO- = 0.0578M

pH = -lg(H+) 4.5 = -lg(H+) H+ = 3.16 x 10-5


Ka = (CH3COO-)(H+) (CH3COOH) 1.8 x 10-5 = 0.25 x (H+) 0.10 H+ = 7.2 x 10-6

pH = -lg H+ pH = -lg(7.2 x 10-6) pH = 5.14

pH = pKa -lg(acid) (salt) pH = 4.74 – lg (0.10) (0.25) pH = 5.14

Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt) Ka = 1.8 x 10-5M

1st method (formula)

1

Convert Ka to pKa

Ka = (1.8 x 10-5) pKa = -lgKa

pKa = -lg(1.8 x 10-5 ) pKa = 4.74

2nd method (Ka)

2

1st method (formula) Convert Ka to pKa

2nd method (Ka)

3


Find conc salt

2nd method (Ka)

Click here for detail explanation from chem guide

https://www.khanacademy.org/science/chemistry/acids-and-bases/v/pka-and-pkb-relationship


http://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html

Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt) Kb = 1.8 x 10-5M

Basic Buffer Calculation

Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base) Kb = 1.8 x 10-5M

Moles, NH4CI = mass/RMM = 4.28 / 53.5 = 0.08mol

Conc of NH4CI = moles/vol = 0.08/0.25 = 0.32M

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3) 1.8 x 10-5 = 0.32 x OH- 0.50 OH- = 0.50 x 1.8 x 10-5 0.32 OH- = 2.81 x 10-5

pOH = pKb -lg(base) (salt) pOH = 4.74 – lg(0.50) (0.32) pOH = 4.55 pH + pOH = 14 pH = 9.45

1

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

pOH = -lgOH-

pOH = -lg 2.81 x 10-5 pOH = 4.55 pH + pOH = 14 pH = 9.45


pOH = pKb -lg(base) (salt) pOH = 4.74 – lg(0.50) (0.32) pOH = 4.55 pH + pOH = 14 pH = 9.45

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3) 1.8 x 10-5 = 0.32 x OH- 0.50 OH- = 0.50 x 1.8 x 10-5 0.32 OH- = 2.81 x 10-5

2

pOH = -lgOH-


2nd method (Kb) Find conc salt

Find mass of CH3COONa added to 500ml, 0.10M CH3COOH(acid) pH = 4.5, Ka = 1.8 x 10-5M, pKa = 4.74

pH = pKa -lg (acid) (salt) 4.5 = 4.74 – lg (0.10) (salt) lg(0.10) = 4.74 -4.5 (salt) lg (0.10) = 0.24 (salt) Conc salt = 0.0578M


Ka = (CH3COO-) (H+) (CH3COOH) 1.8 x 10-5 = CH3COO- x (10-4.5) (0.10) CH3COO- = 0.10 x 1.8 x 10-5 10-4.5 Conc salt = 0.0578M

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

Conc salt = 0.0578M → x RMM (82) → 4.74g in 1000ml 2.37g in 500ml

3

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

Click here addition base to buffer

Click here acidic buffer simulation

Click here addition acid to buffer

http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons7.htm






Basic Buffer Preparation Acidic Buffer Preparation

Prepare Acidic Buffer at pH = 5.2 • Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]

Use same conc of acid/salt but different vol ratio • 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 100ml (salt) 0.1M, 100ml (salt)

Use same vol of acid/salt but different conc ratio • 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 10.0M, 10ml (salt) 1.00M, 10ml (salt)

Buffer capacity • Adding water will not change the pH of acidic buffer • Ratio of acid/salt still the same

Prepare Basic Buffer at pH = 9.5 or pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]

Use same conc of base/salt but different vol ratio • 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74 1M, 100ml (salt) 0.1M, 100ml (salt)

Use same vol of base/salt but different conc ratio • 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74 1.00M, 10ml (salt) 0.10M, 10ml (salt)

Buffer capacity • Adding water will not change the pH of basic buffer • Ratio of base/salt still the same

Buffer solution

Buffer Preparation

http://www.kliva.com/en/catalog/measuring-instruments/electrodes-sensors-buffer/1346/

1 1

2 2

3 Use fix vol, 1dm3 and use different mole ratio (Acid/salt) • 0.35 mole acid + 1 mole salt to 1 dm3 solvent = 0.35 • 0.035 mole acid + 0.1 mole salt to 1 dm3 solvent = 0.35

Use fix vol, 1dm3 and use different mole ratio (base/salt) • 1.74 mole base + 1 mole salt to 1 dm3 solvent = 1.74 • 0.174 mole base + 0.1 mole salt to 1 dm3 solvent = 1.74

3

3 ways to prepare buffer 3 ways to prepare buffer



Use same conc of acid/salt but different vol ratio Buffer A Buffer B • 1M, 35ml (acid ) = 0.35 or 0.1M, 35ml (acid) = 0.35 1M, 100ml (salt) 0.1M, 100ml (salt)


Use same conc of base/salt but different vol ratio Buffer A Buffer B • 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74 1M, 100ml (salt) 0.1M, 100ml (salt)

Buffer solution

Buffering Capacity


1 1

1M, 35ml

(acid ) 1M, 100ml

(salt )

0.1M, 35ml

(acid )

0.1M, 100ml

(salt)

B A

1M, 174ml

(base)

1M, 100ml

(salt)

0.1M, 174ml (base)

0.1M, 100ml (salt)

B A

Buffer A > Buffer B Stronger buffering capacity • Amt of acid/salt is higher to neutralise added H+ or OH-

• Ratio acid/salt same, pH buffer same but buffering capacity diff • Higher buffer conc – Higher buffering capacity

Buffer A > Buffer B Stronger buffering capacity • Amt of base/salt is higher to neutralise added H+ or OH-


Which has greater buffering capacity ? Which has greater buffering capacity ?




Buffer solution

Buffering Capacity


2 2



3.5M, 10ml

(acid )

10M, 10ml

(salt ) 0.35M, 10ml

(acid )

1M, 10ml

(salt)

B A

1.74M, 10ml

(base)

1M, 10ml

(salt)

0.174M, 10ml (base)

0.1M, 10ml (salt)

B A

Use same vol of acid/salt but different conc ratio Buffer A Buffer B • 3.5M, 10ml (acid ) = 0.35 or 0.35M, 10ml (acid) = 0.35 10M, 10ml (salt) 1.00M, 10ml (salt)



Use same vol of base/salt but different conc ratio Buffer A Buffer B • 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74 1.00M, 10ml (salt) 0.10M, 10ml (salt)





Buffer solution

Buffering Capacity


3 3



0.35mol

(acid )

1.00mol

(salt ) 0.035mol

(acid )

0.10mol

(salt)

B A

1.74mol

(base)

1.00mol

(salt)

0.174mol (base)

0.10mol (salt)

B A



Use fix vol, 1dm3 but diff mole ratio (acid/salt) Buffer A Buffer B • 0.35mol (acid ) = 0.35 or 0.035mol (acid) = 0.35 1.00mol (salt) 0.100mol (salt)

1dm3 1dm3 1dm3 1dm3

Use fix vol, 1dm3 but diff mole ratio (base/salt) Buffer A Buffer B • 1.74mol (base) = 1.74 or 0.174mol (base) = 1.74 1.00mol (salt) 0.100mol (salt)





Buffer solution

Buffering Capacity


4 4

Will pH change

pH Buffer A = pH Buffer B • Same pH • Adding water will not change the pH • Amt of acid/salt still the same • Ratio conc acid/salt same, pH buffer same

0.35mol

(acid )

1.00mol

(salt )

0.35mol

(acid )

1.00mol

(salt)

B A

1.74mol

(base)

1.00mol

(salt)

1.74mol (base)

1.00mol (salt)

B A

Same mole ratio (acid/salt) but different total volume Buffer A Buffer B • 0.35mol (acid)= 0.35 in 1dm3 or 0.35mol (acid) = 0.35 in 2dm3

1.00mol (salt) 1.00mol (salt)

1dm3

2dm3 1dm3

Will pH change

Same mole ratio (base/salt) but different total volume Buffer A Buffer B • 1.74mol (base)= 1.74 in 1dm3 or 1.74mol (base) = 1.74 in 2dm3

1.00mol (salt) 1.00mol (salt)

2dm3

pH Buffer A = pH Buffer B • Same pH • Adding water will not change the pH • Amt of acid/salt still the same • Ratio conc base/salt same, pH buffer same

Acidic Buffer Preparation Acidic Buffer Preparation


Buffer solution

Buffering Capacity


5 5

Which has greater buffering capacity ?

Buffer A > Buffer B • Conc ratio [acid]/[salt] equal to 1 •Buffer has highest buffering capacity when pH = pKa • Conc acid = Conc salt → highest buffering capacity

Concentration ratio [acid]/[salt] = 1

1.00mol

(acid)

1.00mol

(salt)

A

1.00mol

(salt)

B

Buffer A > Buffer B • Further the conc ratio [acid]/[salt] from 1, lower buffer capacity

Same conc ratio (acid/salt) in 1dm3 Buffer A • 1.00mol (acid ) = 1.00 1.00mol (salt)

1dm3 1dm3


Different conc ratio (acid/salt) in 1dm3 Buffer B • 0.35mol (acid ) = 0.35 1.00mol (salt)

Which has greater buffering capacity ?

0.35mol

(acid)

Concentration ratio [acid]/[salt] ratio < 1

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

pH = pKa -lg (acid) (salt)

CH3COOH (acid) ↔ CH3COO- + H+

CH3COONa (salt) → CH3COO- + Na+

pH acidic buffer depend on Ka and ratio [acid]/[salt]

2 ways to prepare buffer • 1st – calculate ratio of [acid]/[salt] • 2nd – titration weak acid with strong base 1 Prepare Acidic Buffer at pH = 5.2

• Choose pKa acid closest to pH 5.2 • pKa = 4.74 (ethanoic acid) chosen • pH = pKa -lg [acid] [salt] • 5.2 = 4.74 – lg [acid] [salt] • [acid] = 0.35 [salt] Ratio of [acid] = 0.35 [salt]

Use same conc of acid/salt but diff vol ratio • 1M, 35ml (acid ) = 0.35 1M, 100ml (salt)

Use same vol of acid/salt but diff conc ratio • 0.35M, 10ml (acid ) = 0.35 1.00M, 10ml (salt)

1M, 35ml

(acid )

1M, 100ml

(salt )

0.35M, 10ml

(acid )

1M, 10ml

(salt )

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

2

Click here buffer simulation

Strong base NaOH

Weak acid CH3COOH

CH3COOH + NaOH → CH3COONa + H2O

Initial 1 mol 0.5 mol added 0.

Change (1 – 0.5)mol 0 mol 0.5mol form

Final 0.5mol left 0 mol 0.5mol form

At half equivalent point : • Amt acid = Amt salt : ( 0.5 = 0.5) • 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 equivalent 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]

Same conc Diff Volume

Same Volume Diff Conc

Same conc ratio Same conc ratio

1mol

0.5mol

Buffer region at half equivalent point

Amt acid = Amt salt

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


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

Preparation acidic buffer by titration • 1st – weak acid (burette) with strong base (flask) 2nd – strong base (burette) with weak acid (flask)

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

2


Strong base NaOH

Weak acid CH3COOH





At half equivalent point : • Amt acid = Amt salt : ( 0.5 = 0.5) • 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 equivalent point when amt acid = amt salt



Amt acid = Amt salt

Strong base NaOH

Weak acid CH3COOH


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


Buffer region Amt acid = Amt salt


Initial 2mol added 1.0 mol 0.

Change (2 – 1)mol 0 mol 1mol form

Final 1mol left (excess) 0 mol 1mol form

When excess 1 mol acid added: • Amt acid = Amt salt : ( 1 = 1) • pH = pKa -lg [acid] → pH = pKa → 4.74 = 4.74

[salt] Buffer at pH = 4.74 form when amt of acid added = amt salt form at equivalent point

1mol

0.5mol

1mol

2mol





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

pOH = pKb -lg (base) (salt)

pH basic buffer depend on Kb and ratio [base]/[salt] • NH3 (base) + H2O ↔ NH4

+ + OH_

• NH4CI (salt) → NH4+ + CI_

2 ways to prepare buffer • 1st – calculate ratio of [base]/[salt] • 2nd – titration weak base with strong acid

Prepare Buffer pH = 9.5 /pOH = 4.5 • Choose pKb base closest to pOH = 4.5 • pKb = 4.74 (NH3) chosen • pOH = pKb -lg [base] [salt] • 4.5 = 4.74 – lg [base] [salt] • [base] = 1.74 [salt] Ratio of [base] = 1.74 [salt]

Use same conc of base/salt but diff vol ratio • 1M, 174ml (base) = 1.74 1M, 100ml (salt)

Use same vol of base/salt but diff conc ratio • 1.74M, 10ml (base) = 1.74 1.00M, 10ml (salt)

1

1M, 100ml

(salt )

1M, 174ml

(base )

1M, 10ml

(salt )

1.74M, 10ml

(base )

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

Strong acid HCI

NH3 + HCI → NH4CI + H2O




At half equivalent point : • Amt base = Amt salt : (0.5 = 0.5) • 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 equivalent point when amt base = amt salt

Weak base NH3


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]

Same conc Diff Volume

Same Volume Diff Conc


Amt base = Amt salt

1mol

0.5mol

Same conc ratio Same conc ratio



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

1 2 Titration bet strong acid with weak base NH3 + HCI → NH4CI + H2O

Strong acid HCI





At half equivalent point : • Amt base = Amt salt : (0.5 = 0.5) • 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 equivalent point when amt base = amt salt

Weak base NH3




Amt base = Amt salt

Preparation basic buffer by titration • 1st – weak acid (burette) with strong base (flask) 2nd – strong acid (burette) with weak base (flask)

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

Strong acid HCI

Weak base NH3

1mol

0.5mol



Initial 2 mol added 1.0mol 0.

Change (2 – 1)mol 0 mol 1mol form

Final 1mol left (excess) 0 mol 1mol form

When excess 1 mol base added: • Amt base = Amt salt : (1 = 1 ) • pOH = pKb - lg [base] → pOH = pKb → 4.74 = 4.74 [salt] Buffer at pOH = 4.74 form when amt of base added = amt salt form at equivalent point


Buffer region Amt base = Amt salt

1mol

2mol





Sample 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


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-


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


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.

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



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



Sample 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

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




Find pH buffer - by mixing 200ml, 0.60M NH3 (base) with 300ml, 0.30M NH4CI (salt) Kb = 1.8 x 10-5M, pKb = 4.74

NH3 aft mixing = moles/vol = (200 x 0.60)/0.5 = 0.24M

NH4 aft mixing = moles/vol = (300 x 0.30)/0.5 = 0.18M

pOH = pKb -lg (base) (salt) pOH = 4.74 – lg (0.24) (0.18) pOH = 4.62 pH + pOH = 14 pH = 9.38

5

Find pH buffer by adding 3.20g CH3COONa to 1.00dm3, 0.01M CH3COOH(acid) Ka = 1.75 x 10-5M, pKa = 4.75

Conc salt = 3.20g/dm3 ↓ ÷ RMM (82) ↓ 0.039M

pH = pKa -lg (acid) (salt) pH = 4.75 – lg (0.01) (salt) pH = 4.75 – lg (0.01) (0.039) pH = 4.75 + 0.591 pH = 5.34


Ka = (CH3COO-)(H+) (CH3COOH) 1.75 x 10-5 = 0.039 x (H+) (0.01) H+ = 0.01 x 1.75 x 10-5 0.039 H+= 4.487 x 10 -6

pH = -lg(H+) = -lg(4.487 x 10 -6) pH = 5.34

200ml, 0.60M NH3

300ml, 0.30M NH4CI

Total Volume 500ml or 0.5dm3

Conc before Conc after pH buffer


NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3) 1.8 x 10-5 = 0.18 x OH- 0.24 OH- = 0.24 x 1.8 x 10-5 0.18 OH- = 2.4 x 10-5

pOH = -lgOH-


2nd method (Kb)

6

1 dm3 , 0.01M

CH3COOH

3.20g CH3COONa

pH buffer

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

Click here on titration animation Click here on titration simulation

Click here acidic buffer simulation Click here acidic buffer simulation

Simulation and Animation on Buffer and Titrations

Click here basic buffer simulation

Click here for videos from Khan Academy

Click here acidic buffer animation

Click here acidic buffer animation Click here buffer simulation

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/simDownload/index4.html

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/simDownload/index4.html

http://www.virtlab.com/index.aspx

http://www.virtlab.com/index.aspx









http://www.media.pearson.com.au/schools/cw/au_sch_derry_ibchl_1/int/titrationCurve/1502.html

http://www.media.pearson.com.au/schools/cw/au_sch_derry_ibchl_1/int/titrationCurve/1502.html

http://auth.mhhe.com/physsci/chemistry/animations/chang_7e_esp/crm3s5_5.swf

http://auth.mhhe.com/physsci/chemistry/animations/chang_7e_esp/crm3s5_5.swf



ib chemistry on acidic, basic buffer and ph calculation

Education