Absorption spectrum- Colour Formation

Colour seen BLUE Blue reflect/transmit to eyes - Red/orange absorb (complementary colour)

Complementary colour

Blue

transmitted

Wavelength - absorbed

Visible light

absorbed

Absorption spectrum- Colour Formation

Violet

= 410nm

E = hf = 6.626 x 10-34 x 7.31 x 1014

= 4.84 x 10-19 J

Red

= 700nm

E = hf = 6.626 x 10-34 x 4.28 x 1014

= 2.83 x 10-19 J High Energy

Low Energy

https://www.wou.edu/las/physci/ch462/tmcolors.htm

Absorption spectrum- Colour Formation

Colour seen GREEN GREEN reflect/transmit to eyes - Red/Blue absorb (complementary colour)

Complementary colour

Green

transmitted

Wavelength - absorbed

Visible

light

absorbed

Violet

= 410nm

E = hf = 6.626 x 10-34 x 7.31 x 1014

= 4.84 x 10-19 J

Red

= 700nm

E = hf = 6.626 x 10-34 x 4.28 x 1014

= 2.83 x 10-19 J High Energy

Low Energy

Absorption spectrum- Colour Formation

https://www.wou.edu/las/physci/ch462/tmcolors.htm

Beer Lambert Law Linear relationship bet Abs and conc at fix wavelength

lcI

Io 10logAbsorbanceI

Io 10log

Using UV/vis spectrophotometer

Protein Assay Amt /Conc protein solution

Colour formation Bind protein with complex dye (Coomassie blue/BSA/Biuret)

Blue complex sol formed

Find wavelength complex absorb Max Max 595nm Max absorption at 595nm

Protein sol colourless Complex protein with BSA

Add complex dye BSA

Colour seen BLUE Blue reflect/transmit to eyes - Red/orange absorb (complementary colour)

Wavelength Max 595nm

Blue

transmitted

absorbed

Using UV/vis spectrophotometer

Monochromator Cuvette Detector

Io I

Io = Intensity incident light I = Intensity transmitted = molar absorption coefficient

(constant for absorbing sub I = path length (constant, 1cm)

lcAbsorbance

concAbsorbance ...

and I constant

Beer Lambert Law Linear relationship bet Abs and Conc at fix wavelength

lcI

Io 10logAbsorbanceI

Io 10log

Using UV/vis spectrophotometer

Monochromator Cuvette Detector

Io I

lcAbsorbance

concAbsorbance ...

and I constant

Ab

sorb

an

ce

- Unknown protein conc determine by interpolation - Abs unknown is 1.11, by interpolation, conc is 0.75mg/l - Must be diluted , at high conc , deviation from Law

interpolation

Tube Vol, Protein

BSA

Vol H2O

Vol Buffer

Dil Protein

conc

Abs/ 540nm

1 0.00 2.00 2.00 0.00 0.00

2 0.10 1.90 2.00 0.125 0.34

3 0.30 1.70 2.00 0.375 0.67

4 0.50 1.50 2.00 0.625 1.02

5 0.70 1.30 2.00 0.875 1.35

6 1.00 1.00 2.00 1.25 1.65

Tube 1-6 are diluted protein with complex dye BSA, prepared using initial protein conc of 5.00mg/l

Add BSA Add water/buffer

Protein colourless Complex protein/ BSA Diluted protein/BSA (5.00mg/l)

0.125 0.375 0.625 0.875 1.25

0.34

0.67

1.02

1.35

1.65

Std calibration plot, known protein conc vs Abs

Unknown Absorbance = 1.11 Conc = 0.75mg/l

interpolation

Abs = 1.11

Conc = 0.75 mg/l

protein conc

Beer Lambert Law Linear relationship bet Abs and Conc at fix wavelength

lcI

Io 10logAbsorbanceI

Io 10log

Monochromator Cuvette Detector

Io I

lcAbsorbance

concAbsorbance ...

and I constant

protein conc

Ab

sorb

an

ce

interpolation

0.125 0.375 0.625 0.875 1.25

0.34

0.67

1.02

1.35

1.65

Std calibration plot, known protein conc vs Abs

Beers Lambert Law Apply for diluted solution

Absorbance Conc Absorbance,A = log10 (Io/I) = lc

AbsorbanceI

Io 10log

Abs = lc If and l = constant

Amt light absorb depend on = Molar extinction compound

c = Conc l = path length

Molar extinction of compound, : Measure strength of absorption of sub Higher = Higher Absorbance Sub with high = effective at absorbing light even when low conc is used.

Path Length, l: Longer path length Higher Abs

Concentration, c: Higher conc of analyte Higher Abs

Abs Conc, c

Using UV/vis spectrophotometer

Molar absorptivity, = A/bc = 0.3554 /(2.10 x 7.25 x 10-5) = 2.33 x 103 L mol-1cm-1

Determine conc of unknown using Beer-Lambert Law

7.25 x 10-5M X has absorbance of 0.355 when measured in 2.10 cm cell at wavelength 525nm. Cal Molar absorptivity,

100dm3 contaminated water was reduced by boiling to 7.50dm3. Reduced vol was tested, its absorbance is 2.00. Cal conc Pb2+ (mgdm3) in original sample.

V = 100 dm3

M = ?

V = 7.50 dm3

M = 1.15mg/dm3

Amt bef heating = Amt aft heating Moles bef = Moles aft

M x V = M x V M x 100 = 1.15 x 7.50 M = (1.15 x 7.50)/100 M = 0.0863 mgdm-3

Std calibration plot, Abs vs Conc Pb

Ab

sorb

an

ce

0.3 0.5 0.7 1.1 1.2

Pb conc

interpolation

Conc Pb2+

1.15 mgdm-3

Std calibration curve, Abs vs Conc Conc unknown (Pb2+) by interpolation

Abs = 0.340

Conc = 0.310

1

2

3 Determine unknown conc of Pb2+ using std calibration plot. If unknown sample has Abs 0.34, find conc Pb

Abs

Conc

Determine conc of unknown using Beer-Lambert Law

V = 5 cm3

M = ?

Amt bef dilution = Amt aft dilution Moles bef = Moles aft

M x V = M x V M x 5 = 0.38 x 100 M = (0.38 x 100)/5

M = 7.6 mgcm-3

Std calibration plot, Abs vs Conc protein

Ab

sorb

an

ce

o.1 0.2 0.3 0.4 protein conc

interpolation

Conc protein

0.38 mgcm-3

4 5 cm3 sample protein diluted with buffer to vol of 100 cm3 and analysed with UV. Abs was 1.85 Using std calibration plot, determine conc protein in original sample.

V = 100 cm3

M = 0.38

2 cm3 protein was diluted with buffer to vol of 25 cm3 and analysed with UV. Abs was 0.209 Using std calibration plot, determine conc protein in original sample.

5

Conc protein

0.310 mmoldm-3

V = 2 cm3

M = ?

V = 25 cm3

M = 0.310

Amt bef dilution = Amt aft dilution Moles bef = Moles aft

M x V = M x V M x 2 = 0.310 x 25 M = (0.310 x 25)/2

M = 3.87 mmoldm-3

interpolation

Chromatography Techniques Separation technique of mix into their pure components

Identify sample - mix or pure both quantitatively and qualitatively Interaction of sub bet 2 phase - Stationary phase and Mobile phase

Separation based on Partition or Adsorption

Chromatography Techniques

Separation analysis

Paper Chromatography

Thin Layer Chromatography

Adsorption Chromatography

Chromatography

Partition Chromatography

Column Chromatography

Partition Chromatography Component distribute bet TWO immisible liquid phase Depend on relative solubility bet TWO phase Solutes bond to stationary phase or mobile phase

Adsorption Chromatography Component adsorb on solid stationary phase Depend on polarity of stationary, mobile phase and solutes Stationary phase is polar polar solutes adsorb strongly Stationary phase is non polar non polar solutes adsorb strongly Mobile phase is polar polar solutes stay in mobile phase Mobile phase is non polar non polar solutes stay in mobile phase

Application Detection amino acids in mix Diff dyes in food colouring Separation plant pigments

Y adsorb

strongly

Application Collection of sample of pigments

X adsorb

strongly

Application Detection amino acids in mix Diff dyes in food colouring Separation plant pigments

X adsorb

weakly

Chromatography

Partition Chromatography components distribute bet 2 immisible liquid phase relative solubility in 2 phase bond strongly to mobile phase move faster

Adsorption Chromatography Components adsorp on solid stationary phase

Stationary phase has layer of liq

Mobile

liq phase

containing

X and Y X

X

X

Y

Y

Y

Stationary

Liquid phase

Partition distribution solute X and Y bet 2 liq phase X more soluble in mobile phase (move with mobile liq phase) Y less soluble in mobile phase (stay on stationary liq phase)

Stationary phase has layer of liq

Y

Y

Y

Y

X

X X

X X

X Mobile Iiq phase

containing X

Stationary phase solid AI2O3 SiO2

O- O- O- O-

O-

Adsorption solute X and Y adsorb temporary on solid Y adsorb strongly on solid phase, eluted slower

X in liq mobile phase, eluted faster

Y

Y

Y

Y

Mobile

liq phase

containing

X and Y

X

X

X

O- O- O-

Y Y

Y

Y

X X

X X

Mobile liq phase

containing X

Y adsorb strongly

X X

X

X

Y Y

Y

Y

Separation of X and Y

X X

X Y Y Y

Chromatography Techniques

Paper Chromatography

Partition chromatography Distribution solute bet both liquid phase Depend on relative solubility

Aqueous liq phase on surface of stationary phase (paper)

Mobile liq phase - solvent

Solvent move by capillary action

Stationary phase - Cellulose paper absorb water on its surface

Mobile Liquid phase with solute X and Y

Y

Y

Y

X

X

Chromatography Techniques

Component separated identified using Rf value Rf = Retention factor for given eluent. Measured distance from original spot to centre of particular component to solvent front

solventbyceDis

solutebyceDisfactortention

....tan

....tan.Re

Rf green spot = (3/12) = 0.25

Rf blue spot = 6/12 = 0.5

Separation using TLC TLC techniques step by step

Column Chromatography Column separation

http://www.youtube.com/user/ChemToddlerhttp://www.youtube.com/watch?v=yoYECMP_1iI&feature=relatedhttp://www.youtube.com/watch?feature=player_embedded&v=nCqwZuE4boYhttp://www.youtube.com/watch?feature=player_embedded&v=clx8IfAgtXo

Electrophoresis

Amino acid Zwitterion (Electrically neutral) Amino acid with isoelectric point

Isoelectric point pH when amino acid is electrically neutral Isoelectric point alanine = 6.02 (Ave of pKa)

Separation amino acid based on charges using electric field

At pH 6.02

Alanine is electrically neutral (Zwitterion)

pH = isoelectric

Alanine ( Neutral)

pH < isoelectric

Alanine (+ve)

pH > isoelectric

Alanine (- ve)

Electrophoresis

Isoelectric point pH when amino acid is electrically neutral

Isoelectric point alanine = 6.02 (Ave of pKa)

Separation amino acid based on charges using electric field

At pH 6.02

Alanine is electrically neutral (Zwitterion)

pH = isoelectric

Alanine (Neutral)

pH < isoelectric

Alanine (+ve)

pH > isoelectric

Alanine (-ve)

Mix of amino acid (ala, arg, isoleu, asp acid) Spot at center on gel (polyacrylamide)

Buffer, pH 6 added. Electric field applied

Amino acid separate based on charges. Ninhydrin applied to identified spots.

Separation based on charges pH = iso point = No movement (zwitterion)/ electrically neutral pH < iso point = + ve charge (cation) = move to ve (cathode) pH > iso point = - ve charge (anion) = move to +ve (anode)

pH = 6

+ ve - ve

http://chemwiki.ucdavis.edu/Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map:_Essential_Organic_Chemistry_(Bruice)/16:_The_Organic_Chemistry_of_Amino_Acids,_Peptides,_and_Proteins/16.05:_Separating_Amino_Acids

Electrophoresis

Isoelectric point pH when amino acid is electrically neutral

Isoelectric point alanine = 6.02 (Ave of pKa)

Separation amino acid based on charges using electric field

pH = isoelectric

Alanine (Neutral)

pH < isoelectric

Alanine (+ve)

pH > isoelectric

Alanine (-ve)

+ H+

Acidic (pH < pI) Cation/zwitterion

- H+

+ H+

- H+

Cation (conjugate acid)

Zwitterion (conjugate base)

+ H+ + H+

Zwitterion (conjugate acid)

Anion (conjugate base)

Alkaline(pH > pI) Zwitterion/anion

Cation (+) Zwitterion Anion (-)

+ H+

Acidic (pH < pI) Cation/zwitterion

- H+

+ H+

- H+

Cation (conjugate acid)

Zwitterion (conjugate base)

+ H+ + H+

Zwitterion (conjugate acid)

Anion (conjugate base)

Alkaline(pH > pI) Zwitterion/anion

Acidic Base Buffer Calculation

][

][log1

cation

zwitterionpKpH

][

][log2

zwitterion

anionpKpH

Find pH 0.8M zwitterion and 0.2M anionic form serine

+ H+ + H+

- H+ - H+

Amino acid

pK1 pK2 Isoelectric

Serine 2.2 9.1 5.7

Zwitterion (conjugate acid) Anion (conjugate base)

5.88.0

2.0log1.9

][

][log2

pH

zwitterion

anionpKpH

At pH 7, alanine contain zwitterion and anionic form i. Deduce structural formula ii. State eqn of buffer when small amt acid and base added

Zwitterion (conjugate acid)

Anion (conjugate base)

Alkaline (pH > pI) Zwitterion/anion

Zwitterion (conjugate acid)

Anion (conjugate base)

+ OH- + H+

Acid added Base added

2

1

Zwitterion (conjugate base)

Cation (conjugate acid)

Acidic (pH < pI) Cation/zwitterion

Amino acid

pK1 pK2 Isoelectric

Glycine 2.3 9.6 6.0

Amino acid prepared by mixing 0.6 dm3, 0.2M HCI and 0.4dm3, 0.5M glycine Find pH original buffer

n (HCI) = 0.6 x 0.2 = 0.12 mol n (glycine) = 0.4 x 0.5 = 0.20 mol Total vol = 1 dm3

Conc (HCI) = Mol/Vol = 0.12/1 = 0.12M Conc (glycine) = Mol/Vol = 0.20M

][

][log1

cation

zwitterionpKpH

12.2

]12.0[

]08.0[log3.2

pH

pH

2

][

][log1

cation

zwitterionpKpH

Zwitterion (conjugate base)

Cation (conjugate acid)

Acidic (pH < pI) Cation/zwitterion

Amino acid

pK1 pK2 Isoelectric

Glycine 2.3 9.6 6.0

Amino acid prepared by mixing 0.4 mol zwitterion and 0.16mol cationic glycine in 1 dm3

Find pH of buffer

Total vol = 1 dm3

Conc (cation) = Mol/Vol = 0.16/1 = 0.16M Conc (zwitterion) = Mol/Vol = 0.4/1 = 0.4M

][

][log1

cation

zwitterionpKpH

7.2

]16.0[

]4.0[log3.2

pH

pH

Find pH buffer when - 0.025M of NaOH added to buffer

3

4

0.16 0.025 0.4

- 0.025 - 0.025 + 0.025

0.135 0.425

OH - - -COOH +

Conc (zwitterion) = Mol/Vol = 0.4/1 = 0.4M Conc (cation) = Mol/Vol = 0.16/1 = 0.16M

8.2

]135.0[

]425.0[log3.2

pH

pH

NH3 NH4+

Buffer Solution

Acid part

Neutralize

each other

Salt part

Base part

- NH3(weak base) + NH4CI (salt) - NH3 + H2O NH4+ + OH NH3 neutralise added H+ - NH4CI NH4+ + CI NH4+ neutralise added OH

- Effective buffer equal amt weak base NH3 and conjugate acid NH4+

Acidic Buffer Basic Buffer

Resist a change in pH when small amt acid/base added.

CH3COOH + H2O CH3COO- + H3O

+

Acidic Buffer - weak acid and its salt/conjugate base

CH3COOH CH3COO-

Conjugate acid base pair

CH3COOH

CH3COO-

Weak Acid Conjugate Base

BUFFER

Dissociate fully

HCOOCHCOOHCH 33

COOHCH3 COONaCH3

NaCOOCHCOONaCH 33Dissociate partially

- CH3COOH (weak acid) + CH3COONa (salt) - CH3COOH CH3COO- + H+ CH3COOH neutralise added OH - CH3COONa CH3COO

- + Na+ CH3COO- neutralise added H+

- Effective buffer equal amt weak acid CH3COOH and base CH3COO-

COOHCH3COOCH3

BUFFER

Add acid H+ Add alkaline OH-

Neutralize

each other

Basic buffer - weak base and its salt/conjugate acid

OHNHOHNH 423

NH3 + H2O NH4+ + OH-

NH3

Weak Base

NH4+

Conjugate acid

CINH 43NH

BUFFER

Conjugate acid base pair

Add acid H+ Add alkaline OH-

Neutralize

each other

Neutralize

each other

Dissociate partially

CINHCINH 44

3NH

4NH

Base part Salt part

Acid part

Dissociate fully

BUFFER

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 Mix Weak acid + Salt/Conjugate base CH3COOH CH3COO

- + H+ (dissociate partially)

CH3COONa CH3COO- + Na+ (dissociate fully)

Basic Buffer Formula Mix 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 Eqn

multiply -lg both sides

Henderson Hasselbalch Eqn

Acidic Buffer Calculation

How much 0.10M butanoic acid and solid potassium butanoate needed to make 1.0 dm3, pH 5.00 buffer solution. State assumption used. pKa acid = 4.83

Need 0.15 mol in 1 dm3

Mass salt = mol x RMM = 0.15 x 126.12 = 19 g salt

Click here video Khan Academy

Find pH buffer made with 0.20 mol CH3COONa(salt) in 500cm3 of 0.10M CH3COOH(acid)

Ka = 1.8 x 10-5

Find pH buffer - adding 25 ml, 0.10M CH3COOH(acid) 25ml, 0.10M CH3COONa(salt) Ka = 1.8 x 10

-5

1st method (formula)

1

Convert Ka to pKa

2nd method (Ka)

2

1st method (formula)

3

1st method (formula)

Molar mass salt = 126.12 gmol-1

2nd method (Ka)

Click here explanation from chem guide

74.4

]05.0[

]05.0[lg74.4

][

][lg

pH

pH

salt

acidpKpH a

74.4

)108.1lg(

)lg(

108.1

05.0

))(05.0(108.1

)(

))((

5

5

5

3

3

pH

pH

HpH

H

H

COOHCH

HCOOCHK a

3/15.0][

][

]10.0[lg83.400.5

][

][lg

dmmolsalt

salt

salt

acidpKpH a

74.4

)108.1lg(

lg

108.1

5

5

a

a

aa

a

pK

pK

KpK

K

35.5

105.4][

10.0

))(40.0(108.1

)(

))((

6

5

3

3

pH

MH

H

COOHCH

HCOOCHK a

35.5

]40.0[

]10.0[lg74.4

][

][lg

pH

pH

salt

acidpKpH a

3/40.0

50.0

20.0

dmmolconc

conc

vol

molconc

Conc salt

Equal vol and conc Ratio acid/salt = 1

Assumption used

[butanoic acid]eq = [salt] used [acid]eq = [acid] used No vol change during mixing

https://www.khanacademy.org/science/chemistry/acids-and-bases/v/pka-and-pkb-relationshiphttps://www.khanacademy.org/science/chemistry/acids-and-bases/v/pka-and-pkb-relationshiphttp://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html

Acidic Buffer Calculation

Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid)

Ka = 1.8 x 10-5

Conc CH3COO- = Moles/Vol

= 0.20/0.5 = 0.40M

Click here video0 Khan Academy

Find conc CH3COONa(salt) added to 1.0dm3 of 1.0M CH3COOH(acid)

Ka = 1.8 x 10-5, pKa = 4.74 , pH 4.5

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

-5

1st method (formula)

4

Convert Ka to pKa

2nd method (Ka)

5

1st method (formula) Convert Ka to pKa

2nd method (Ka)

6

1st method (formula)

Conc salt

2nd method (Ka)

Click here explanation from chem guide

14.5

]25.0[

]10.0[lg74.4

][

][lg

pH

pH

salt

acidpKpH a

14.5

)102.7lg(

)lg(

102.7

10.0

))(25.0(108.1

)(

))((

6

6

5

3

3

pH

pH

HpH

H

H

COOHCH

HCOOCHK a

34.5

]40.0[

]10.0[lg74.4

][

][lg

pH

pH

salt

acidpKpH a

74.4

)108.1lg(

lg

108.1

5

5

a

a

aa

a

pK

pK

KpK

K

74.4

)108.1lg(

lg

108.1

5

5

a

a

aa

a

pK

pK

KpK

K

MCOOCH

COOCH

COOHCH

HCOOCHK a

0578.0

0.1

)1016.3)((108.1

)(

))((

3

5

35

3

3

Msalt

salt

salt

salt

acidpKpH a

0578.0][

24.0][

]0.1[lg

][

]0.1[lg74.45.4

][

][lg

34.5

)105.4lg(

)lg(

105.4

10.0

))(40.0(108.1

)(

))((

6

6

5

3

3

pH

pH

HpH

H

H

COOHCH

HCOOCHK a

51016.3

)lg(5.4

)lg(

H

H

HpH

Conc [H+]

https://www.khanacademy.org/science/chemistry/acids-and-bases/v/pka-and-pkb-relationshiphttps://www.khanacademy.org/science/chemistry/acids-and-bases/v/pka-and-pkb-relationshiphttp://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html

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

-5

Basic Buffer Calculation

Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base)

Kb = 1.8 x 10-5

Mole NH4CI = mass/RMM = 4.28 / 53.5 = 0.08 mol

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

7

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

1st method (formula)

8

2nd method (Kb) Conc salt

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

-5, pKa = 4.74

Conc CH3COO- = 0.0578M x RMM (82) 4.74g in 1000ml 2.37g in 500ml

9

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

Click here addition base to buffer

Click here addition acid to buffer

45.955.414

55.4

]32.0[

]50.0[lg74.4

][

][lg

pH

pOH

pOH

salt

basepKpOH b

45.955.414

55.4

)1081.2lg(

)lg(

5

pH

pOH

pOH

OHpOH

5

5

3

4

423

1081.2

50.0

))(32.0(108.1

)(

))((

OH

OH

NH

OHNHK

OHNHOHNH

b

45.955.414

55.4

]32.0[

]50.0[lg74.4

][

][lg

pH

pOH

pOH

salt

basepKpOH b

45.955.414

55.4

)1081.2lg(

)lg(

5

pH

pOH

pOH

OHpOH

5

5

3

4

423

1081.2

50.0

))(32.0(108.1

)(

))((

OH

OH

NH

OHNHK

OHNHOHNH

b

0578.0][

24.0][

]10.0[lg

][

]10.0[lg74.45.4

][

][lg

3

3

3

3

COOCH

COOCH

COOCH

COOCH

acidpKpH a

5.410

)lg(5.4

)lg(

H

H

HpH

MCOOCH

COOCH

COOHCH

HCOOCHK

HCOOCHCOOHCH

a

0578.0][

)10.0(

)10)((108.1

)(

))((

3

5.4

35

3

3

33

Conc [H+]

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Given 100ml of 0.05M HCOOH , what vol 0.05M NaOH needed to make pH buffer 4.23. pKa = 3.75

Buffer Calculation

100ml buffer contain 0.10M butanoic acid and 0.20 M sodium butanoate. What pH change when 2.0cm3, 0.10M HCI added. pKa = 4.82

10

11

After adding acid

Click here addition base to buffer

Click here addition acid to buffer

375

][

][lg75.323.4

][

][lg

cmvol

salt

acid

salt

acidpKpH a

1000

05.0....

vsaltmolNaOHmol

NaOH + HCOOH HCOONa + H2O All NaOH react to form salt Mol NaOH react = Mol salt form

Mol acid remain = Mol initial mol react

1000

05.0

1000

10005.0..

vacidmol

12.5

]020.0[

]010.0[lg82.4

][

][lg

pH

pH

salt

acidpKpH a

How would adding 100ml DI water affect pH.

Before adding acid

0002.01000

10.00.2....

addedacidmol

0102.00002.001.0.. acidmolTotal

0198.00002.002.0..

0002.01000

10.00.2....

leftsaltmol

reactedsaltmol

After adding acid

10.5

]0198.0[

]0102.0[lg82.4

][

][lg

pH

pH

salt

acidpKpH a

Buffer do not change pH on dilution

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