Option B Chemistry UV Spectroscopy, Electrophoresis and buffer calculation

1. Absorptionspectrum- 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
λ = 410nm
E = hf
= 6.626 x 10-34 x 7.31 x 1014
= 4.84 x 10-19 J
Red
λ = 700nm
E = hf
= 6.626 x 10-34 x 4.28 x 1014
= 2.83 x 10-19 J
High Energy
Low Energy

2. Absorptionspectrum- 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
λ = 410nm
E = hf
= 6.626 x 10-34 x 7.31 x 1014
= 4.84 x 10-19 J
Red
λ = 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

3. Beer Lambert Law
Linear relationshipbet Abs and conc at fix wavelength
lc
I
Io
10logAbsorbance
I
Io
10log
Using UV/vis spectrophotometer
ProteinAssay – Amt /Conc protein solution
Colour formation – Bind protein with complex dye
(Coomassie blue/BSA/Biuret)
↓
Blue complex sol formed
↓
Find wavelength complex absorb – λMax
λMax – 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 – λMax 595nm
Blue
transmitted
absorbed
Using UV/vis spectrophotometer
Monochromator Cuvette Detector
Io I
Io = Intensity incident light
I = Intensity transmitted
ε = molar absorption coefficient
(constant for absorbing sub
I = path length (constant, 1cm)
lcAbsorbance 
concAbsorbance ...
ε and I constant

4. Beer Lambert Law
Linear relationshipbet Abs and Conc at fix wavelength
lc
I
Io
10logAbsorbance
I
Io
10log
Using UV/vis spectrophotometer
Monochromator Cuvette Detector
Io I
lcAbsorbance 
concAbsorbance ...
ε and I constant
Absorbance
- 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

5. Beer Lambert Law
Linear relationshipbet Abs and Conc at fix wavelength
lc
I
Io
10logAbsorbance
I
Io
10log
Monochromator Cuvette Detector
Io I
lcAbsorbance 
concAbsorbance ...
ε and I constant
protein conc
Absorbance
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
Beer’s Lambert Law
• Apply for diluted solution
• Absorbance α Conc
• Absorbance,A = log10 (Io/I) = έlc
Absorbance
I
Io
10log
Abs = έlc
If έ and l = constant
Amt light absorb depend on
• έ = Molar extinction compound
• c = Conc
• l = path length
Molar extinctionof compound, έ :
• Measure strength of absorption of sub
• Higher έ ↑ = Higher ↑ Absorbance
• Sub with high έ = effectiveat 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 α Conc, c
Using UV/vis spectrophotometer

6. Molar absorptivity,έ
έ = 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, έ
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 = Molesaft
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
Absorbance
0.3 0.5 0.7 1.1 1.2
Pb conc
interpolation
Conc Pb2+
1.15 mgdm-3
Std calibrationcurve, 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

7. Determine conc of unknown using Beer-Lambert Law
V = 5 cm3
M = ?
Amt bef dilution= Amt aft dilution
Moles bef = Molesaft
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
Absorbance
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 = Molesaft
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

8. ChromatographyTechniques
• 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 Partitionor 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 stationaryphase
• 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
• Detectionamino acids in mix
• Diff dyes in food colouring
• Separation plant pigments
Y adsorb
strongly
Application
Collection of sample
of pigments
X adsorb
strongly
Application
• Detectionamino acids in mix
• Diff dyes in food colouring
• Separation plant pigments
X adsorb
weakly

9. 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 stationaryphase
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
XX
XX
XMobile
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
XX
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

10. 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
Componentseparatedidentifiedusing Rf value
• Rf = Retention factorfor given eluent.
• Measured distance from original spot
to centre of particularcomponent to solvent front
solventbyceDis
solutebyceDis
factortention
....tan
....tan
.Re 
Rf green spot
= (3/12) = 0.25
Rf blue spot
= 6/12 = 0.5
Separation using TLC TLC techniquesstep by step
Column Chromatography Column separation

11. 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)

12. 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

13. 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
(conjugateacid)
Zwitterion
(conjugatebase)
+ H+ + H+
Zwitterion
(conjugateacid)
Anion
(conjugatebase)
Alkaline(pH > pI)
Zwitterion/anion

14. Cation (+) Zwitterion Anion (-)
+ H+
Acidic (pH < pI)
Cation/zwitterion
- H+
+ H+
- H+
Cation
(conjugateacid)
Zwitterion
(conjugatebase)
+ H+ + H+
Zwitterion
(conjugateacid)
Anion
(conjugatebase)
Alkaline(pH > pI)
Zwitterion/anion
Acidic Base Buffer Calculation
][
][
log1
cation
zwitterion
pKpH 
][
][
log2
zwitterion
anion
pKpH 
Find pH 0.8M zwitterionand 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)
5.8
8.0
2.0
log1.9
][
][
log2


pH
zwitterion
anion
pKpH

15. At pH 7, alanine contain zwitterion and anionic form
i. Deduce structuralformula
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 addedBase added
2
1

16. 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
zwitterion
pKpH 
12.2
]12.0[
]08.0[
log3.2


pH
pH
2

17. ][
][
log1
cation
zwitterion
pKpH 
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
zwitterion
pKpH 
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

18. NH3 ↔ NH4
+
Buffer Solution
Acid part
Neutralize
each other
Salt part
Base part
- NH3(weakbase) + NH4CI (salt)
- NH3 + H2O ↔ NH4
+ + OH− → NH3 neutraliseadded H+
- NH4CI → NH4
+ + CI− → NH4
+ neutralise added OH−
- Effectivebuffer 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/conjugatebase
CH3COOH ↔ CH3COO-
Conjugate acid base pair
CH3COOH CH3COO-
Weak Acid Conjugate Base
BUFFER
Dissociate fully

 HCOOCHCOOHCH 33
COOHCH3 COONaCH3

 NaCOOCHCOONaCH 33
Dissociate partially
- CH3COOH (weakacid) + CH3COONa (salt)
- CH3COOH ↔ CH3COO- + H+ → CH3COOH neutraliseadded OH−
- CH3COONa → CH3COO- + Na+ → CH3COO- neutraliseadded H+
- Effectivebuffer equal amt weak acid CH3COOH and base CH3COO-
COOHCH3

COOCH3
BUFFER
Add acid H+Add alkaline OH-
Neutralize
each other
Basic buffer - weak base and its salt/conjugateacid

 OHNHOHNH 423
NH3 + H2O ↔ NH4
+ + OH-
NH3
Weak Base
NH4
+
Conjugate acid
CINH43NH
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

19. How to prepare acidic/ basic buffer
Acid Dissociationconstant
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 Dissociationconstant
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 BufferAcidic Buffer
salt salt
acid base
Henderson Hasselbalch Eqn
multiply -lg
both sides
Henderson Hasselbalch Eqn

20. Acidic BufferCalculation
How much 0.10M butanoic acid and solid potassium butanoate neededto 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
acid
pKpH 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
HCOOCH
Ka
3
/15.0][
][
]10.0[
lg83.400.5
][
][
lg
dmmolsalt
salt
salt
acid
pKpH 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
HCOOCH
Ka
35.5
]40.0[
]10.0[
lg74.4
][
][
lg



pH
pH
salt
acid
pKpH a
3
/40.0
50.0
20.0
dmmolconc
conc
vol
mol
conc



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

21. Acidic BufferCalculation
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
acid
pKpH 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
HCOOCH
Ka
34.5
]40.0[
]10.0[
lg74.4
][
][
lg



pH
pH
salt
acid
pKpH 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
HCOOCH
Ka
0578.0
0.1
)1016.3)((
108.1
)(
))((
3
5
35
3
3








Msalt
salt
salt
salt
acid
pKpH 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
HCOOCH
Ka
5
1016.3
)lg(5.4
)lg(






H
H
HpH
Conc [H+]

22. 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
base
pKpOH 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
OHNH
K
OHNHOHNH
b
45.955.414
55.4
]32.0[
]50.0[
lg74.4
][
][
lg




pH
pOH
pOH
salt
base
pKpOH 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
OHNH
K
OHNHOHNH
b
0578.0][
24.0
][
]10.0[
lg
][
]10.0[
lg74.45.4
][
][
lg
3
3
3
3








COOCH
COOCH
COOCH
COOCH
acid
pKpH a
5.4
10
)lg(5.4
)lg(






H
H
HpH
MCOOCH
COOCH
COOHCH
HCOOCH
K
HCOOCHCOOHCH
a
0578.0][
)10.0(
)10)((
108.1
)(
))((
3
5.4
35
3
3
33









Conc [H+]

23. 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
3
75
][
][
lg75.323.4
][
][
lg
cmvol
salt
acid
salt
acid
pKpH a



1000
05.0
....
v
saltmolNaOHmol


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
..
v
acidmol




12.5
]020.0[
]010.0[
lg82.4
][
][
lg



pH
pH
salt
acid
pKpH a
How would adding 100ml DI water affect pH.
Before adding acid
0002.0
1000
10.00.2
.... 

addedacidmol
0102.00002.001.0.. acidmolTotal
0198.00002.002.0..
0002.0
1000
10.00.2
....




leftsaltmol
reactedsaltmol
After adding acid
10.5
]0198.0[
]0102.0[
lg82.4
][
][
lg



pH
pH
salt
acid
pKpH a
Buffer do not change pH on dilution