Option B Chemistry UV Spectroscopy, Electrophoresis and buffer calculation

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Transcript of Option B Chemistry UV Spectroscopy, Electrophoresis and buffer calculation

  • 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 ads