Lab Manual Part - I - Lab Course M.Sc.Biochemistry

Lab Course-II Date:…………………

Experiment No. 01AIM: Preparation of normal, Molar & percent solutions. Molarity (M) :

This is the most common method for expressing the concentration of a solution in biochemical studies. The molarity of a solution is the number of moles of the solute dissolved per L of the solution. A solution which contains 1 mole of the solute in one L of the solution is called a molar solution. Molarity of a solution can be calculated as follows: Weight of a solute in g/L of solution Molarity = Mol. Wt. of solute

It may be noted that in case of molar solutions, the combined total volume of the solute and solvent is one L. Thus for preparing 0.1 M NaOH, one may proceed as follows: Mol. Wt. of NaOH = 40 Required molarity of solution = 0.1MAmount (in g) of NaOH per L of solution = Mol. Wt.of NaOH x molarity = 40 x 0.1= 4 gThus, weigh 4 g of NaOH, dissolve it in a small volume of solvent (water) and make the final volume to 1 L with the solvent. Sometime it is desirable to know number of moles of a substance in a reaction mixture. This can be calculated using a simple relationship:

1 M solution = 1 mole of the substance/L of solution. = 1 mmole/ml of solution = 1 µmole/µl of solution 1 mM solution = 1 mmole/L of solution = 1 µmole/ml of solutionNormality (N):The normality of a solution is the number of gram equivalents of the solute per L of the solution. Therefore, Amount of a substance in g/L of solution Normality = Eq. wt. of substanceFor preparing 0.1 N Na2CO3 (Eq.wt. of Na2CO3= 53) solution, dissolve 5.3g Na2CO3 in a final volume of 1 L of solution.Percentage by Mass or % (w/w):

DEPARTMENT OF BOCHEMISTRY1


It is the weight of the component present in 100 parts by weight of the solution. In a solution containing 10g sugar in 40g of water, then 10x100 Mass % of sugar = = 20% (10+40)Percentage by volume or % (v/v) :It is the volume of the component in 100 parts by volume of the solution. In a solution containing 20 ml alcohol in 80 ml of water, the % volume of alcohol will be 20 x 100 = 20% (20 + 80)



Experiment No. 02AIM: Preparation of different buffers.

(A) Acetate Buffer: Take 10 ml of 0.1N acetic acid in a beaker. Measure the pH. Add 1 ml

of 0.01 N HCL. Mix well, measure the pH. Take 10 ml of 0.1 N sodium acetate solutions in a beaker. Measure the

pH. Add 1 ml of 0.01 N HCL. Mix and measure the pH. Mix equal volumes of the acetic acid and sodium acetate solution.

Measure the pH. Now add 0.1 N HCl, mix and measure the pH. Mix acetic acid and sodium acetate. After measuring the pH. Add 1

ml of 0.01 N NaOH solution and check the pH

(B) Phosphate Buffer:Prepare a solution of H3PO4 [0.1N] and carry out the titration with NaOH as before. Draw the graph and comment on the observations. H3PO4 having three hydrogen atoms have three stages of ionizations as follows:

H3PO4 H+ +H2P04

-

H2PO4 H+ HPO4

HPO4 H+ + P0

From the graph, predict the buffer ranges possible for phosphoric acid. The buffer range at pk2 is closer to 7.0, the physiological range. Prepare 0.1 N solution of NaH2PO4 . Mix them in various proportions and measure the pH. Calculate, using the Hasselbalch equation, what the pH should be and compare with the observed value.

(C) Construct a similar buffer system with Na2HCO3 and NaHCO3

mixture.(D) Using pKa values given in literature, prepare buffers in various pH

ranges by measuring out the components. Check at the end with measurement of pH.


pK1

pK2

pK3


Experiment No. 03

AIM: Find the strength in g/L of given NaOH solution with the help of standard sodium-carbonate solution (N/10) and intermediate solution of an acid.REQUIREMENT: A standard solution of 0.1N sodium carbonate and HCl solution having approx. normality N/10.APPARATUS: Burette, pipette, 2 beakers, titration flask and stand.REAGENT PREPARATION: COOH

1) 0.1N Oxalic acid solution: Dissolve 6.3g of oxalic acid .2H2O in distilled water and make up the volume to 1000mL. COOH

2) NaOH solution : Dissolve about 4gm of NaOH in distilled water and make up the volume to 1000mL

3) Phenolphthalein indicator: Dissolve 1g of phenolphthalein in 100mL of distilled ethyl alcohol and add 100ml of distilled water with constant stirring.

THEORY: The experiment involves two titrations.I. Titration: In the first titration, HCl (which is a secondary standard), is

titrated against Na2CO3 to determine its exact normality. The titration is carried out using methyl orange as indicator.2HCl + Na2CO3 2NaCl + H2O + CO2

II. Titration: Once the normality of HCl is known, the normality of given NaOH can be determined by titrating it with HCl using phenolphthalein as indicator.HCl + Na2CO3 NaCl + H2O

INDICATOR: Methyl orange (Titration I)

Phenolphthalein (Titration II)END POINT: Orange to red (I titration) Disappearance of pink colour (II titration)PROCEDURE: Titration I

1) Rinse and fill the burette with the given HCl.2) Pipette out 10ml of Na2CO3 in a titration flask and a few drops of methyl

orange.3) Titrate it with the acid till a permanent red colour is obtained.4) Repeat to get 3 concordant values.Titration II5) Pipette out 10ml of the given NaOH in a clean conical flask.



6) Add a few drops of phenolphthalein. The solution will turn pink.7) Titrate the above solution with the HCl taken in the burette till the pink

colour disappears.8) Repeat to get three concordant readings.

OBSERVATIONS:Titration I:

S.No.

Volume of the solution Taken in the titration flask.

Burette readings Volume of the HCl used.(Final – Initial) ml

Initial reading

Final reading

1 2 3 Titration II: Same as titration I.CALCULATION:I Titration: Let the volume of acid used = x mL Acid Na2CO3

N1V1 = N2V2

N1 X x = 1/10 X 10 N1 = 1/10 X 10/x = 1/xII Titration: Let the volume of acid used = y ml Acid NaOH N’1V’1 = N’2V’2

1/x X y = N’2 X 10 N’2 = 1/x X y/10Strength of NaOH solution = N’2 X Eq. wt. = N’2 X 40 g/LRESULT: The strength of the given NaOH solution = ……. g/LPRECAUTIONS:

1) Wash the apparatus with water and then with detergent followed by plenty of water.

2) Rinse the burette and pipette with the respective solutions to be taken in.3) Do not rinse the conical flask with the solution to be taken in it.4) Use anti parallax card to avoid error in burette reading.



Experiment No. 04AIM: Find the strength in grams per liter of the given sodium hydroxide solution with the help of standard oxalic acid solution.REQUIREMENT: A standard solution of 0.1N oxalic acid.APPARATUS : Burette, Pipette, 2beakers, titration flask, stand.REAGENT PREPARATION: COOH

1) 0.1N Oxalic acid solution : Dissolve 6.3g of oxalic acid .2H2O in distilled water and make up the volume to 1000mL. COOH

2) NaOH solution : Dissolve about 4gm of NaOH in distilled water and make up the volume to 1000mL

3) Phenolphthalein indicator: Dissolve 1g of phenolphthalein in 100mL of distilled ethyl alcohol and add 100ml of distilled water with constant stirring.

THEORY : A known amount of oxalic acid is titrated with NaOH solution using

phenolphthalein as indicator. Using normality equation, the normality and hence the strength of given NaOH is determined. COOH COONa 2H2O + 2NaOH + 4 H2O COOH COONa INDICATOR : Phenolphthalein.END POINT : Pink to colourless.PROCEDURE :

1) Rinse and fill the burette with the given oxalic acid solution (N/10).2) Pipette out 10ml of the given NaOH solution in the flask.3) Add a drop of phenolphthalein to the solution. The solution will turn pink.4) Run the oxalic acid solution from the burette into the titration flask till the

pink colour of NaOH solution disappears.5) Repeat to get three concordant values.

OBESRVATIONS :Normality of oxalic acid solution = 0.1N

S.No. Volume of the solution taken in the titration flask (ml)

Burette Reading Volume of oxalic acid used (Final - Initial)

Initial reading Final Reading

1 2 3 x mL

CALCULATION:



Acid NaOH Solution N1V1 = N2V2

1/10 X x = N2 X 10 N2 = x/100Strength = Normality X Eq. Wt = x /100 X 40g/L = y g/L ( Eq. wt of NaOH = 40 g)RESULT: The amount of NaOH present in the given mixture = …….. g/LPRECAUTIONS :

1) Wash the apparatus with water and then with detergent followed by plenty of water.

2) Rinse the burette and pipette with the respective solutions to be taken in.3) Do not rinse the conical flask with the solution to be taken in it.4) Use anti parallel card to avoid error in burette reading.



Experiment No. 05

AIM: To find out the strength of the given hydrochloric acid solution (approx strength N/10) by titrating it against sodium hydroxide using pH meter.APPARATUS: pH meter with glass electrode, reference electrode, beaker, burette, stirrer etc.CHEMICALS: HCl, NaOH.THEORY:

When an alkali is added to an acid solution, the pH of the solution increases slowly. But at the equivalence point, the rate of change of the solution is very rapid. A plot is drawn between volume of the alkali added and the pH of the solution. The sharp break in the curve gives the equivalence point, from which the strength can be calculated using normality equation.INSTRUMENTATION:

In pH meter the glass electrode is incorporated in an ordinary potentiometric circuit. The potentiometric pH meter differs from a simple potentiometer to the extent that the galvanometer is replaced by an electronic circuit that amplifies the current in the cell circuit by a factor of 109 or more.

Before using, the pH meter is first standardized with a buffer solution of known pH. Then the glass and reference electrode are immersed in an unknown solution and the pH is read directly on pH scale.PROCEDURE:

1) Calibrate the pH meter with the glass electrode in the buffer solution of known pH.

2) Wash the glass electrode and the reference electrode with distilled water and then rinse with the acid solution.

3) Take 5ml of HCl solution in a beaker. Add sufficient water so as reference and glass electrodes are completely dipped.

4) Note down the pH of the pure acid solution.5) Now add 10ml of N/10 NaOH from the burette and note down the pH after

each addition.6) Continue adding NaOH solution from the burette and note down the pH after

each addition.7) Near the equivalence point the change in pH is much more rapid than in any

other region.OBSERVATION: Volume of acid taken = 5mlVol. of alkali added 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, and 10.0CALCULATION:

Plot a curve with pH values as ordinate and volume of alkali added as abscissa. The sharp break in curve corresponds to the equivalence point.



Volume of alkali added (ml)Let the volume of alkali at equivalence point = x mL Acid alkali N1V1 = N2V2

N1 X 5 = N/10 X x

N1 = N/10 X Strength of HCl solution = 36.5 X x / 10 X 5 g/LRESULT: The strength of given acid solution is ……. g/lPRECAUTIONS:

1) The pH meter should be calibrated before use.2) After addition of alkali, the solution should be thoroughly stirred.3) Electrodes must be immersed properly in the solution.



Experiment No. 06AIM: Determination of pKa value of acetic acid, glycine & aspartic acid.Henderson – Hasselbalch Equation

Henderson – Hasselbalch equation is important for understanding buffer action and acid base balance in the blood and tissues of the mammalian system. The equation is derived in the following way. Let us denote a weak acid by the general formula HA, and its salt by the general formula BA (B- being the metal ion and A- being the conjugate base). The salt dissociates completely, while the weak acid dissociates only partly. We can write the equilibrium reactions for the dissociation of HA and BA in the buffer solution as follows. HA H+ + A-

BA B+ + A-

We well soon find that Henderson- Hasselbalch equation is simply another way or writing the expression for the dissociation constant of a weak acid.

[H+] [A

-] Ka =

[HA]

Solving for [H+], we get

[HA] [H+]= Ka [A

-]Taking the negative logarithm of both sides, the equation becomes

[HA] -log [H]= -log Ka –log

[ A-]



However, - log [H+] = pH, and Ka = pKa. Therefore

[HA] pH= pKa – log

[A-]

To change the negative sign, we invert –log [HA]/ [A-] and obtain

[A-]

pH= pKa + log

[HA]

This is Henderson- Hasselbalch equation. Now, the weak acid, HA, is only slightly dissociated even in the absence of the salt. Thus very little of the A - ions

come through the dissociated and gives a high concentration o A- ions. It can therefore be safely assumed that the concentration of the undissociated acid [HA] is equal to the total acid concentration. We can also assume that all A - has dissociated from BA and therefore the concentration of the conjugate base, [A -] is equal to the concentration of the slat, [BA]. Taking in to consideration these assumptions, the Henderson – Hasselbalch equation can take many different forms.

[Salt] pH= pKa + log

[acid]

or [conjugate base] pH= pKa + log

[acid]or [proton acceptor] pH= pKa + log

[proton donor] As with all the equations considered so far, the Henderson-Hasselbalch

equation also applies more accurately when concentrations are converted to activities by multiplying with appropriate activity coefficients. This is necessary because the values of pKa and activities vary with ionic strength. The value of pKa on the basis of activities can be calculated with the help of the following relationship:



pKa (activity) = pKa (concentration) – 1.018 µWhere µ is the ionic strength of the solution For most calculations, however, concentrations can provide fairly accurate results. Now, that we have derived an equation which relates pH to the ratio of salt concentration (conjugate base concentration) and the weak acid concentration, let us see the quantitative basis of buffer solutions resisting a large change in pH. We have seen that addition of 10 ml 0.1 NH4Cl to 990 ml of 0.1 N acetic acid and 0.1 M sodium acetate buffer solution. The H+ ions dissociating from HCl are neutralized by the acetate ions.

CH3COO- + H+ CH3COOH

The addition of HCl therefore lowers the concentration of the acetate ion slightly and raises the concentration of acetic acid by the same amount. If we assume that all H+ ions have been neutralized, the drop in acetate ion concentration will be 10-3 mole/liter. The concentration of acetic acid would rise by the same amount.

Mole Mole Mole

[CH3COO-]FINAL = 0.1 -0.001 = 0.0999 Litre litre litre

Mole mole mole

[CH3COOH]FINAL = 0.1 + 0.001 = 0.101 Litre litre litre

Substituting the final salt and acid concentration in the Henderson-Hasselbalch equation we get

0.0999 pH = pKa + log 0.101pKa of acetic acid acid is 4.76. Therefore, 0.0999 pH = pKa + log = 4.75 0.101 The pH of the buffer solution after addition of 10 ml of 0.1 N HCl changes from 4.76 to 4.75; a drop of merely 0.01 units of pH.



Henderson – Hasselbalch equation gives a very important relationship which makes it possible to calculate the pKa of any given acid with extreme ease. The relationship is, that if the molecular ration of salt to acid is unity in a solution, the pH of that solution will be equal to the pKa of the acid used.

[salt] pH= pKa + log

[acid] pH = pKa + log 1.0 pH = pKa + 0 pH = pKa

Thus to calculate the pKa of any acid one only needs to dissolve that acid and its salt in equal concentration and then experimentally determine the pH of the solution. It will be equal to the pKa of the acid. Some extremely important problems about buffers which can be solved using Henderson – Hasselbalch equation are provided.

Henderson – Hasselbalch equation makes it clear that the pH of a buffer solution depends upon the pKa of the acid and upon the salt to acid concentration ratio. The lower the pKa of the acid the lower will be the pH. The buffer ph will increase with increasing salt concentration. Again, according to Henderson – Hasselbalch relationship, the actual salt and acid concentrations can be varied widely without and change in Ph if the ratio between the two is unity. Thus, a lactate buffer containing 0.01 M lactate and 0.01N lactic acid will have the same Ph even if the buffer is diluted 10 times or even 20 times. In actual cases, however, the pH of the diluted buffer increases slightly. This increase is not significant enough.For example

Calculate the pKa of acetic acid, given the fact that the concentration of free acetic acid is 0.1 N and that of sodium acetate is 0.2 M. The pH of the solution is 5.06Ans. [Acetate] pH= pKa + log [Acetic acid] [Acetate] pKa = pH – log [Acetic acid]



0.2 = 5.06 – log = 5.06 – 0.301 0.1 = 4.76 pKa of acetic acid is 4.76 pKa valve of some amino acids

Compound Conjugated acid pKaAsparatic acid α-COOH

β-COOHα-NH3+

2.98

Glycine α-COOHα-NH3+

5.97



Experiment No. 07

AIM: Separation and identification of amino acids / sugars by paper chromatography.PRINCIPLE:

Amino acids in a given mixture or sample aliquot are separated on the basis of differences in their solubilities and hence differential partitioning coefficients in a binary solvent system. The amino acids with higher solubilities in stationary phase move slowly as compared to those with higher solubilities in the mobile phase. The separated amino acids are detected by spraying the air dried chromatogram with ninhydrin reagent. All amino acids give purple or bluish purple colour on reaction with ninhydrin except proline and hydroxylproline which give a yellow coloured. The reactions leading to the formation of purple complexes are given below:Ninhydrin + Amino acid Hydrindantin + RCHO + NH3 + CO2

Ninhydrin + Ammonia + Hydrindantin Purple coloured product + 3H2OMATERIALS AND REAGENTS

1) Whatman No. 1 filter paper sheet.2) Micropipette / micro syringe.3) Hair drier.4) Sprayer.5) Oven set at 105oC.6) Chromatographic chamber saturated with water vapours.7) Developing solvent: Take butanol, acetic acid and water in the ratio of 4:1:5

in a separating funnel and mix it thoroughly. Allow the phases completely. Use the lower aqueous phase for saturating the chamber



Experiment No. 08AIM: (a) Verify Beer’s law and apply it to find the concentration of the given unknown solution. (b) To determine λmax (wave length of maximum absorption) of solution of KMnO4 using a spectrophotometer.THEORY:

When an electromagnetic radiation is passed through a sample, certain characteristic wavelengths are absorbed by the sample. As a result the intensity of the transmitted light is decreased. The measurement of the decrease in intensity of radiation is the basis of spectrophotometer. Thus the spectrophotometer compares the intensity of the transmitted light with that of incident light.The absorption of light by a substance is governed by certain laws.According to the Beer Lambert’s law the intensity of the incident light is proportional to the length of thickness of the absorbing medium and the concentration of the solution, Log Io/I = A = ε . cIo = Intensity of incident lightI = Intensity of transmitted lightA = AbsorbanceL = Thickness of the mediumc = Concentration in mol L-1

ε = Molar absorption coefficient The molar absorption coefficient is the absorbance of a solution having unit concentration (C = 1M) placed in a cell of unit thickness (l= 1cm). Absorbance is also called Optical Density (OD)The absorbance (OD) of a solution in a container of fixed path length is directly proportional to the concentration of a solution. i .e

A = ε.c A plot between absorbance and concentration is expected to be linear. Such a straight line plot, passing through the origin, shows that Beer – Lambert’s law is obeyed. This plot, known as calibration curve can be employed in finding the concentration of a given solution.CHEMICALS: Distilled water, standard solution of KMnO4, tissue paper.APPARATUS: UV – visible spectrophotometer, beaker.SPECTROPHOTOMETER: A spectrometer is a device which detects the percentage transmittance of light radiation when light of certain intensity and frequency range is passed through the



sample. Thus the instrument compares the intensity of the transmitted light with that of the incident light.There are many spectrophotometers available for the visible range extending from 3800- 7800 Ao.Setting of the Spectrophotometer

1) Spectrophotometer should initially read zero on transmittance scale (T). If it does not read zero, set it mechanically with adjusting knob.

2) Connect the instrument to the mains and put on the power switch.3) Adjust the wavelength knob to the required wavelength region on scale.4) Choose the position of wavelength switch correspondingly either to 340 –

400 nm or 400-960nm.5) Adjust the meter needle on zero transmittance scale and 100 on O.D scale.

Working of the Spectrophotometer6) Open the lid of the cell compartment and insert a cuvette containing the

blank solvent (distilled water). Close the lid.7) Adjust the needle to 100% transmittance or zero optical density.8) Remove the cuvette and close the lid tightly again. Empty the cuvette and

rinse it with the standard solution of KMnO4 (0.01IM). Fill it with standard solution.

9) Now place the cuvette containing the standard solution in the cell compartment. Note the O.D and transmittance.

10) Now change the wavelength by 20nm and note absorbance (OD) and transmittance for each wavelength.

11) Plot a graph between wavelength measurement on the x-axis and absorbance (OD) on the y-axis.

Verification of Beer’s law12)Fix the wavelength at λmax position.13)Prepare KMnO4 solution with concentration 0.2%, 0.5%, 1.0%, 1.5%, 2.0%, 2.5%, and 3.0% etc. (20ml each)14) Note down the absorbance (OD) of series of solution of KMnO4 prepared above by the method described above.15)Plot a graph between OD against concentration. (If a straight line is obtained Beer’s law is verified)16) Now find out the OD of the unknown solution of the KMnO4. Find

out the concentration of this solution from the graph.OBSERVATION:

(i) Determination of λmax

Wavelength (nm) Absorbance (OD)



(ii) Verification of Beer’s law S.NO. Concentration (C)

(moles / L) Absorbance (OD)

CALCULATION: (i) A curve is plotted between wavelength and absorbance (OD).(ii) A curve is plotted between O.D and concentration and if a straight line is

obtained as shown by equation (i), Beer’s law is verified.(iii) From the graph of O.D versus concentration, the concentration of the

unknown solution can be found out. For example, in the fig x is the O.D of unknown solution then its concentration will be 1.0%.

OBSERVATION AND RESULT: (i) λmax for KMnO4 = …..nm(ii) KMnO4 solution obeys Beer’s law(iii) Concentration of the unknown solution = …..mg/L

PRECAUTIONS:i) Always use dilute solutions for getting calibration curve.ii) Cuvette should be cleaned properly and must be wiped with tissue paper.iii) Do not leave any finger marks on the cuvette.


Lab Manual Part - I - Lab Course M.Sc.Biochemistry

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