CH4304EXPTS

Errors and Significant figures

Systematic (determinate) -consistent error. Can be detected & corrected

Errors

Random (indeterminate)-due to limitations of physical measurement. Cannot be eliminated.

Systematic error

(i) operational & personal e.g. mechanical loss of material during analysis.or inability to detectcolour changes at end-point of titration.

(ii) instrumental or reagent errors- faulty calibration of balances, attack of reagents upon glassware.

(iii) errors of method - incorrect sampling & from incompletenes of a rx.

(iv) additive & proportional errors.

Systematic error can be reduced by(a) calibration of apparatus & application of correction factors(b) running a blank determination – separate determination with a blank under identical experimental conditions(c) running a control - with a quantity of a standard substance with approx the same weight of constituent as contained in sample.

(d) use of independent method of analysis(e) running parallel determinations - serves as a check on the precision of result-does not guarantee accuracy(f) use of standard addiion/internal standard or

isotopic dilution.(g) amplification methods - useful when small amount of material is to be analysed.

Significant figures and computations.

1. The number of significant figures in a value is the minimum number

of digits required to write the value in scientific notation. e.g.

CH4304 Laboratories 1

0.00035 = 3.5 x 10-4 i.e. two significant figures. 0.000 0045 = 4.5

x 10-6, two significant figures.

2. The first uncertain digit in a calculated result should be the last

significant figure. e.g.

3. For addition and subtraction the last significant figure is determined

by the decimal place of the least certain number 21.29965 + 12.2

= 33.5

4. For multiplication and division the number of figures is usually

limited by the factor with the least number of digits. Retain in each

factor one more significant figure than is contained in the factor

having the largest uncertainty e.g. 1.26 x 1.236 x 0.6834 x 24.8652

should be carried out as 1.26 x 1.236 x 0.688 x 24.87 and the

result expressed to three significant figures

5. For values expressed as logs the number of figures in the mantissa

should equal the number of significant figures in the original value.


Calculating Uncertainty

1. For addition and subtraction use absolute uncertainty

e.g. 1.76 ( ± 0.03) + 1.89 (±0.09) - 0.59 (± 0.02)

= 3.06 ± e4 = 3.06 ± 0.041

1. For multiplication and division use relative uncertainties

relative uncertainty (0.03)/(1.76) x 100 = 1.70%


Mean & Standard Deviation

When a quantity is measured with the greatest exactness of which the

instrument, method and observer are capable of it is found that the

results of successive determinations differ among themselves to a

greater or lesser degree the average value is accepted as the most

probable – this may not always be he true value

Difference between the measured and the true value = absolute

error and this is a measure of the accuracy of a result.

The true or absolute value of a quantity cannot always be established

experimentally so the observed value must be compared to the most

probable value

The spread of values is measured most efficiently by the sample

standard deviation s

s2 = variance

R.S.D. relative standard deviation

Sometimes referred to as the coefficient of variation C.V.

Example 1


n

x

x mean Sample ii

1

x-x

s i

2i

n

x

s

x

x100s C.V.

Analysis of a sample of iron ore gave the following % values for the

iron content:

7.08 7.21 7.09 7.16 7.14 7.07 7.14 7.18 7.11

The greater the number of measurements (n) the closer the sample

mean (`x) approach the population mean (m). The standard error of

the mean sx

For example 1 sx = ± 0.014. If 100 measurements were made

denominator =1001/2 so

sx = ± 0.0047 so the precision of a measurement may be improved

by increasing the number of measurements.

Distribution of standard errors

If a large number of replicate observations are made (>50) of a

continuous variable the results obtained will usually be distributed

about the mean in a roughly symmetrical manner-Gaussian Error

Curve.


0.66% C.V. 0.04s 7.13x 7

n

ssx ±

2)/2-x(-e2

1y m

68% of all values fall within one S.D. of the mean, 95% will fall within 2

S.D (1.96). and 99.7 will fall within 3 S.D of mean.

From e.g. 1 S.D. = 0.047 so 68% of values will lie ± 0.047 of the mean.

95% will lie between ± of the mean value i.e. there is a 5% (1 in 20) os

a resultdiffering from the mean by more than ± 0.094%.

Reliability of results


Statistical figures obtained from a set of data are of limited use by

themselves

a) reliability of results

Analysis of results

b) comparison of results with true value

or with other sets of data

Rejection of results should be considered only when a suitable

statistical test has been applied or when there is an obvious chemical

or instrumental reason

E.g. 2. The following data were obtained for the determination of Zinc

in a sample of dust:

4.3 mg g-1, 4.1mg g-1 , 4.0mg g-1 , 3.2mg g-1 . Should the value 3.2 be

rejected. Apply Q test

If Q (observed ).>Q (tabulated) discard the questionable point. Q for 4

points from table = 0.831 so result should be retained.


727.01.1

8.0

2.33.4

4.0-3.2

aluesmallest v - luelargest va

luenearest va - valuelequestionab Q

Correlation and regression

calibration: construct calibration curves by measuring analytical

signal for each standard & plotting against concentration

two statistical operations should be performed on calibration

curve

Correlation coefficient {r}: to determine if graph is

linear or in the shape of a curve

Least square error associated with “best fit” line/curve

n: number of data points

r must be close to 1 for good fit

Linear Regression-“ least squares”

2 assumptions

errors in y values substantially greater than errors in x values

uncertainties (standard deviations) in all y values are similar

eqn of line y = mx + b

point (xi, yi) vertical deviation (di) = yi - y

di = yi-y = y-(mx + b)

di2 = (yi - mx - b)2 : least squares


Standardised Acid and Standardised Base


Standard HCl

Procedure

1. Use the information provided in Table 1 to calculate the volume of

concentrat (37 wt %) HCl that should be added to 1 L of distilled water to produce 0.1 M HCl, and prepare this solution.

2. Dry primary grade sodium carbonate for 1 hr at 110oC and cool it in a

desiccator.

3. Weigh four samples containing enough Na2CO3 to react with 25 mL

of 0.1M HCl and place each in a 125 mL flask. As you are ready to titrate each one, dissolve it in about 25 mL of distilled water.

2HCl + Na2CO3 CO2 + 2NaCl

FW = 105.989

Add 3 drops of bromocresol green indicator to each and titrate one rapidly (to a green colour) to find the approximate end-point.

4. Carefully titrate each sample until it just turns from blue to green. Then boil the solution to expel CO2. The solution should return to a blue colour.

5. Carefully add HCl from the buret until the solution turns green again. A blank titration can be performed by using 3 drops of indicator in 50 mL of 0.05 M NaCl. Subtract the volume of HCl needed for the blank solution from that required to titrate Na2CO3.

6. Calculate the mean HCl molarity, the standard deviation and the relative standard deviation.

Standard NaOH


1. A 50 wt % aqueous NaOH is provided. Normally this solution is prepared in advance and allowed to settle. Na2CO3 is insoluble in this solution and precipitates. The solution should be stored in a tightly sealed polyethylene bottle and handled gently to avoid stirring the precipitate when the supernate is taken. The density is close to 1.50g of solution per mL.

2. Primary standard-grade potassium hydrogen phthalate should be dried

for 1Hr at 110oC and stored in a dessicator.

F.W. = 204.22

3. Boil 1.0L of water for 1Hr to expel CO2. Pour the water into a polyethylene bottle which should be tightly capped. Calculate the volume of 50% aqueousNaOH needed to produce 1.0L of approximately 0.1 M NaOH. transfer this to the water bottle, mix well and allow to cool.

4. Weigh four samples of the solid KH(C8H4O4) (MW 204.23)and dissolve each in approx. 25 ml of distilled water in a 125ml flask. Each sample should contain enough solid to react with about 25ml of 0.1M NaOH. Add 3 drops of phenolphthalein ndicator to each and titrate one of them rapidly to find the approximate end-point. The burette should be fitted with a loose fitting cap to minimise entry of CO2.

5. Calculate the volume of NaOH required for each of the other three samples and titrate them carefully swirling the flask occasionally. The end point is the firstappearance of a faint pink colour. This colour will disappear as CO2 from the air dissolves in the solution.

6. Calculate the average molarity, the standard deviation, and relative standard deviation.

Name Approxim Approxim mL of Reagent


ate weight percent

ate molarity

needed to prepare 1 L of 1.0 M solution

Acetic acid 99.8 17.4 57.5Hydrochloric acid 37.2 12.1 82.6Nitric acid 70.4 15.9 62.9Perchloric acid 70.5 11.7 85.5Phosphoric acid 85.5 14.8 67.6Sulphuric acid 96.0 18.0 55.6Ammonia* 28.0 14.5 69.0Sodium Hydroxide 50.5 19.4 51.5Potassium hydroxide

45.0 11.7 85.5

*28wt% ammonia is the same as 56.6 wt% ammonium hydroxide

Preparation of buffersBuffer solution can be defined as an equilibrium mixture of a conjugate acid-base pair which is capable of resisting substantial; changes in pH upon the addition of small amounts of acidic or basic


substances. With some exceptions, either the acid or the base of the conjugate pair is a weak electrolyte.

The Henderson-Hasselbalch equation is a mathematical statement which defines the pH of a solution of a conjugate acid-base pair in terms of the dissociation constant of the weak acid-base pairs and the equilibrium concentrations of the acid and its conjugate base. This equation is extremely useful in calculating the pH of the buffer.

Figure 1: Preparation of a 0.02M Tris buffer at pH 8.0.

Buffer A and B are prepared in the same way except 0.5 mole of NaCl is added to flask B before the volume is brought to 1 L for buffer B.

Ion-exchange chromatography - spectrophotometric determination of the composition of a sample of VOSO4

VOSO4 supplied commercially contains impurities H2SO4 and H2O. A solution will be prepared from a known mass of reagent. The VO2+ can be assayed spectrophotometrically and the total cation (VO2+ and H+) can be assayed by ion-exchange. Together these measurements enable the determination of the quantities of impurity in the sample.

Ion exchange resins are polymer matrices containing charged site which are used to separate charged species .


Anion exchangers contain bound positive groups e.g. –NR3+

Cation exchangers contain bound negative groups e.g. –SO3-

Procedure

It is initially necessary to determine the ion-exchange capacity of the resin (moles of Na+/gram of resin). Weigh accurately about 2.000g of Amberlite 120A resin in the Na form. Make a slurry of the resin in distilled water and pour into the glass chromatography column. The level of the water should always cover the resin. Never let the resin run dry


2. Conversion of the resin to the H+ form. Pipette 5.0mL of 1.0M acid (H2SO4 or HCl) onto the column and allow to equilibrate. Wash off excess acid with the distilled water until a neutral eluate is obtained (use litmus paper).

3. Determine ion exchange capacity of resin in moles Na+/g resin. Pipetted 2.0mL of standardised 0.500M NaCl solution onto the column and allow to equilibrate for 5 minutes. Wash with about 25 mL of distilled water and collect the eluate in a 150mL flask. Add three drops of phenolphthalein indicator and titrate with standard 0.100N NaOH.


4. VOSO4 solution: weigh accurately, approximately 0.50g of VOSO4 and make up to 100mL with distilled water. Using the molar absorptivity () of the vanadyl ion at 750nm, VO2+ = 18M-1cm-1 it is possible to calculate the amount of pure VOSO4 in the sample.

5. Following regeneration of the resin as previously described, (2 above) the

procedure is repeated for the vanadyl sulphate solution to determine the total cation content, i.e. VOSO4 + H2SO4 .Care should be taken not to exceed the exchange capacity of the resin.

4. The water in the sample can be calculated from a accurate measurement of the initial weight. i.e. original weight - (VOSO4 + H2SO4)

5. Express your answer as (VOSO4)x(H2SO4)y(H2O)z

Example

(a) In an experiment to determine the purity of Vanadyl Sulphate

2.000 g of ion-exchange resin in the H+ form were used.

20.00 ml of 0.100M NaCl was put on the column and allowed

to equilibrate. The eluate was collected and titrated against

0.010M NaOH, the titre was 32.0 ml . What is the ion-

exchange capacity in mols/g of resin.

Ans

32.0 x 0.010 M = 3.2 x 10-4 mols NaOH = 1.6x10-4 mols /g resin

(b) b) Assuming the VOSO4 was absolutely pure what is the

maximum volume of solution that can be put on the column

such that the ion-exchange capacity is not exceeded given

that a solution of Vanadyl sulphate was prepared by

dissolving 0.1420g in 100mls of H2O. .

Ans: VOSO4 mol wt = 163 g/mol so if 100% pure the 0.1420g of

VOSO4 = 8.7 x 10-4 mol. Resin capacity for VO2+ ion is 1.6x10-4

mols, so the maximum capacity of the resin for VOSO4 (if 100%

pure ) is 18.37 ml of solution

(c) c) A solution of Vanadyl sulphate was prepared by dissolving


0.1420g in 100mls of H2O. 10.0 ml of the solution was put on

the regenerated resin (H+ form) and allowed to equilibrate.

The eluate was collected and titrated against 0.010M NaOH.

The titre was 18.0 ml. The absorbance of the solution in a

1cm cuvette was 0.126 at 750nm when =18cm-1M-1. .

Calculate the molar ratio as (VOSO4)1.00 (H2SO4)y(H2O)z

Ans: A = c so c= A/ = 0.126/18 = 7.0 x10-3M in 10 mL

7.0 x10-5 mol VOSO4 = 0.01141 g

Titre 18.0 ml mol of (VOSO4 + H2SO4)

= 9 x 10-5 mol 2.0 x 10-5 mol H2SO4 = 1.96 x10-3 g H2SO4

wt of H2O = 0.01420 (in 10 mls) – (0.01141 + 1.96 x10-3)g

=

8.3 x 10-4 g H2O = 4.6 x10-5 mol

(VOSO4) 7.0 x 10-5 (H2SO4) 2.0 x 10-5 (H2O) 4.6 x10-5

(d) = (VOSO4)1.00 (H2SO4)0.29(H2O)0.66

Gas Chromatography

Gas chromatography is an instrumental technique for the separation of volatile molecules. The mobile phase is an inert gas generally nitrogen or helium and the stationary phase is a nonvolatile liquid chemically bonded onto a quartz capillary column of very narrow diameter.. The column is placed in a temperature programmable oven and the Nitrogen passes through it. Gas chromatography involves a sample being vapourised and injected onto the head of the chromatographic column. Samples are injected onto the column via a heated injection port and are transported through the column by the flow of inert, gaseous mobile phase; and are separated on the basis of having different affinities for the stationary phase. As each component exits (or elutes) from the column it enters a detector and produces a signal peak. The peak area is proportional to the concentration or mass of that component in the sample. Qualitatively, the identity of unknown components can be determined by comparing their retention times with known standards. Quantitatively, the concentration of a given component can be


determined by measuring the peak area. The samples are detected as they exit the column. The integrator gives an output of peak height versus time, called a chromatogram from which samples may be identified and concentrations evaluated.

Background

Chromatographic techniques are used to perform qualitative and quantitative analysis on complex mixtures. As the mixture of components to be separated is carried through thecolumn, separation is affected because of the varying attraction each component of the mixture has for the stationary phase. Substancesthat are more strongly attracted to the stationary phase will becarried down the column at a slower rate than those with a lesser attraction.. The temperature of thecolumn and, to a lesser degree, the flow rate of the mobile phase alsoaffect the degree of separation. The time required for a particularcomponent of a mixture to pass from the injection port, through the column, to the detector is its retention time. Under identicalconditions, the same substance will have the same retention time inrepeated analyses. For this reason, retention time can be usedqualitatively to identify a substance in a mixture.

If a quantitative determination is desired, the method of internalstandards is normally employed. An internal standard is neededwhen conditions such as sample size of gas flow cannot be replicatedexactly between trials. A compound is chosen to serve as aninternal standard that has physical properties similar to those of thesubstance you wish to measure. A calibration curve is constructedin which the ratio of the peak area of the compound of interest (theanalyte) to that of the internal standard is plotted on the y-axis


versus the concentration of the analyte on the x-axis. The peak arearatios will be proportional to the analyte concentration, whereas thepeak area of the analyte alone will not be. Peak areas are calculatedby the electronic integrator.

In this experiment you are provided with an unknown alcohol in water and you are required to

1. identify the unknown on the basis of its retention time using the

standards provided

2. quantify the concentration of butanol in water by internal standardisation using one of the standards provided.

3. calculate N (the number of theoretical plates required to elute the unknown) and the resolution wrt the internal standard

EXPERIMENTAL PROCEDURE: A: Identification of unknown in alcohol mixtureB: quantitative analysis of butanol in a butanol:water mixture.

A: identification of unknown.

In a 10mL volumetric flask make up 3 standards containing 0.20mL each of butanol, ethanol, and propanol in distilled water. Inject each standard onto the GC and record the retention times. Inject the unknown provided. Comparison of the Rt of each of standards against the components of the unknown should allow direct identification of the unknown(s)

The Quantitation of ButanolPrepare a series of standards ranging from 2.0% to 8.0% of butanol asindicated below using pipettes to deliver each liquid. Use the 50ml flasks and make up to the mark with water

After the four standards have been prepared, add 2.0mL of propanol toeach mixture via pipette. Propanol will serve as the internal standard inthis analysis.

standard volume of butanol CH3CH2CH2CH2OH

volume of propanolCH3CH2CH2OH

Total volume


1.0mL 2.0 mL 50.02.0mL 2.0 mL 50.03.0 mL 2.0 mL 50.04.0 mL 2.0 mL 50.0

Swirl each mixture thoroughly to ensure complete mixing, then covereach solution to avoid evaporation.

Obtain 50.0 mL of an unknown butanol:Water mixture from yourinstructor. Using a 1.0 mL pipet, add exactly 2.0 mL propanol to theunknown. Swirl the mixture thoroughly and cover to retard evaporation..

SETTING UP THE Gas ChromatographYou will use the HP5890gas chromatograph in conjunction with the integrator

The integrator output will list the peak number

Peak # Retention time (mins)

Area counts % Area

1 x.xx xxxxxx xx.x23

Experiment should be run isothermally. Your instructor will select the temperature

C ALCULATIONS : 1. Calculate the ratio of peak area values of butanol to propanol.2. Using EXCE L prepare a calibration curve by plotting theratios of concentration of butanol/propanol on the x-axis versus the average of the ratios of the peak area of butanol to peak area of propanol.3. Use EXCELto perform a regression analysis on the calibrationcurve .4. Calculate the concentration of butanol in the unknown sample fromthe regression analysis basing your calculation on the straight lineequation form, y = m x + b.5. Turn in the graph of the calibration curve, the regression analysis,the calculation of the unknown and the Report Sheet.


A major source of precision error in GC is from poor injection technique. Both manual and automatic injection methods are available. If automatic equipment is available use it. If not several approaches can be used to try and reduce error.The first step is to ensure that the syringe is properly filled containing no air bubbles. The instructors will demonstrate how air bubbles can be eliminated .The sample should be injected rapidly as a “plug” to avoid band broadening and poor resolution.

a) draw the sample into the syringe barrelb) draw 2-3ml of air into the barrelc) insert needle into injection port and allow to heat for a few

secondsd) Inject quickly but without excessive force as the syringes

are fragilee) When the syringe has been removed check for any un-

injected sample and subtract from nominal injection volume

f) Clean the syringe using diluent (water in this case) six times before the next injection.

g)


High Performance Liquid Chromatography

HPLC is a technique suitable for separation of most organic compounds and is the most often encountered instrumental quality control method. Separation occurs between a liquid mobile phase consisting of degassed very high purity organic/aqueous solvents which may or may not be buffered and a stationary phase which is chemically bonded to a support material. The most frequently used stationary phase is Octadecylsilane ODS which consists of a long hydrocarbon chain, it is a non-polar material.

.The purpose of this experiment is to determine the concentration of caffeine in common beverages using a reversed-phase column,Equipment and Materials

1. HPLC System with variable wavelength detector.2. Reversed-phase column (C-18); Ultrasphere ODS, 4.6 x 150 mm, 5 micron particles coated with octadecylsilane groups.3. 25-µL flat nose syringe.B. Solutions the mobile phase is is 40:60 CH3OH:H2O


Part A:

. Identify caffeine in beverages on the basis of its retention time (tr)and determine its concentration by standard curve. Stabilize the system with the methanol/water solvent at a flow rate of about 1.0 mL/min. Flush the injection port with at least 3-4 syringe volumes of mobile phase.

(i) Using the pure caffeine provided make up a stock solution of approx 0.1%

(ii) Prepare a series of standards by serial dilution of 0.01%, 0.005%, 0.001%, 0.0005% caffeine in water.

(iii) Load 50 mL of each onto the column, When the injection loop (20 µL) is properly flushed and filled, injected onto column.. Begin recording the chromatograms at an absorbance range (AUFS) such that the entire peak appears on the chromatogram.

(iv) note the retention time and the area associated with each peak. Construct a calibration graph, and find the best fit regression line

(v) Take a sample of beverage. Filter to remove any undissolved material and inject onto the chromatograph. The caffeine can be identified by its retention time and the concentration calculated from its peak area.

..

Part B

You are required to calculate certain performance characteristics of your separation.

(vi) Prepare solutions of 0.1% NaNO3 , 0.1% benzoic acid and 0.1% caffeine as well a mixture of 0.1% caffeine and 0.1% phenol. NaNO3 will be used to estimate tm as it will not be retained by the stationary phase as it is ionic and will elute with the mobile phase.

(vii) Inject 50 mL of each onto the column, note the retention time. The chromatograms for caffeine, phenol and the caffeine/phenol mixture should be measured at 254 nm and the retention times correspond to tr. for each compound. The chromatogram for NaNO3 should be recorded at 210 nm. The retention time measured is tm.

(viii) calculate t’r(the adjusted retention time), k’ (the capacity factor), (the relative retention), N (the number of theoretical plate), H (the height equivalent to a theoretical plate) and R the resolution using the information supplied in the table.


Quantity equation Parameters

Partition coefficient Cs = conc of solute in stationary phase

Cm= conc of solute in mobile phase


Adjusted retention

time

tr= retention time of solute of interest

tm = retention time of unretained solute

Vs = volume of stationary phase

Capacity factor Vm = volume of mobile phase

ts = time solute spends in stationary

phase

tm = time solute spends in mobile phase

subscripts 1 and 2 refer to two solutes

Relative retention w = width at base

w1/2 = width at half height

= std deviation of band

Number of Plates x = distance travelled by centre of band

L = length of column

N = number of plates on column

Plate height tr = difference in retention times

Vr = difference in retention volumes

Resolution wav = average width at baseline

N = number of plates

= relative retention

= capacity factor for second peak

= average capacity factor

DETERMINATION OF NITROGEN BY KJELDAHL METHOD

The methodThe Kjeldahl digestion converts nitrogen compounds (proteins, amines, organic compounds) into ammonia compounds. Free ammonia is released by the addition of caustics, which is then expulsed by distillation and subsequently titrated.


Kjeldahl tablets or copper sulphate/selenium salts accelerate the reaction. The tablets consist of sulfate and metal salts. The sulfate salt raises the boiling point of sulfuric acid, while the selenium, coppers, titanium, or mercury salts shorten the digestion time.

Digestion

Weigh out accurately part of the organic sample (1.5 - 2.0g). Add 10g

of potassium sulphate, 0.2g selenium or 0.5g anhydrous copper

sulphate and pour 25ml of concentrated sulphuric acid into the flask in

such a way as to wash down any solid adhering to the neck. Prepare

three sample flasks and one blank flask. Allow to digest under vacuum

for 2 to 4 hours. The addition of some glass beads to the flasks before

boiling speeds up the digestion process.

When cool carefully add 100ml of water to each flask and transfer to

the distillation apparatus.

Neutralisation

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

(NH4)2SO4 + 2 NaOH Na2SO4 + 2 NH3 + 2 H2O

Add excess of 50% sodium hydroxide solution. Distill off the ammonia

into excess standard acid or a saturated solution of boric acid.


Collection of NH3 NH3 + H3BO3 NH4

+ + H2BO-

3

{H[B(OH)4] + NH3 NH4[B(OH)4]}

The borate formed is determined by titration with standard 0.1N HCl; one mole of HCl

being required for each mole of NH3

Titration H2BO3

- + H

+ H3BO3

{2 NH4[B(OH)4] + H2SO4 (NH4)2SO4 + 2 H[B(OH)4]}

The solution at the equivalence point contains H3BO3 and NH4CL, so an

indicator in the pH range 5-6 is necessary. Bromocresol green or

bromophenol blue.

Calculate the % Nitrogen in the sample.

What methods would you use to reduce or eliminate determinate error

from this experiment.

Karl Fischer Titration of H2O

Water can affect the chemical and physical properties of substances and influence the quality, price and the shelf-life of a product.

Using the Karl Fischer titration, the water content of gases, liquids and solids can be determined easily and with a high degree of accuracy. Determination of water content according to Karl Fischer is rapid,


accurate and reliable; it is often the method of choice in quality and in-process control, production, research and development. It is a very widely used procedure for measuring water in food and pharmaceutical samples.

The Karl Fischer Reagent is comprised of iodine sulphur dioxide, a base and an alcohol.

What is a Karl Fischer chemical reaction?

I2 + 2H20 + SO2 2Hl + H2SO4

This reaction occurs in the presence of a base and a solvent (a typical solvent could be methanol or diethylene glycol, and a base imidazole)

A chemical reaction takes place between iodine (I) and water (W) with the reactants being in a 1:1 ratio between (I) and (W).

1x (I) + 1x(W) --> Reaction The end-point is usually detected coulometrically using an auto-titrator.

The anode solution contains an alcohol, a base, SO2, I-, and possibly another organic. Pyridine was the primary ingredient in Karl Fischer solutions but has mostly been replaced. The alcohol is diethylene glycol and the base imidazole. The anode generates I2 by oxidation of I- in the presence of H2O, A stoichiometric reaction occurs between the alcohol, (ROH) base (B), SO2, and I2.

B.I2 + BSO2 + B + H2O BH+I- + +B-SO3-

+B-SO3- + ROH BH+ROSO3

-

The net rx is oxidation of SO2 by I2 with the formation of alkyl sulphate. One mole of I2 is consumed for each mole of H2O.

Once the iodine in the Karl Fischer reagent is determined, the unknown concentration of water in the sample can be determined. ie, as 1 molI2 reacts with 1 mol H2O, the amount of (I2) used during the titration must be equal to the unknown amount of water present in the sample.

Coulometric Karl Fischer: - With a coulometric Karl Fischer titration, the amount of water present is determined by measuring the amount of current generated during the titration (coulombs) Coulombs are a measurement of current (amps) multiplied by the titration time in


seconds. There is a relationship between the iodine (I2) used in the titration, the samples water content (W) and the current. ie, Current (A) x time (secs) = Coulombs (C) ---> I + W ---> KF reaction According to Faradays Law :- 2 x 96,485 Coulombs are needed to generate 1 mole of iodine and this iodine subsequently reacts 1 : 1 with the water in the KF reaction. How does a coulometric titration work?

Coulometric KF titrations are preferably carried out within the pH range 4-7 Stage 1 Iodine generation Instead of dispensing KF reagent as in Volumetric KF titration, the Metrohm KF instrumentation actually generates the reagent inside the reaction cell. A current flows through the reagent generating iodine at the anode electrode.Stage 2 The instrument detects the end of the titration (end-point) Stage 3 T he instrument subsequently calculates the moisture content.

Determine the moisture content of the sample provide quotingthe average and standard deviation.


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