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Friday, April 1, 2011

DETERMINATION OF POTASSIUM BY FLAME PHOTOMETRY


DETERMINATION OF POTASSIUM BY FLAME PHOTOMETRY
Aim:
To determine the amount of potassium present in the given sample solution.
Principle:
Flame emission spectroscopy is a type of atomic emission spectroscopy. It is mostly applicable for analysis of alkali and alkali earth metals. In this spectroscopy, the sample solution of sodium salt is nebulized in to flame, which may produce solid residue upon solvent evaporation. This solid residue undergoes atomization and gives neutral atoms which may acquire thermal energy from flame and undergoes electronic excitation. Due to unstable nature of excited state, excited atoms come back to ground state by emission of absorbed energy as visible radiation. By measuring the wavelength and intensity of emitted radiation, we can do qualitative and quantitative analysis respectively.
Chemicals required:
1.      Potassium chloride
2.      Distilled water
Apparatus required:
1.      Flame photometer
2.      Volumetric flasks
3.      Pipette
Procedure:
Preparation of standard solutions for calibration curve:
Dissolve exactly 1.090gm of potassium chloride in water and make up to 1 liter. This contains1mg per ml (1000 ppm).
Estimation of potassium by flame photometer:
First, switch on the digital flame photometer followed by the air compressor with the required value (10 bar)
Open the gas from the gas cylinder (after the instrument is warmed up for 10 minutes). Initially allow the ion-free water (distilled water) to aspirate in to the flame and set the digital value as 100. Now the instrument is said to be calibrated. After this calibration of the instrument, no adjustment should be made .Introduce the solutions containing different concentrations of potassium chloride (2, 4, 6, 8, 10µg) to the flame and find out the intensity of emitted light of each solution.
Plot a calibration graph between concentration and intensity of KCl solution which passes through the origin. Finally, introduce the sample of unknown solution containing sodium into the flame and find out the intensity of emitted radiation. From the intensity, the concentration of unknown solution can be determined.
Result:
The amount of potassium in the given sample………..ppm

Determination of Potassium
S. No
Concentration of KCl Solution(ppm)
Flame intensity
1
10

2
8

3
6

4
4

5
2

6
Unknown


The concentration of unknown sample =













Glossary:
Analysis: Analysis deals with methods for determining the chemical composition of samples of matter.
Accuracy: Accuracy describes the correctness of an experimental result. Accuracy is expressed in terms of either absolute error or relative error.
Absolute error= Observed value – True value
Analyte: A specific chemical moiety being measured, this can be intact drug, bimolecular or its derivative, metabolite, and/ or degradation product in a biologic matrix.
Absorbance: when light passes through a solution, light of certain wavelength is absorbed by the colored chemical substance and it is referred to as the absorbance, and the remaining light is transmitted which is referred to as transmittance; absorbance is also known as optical density (OD) or extinction.
Biological matrix: a discrete material of biological origin that can be sampled and processed in a reproducible manner. Examples are blood, serum, plasma, urine feces, saliva, sputum and various discrete tissues.
Blank: a sample of a biological matrix to which no analytes have been added that is used to assess the specificity of the bioanalytical method.
Beer’s Law: the intensity of a beam of monochromatic light decreases exponentially with increase in the concentration of absorbing species arithmetically.
Band Pass: Band pass is the difference in wavelength between the points where the transmittance is one – half the maximum.
Bias: bias provides a measure of the systemic, or determinate, error of an analytical method. Bias is defined by the equation
Bias = µ - xt
Where µ is the populations mean for the concentration of an analyte in a sample that has a true concentration of xt.
Calibration standard: a biological matrix to which a known amount of analyte has been added or spiked. Calibration standards are used to construct calibration curves from which the concentration of analytes in QCs and in unknown study samples are determined.
Chromatography: is the separation of a mixture in to individual components using stationary phase and a mobile phase.
EMR: electro magnetic radiation is made up of discrete particles called photons. 
Error: the error of a measurement is an inverse measure of its accuracy. The similar, the error, greater is the accuracy of the analysis.
Errors can be expressed as either absolute error or relative error
Absolute error= observed error –true value.
Relative error =  x 100%
Frequency: frequency is the number of complete wavelength units passing through a given point in unit time. Frequency is measured in HZ (Hertz) or cps (cycles per second).
Lamberts Law: the rate of decrease of intensity (monochromatic light) with the thickness of the medium is directly proportional to the intensity of incident light.
Limit of detection (LOD): the lowest concentration of an analyte that the bioanalytical procedure can reliably differentiate from background noise.
Lower limit of quantification (LLOQ): the lowest amount of an analyte in a sample that can be quantitatively determined with suitable precision and accuracy.
Monochromatic light: light have single wavelength.
Nephlometry: is the measurement of scattered light as a function of concentration of suspended particles (less than, approximately 100mg/liter)
Polychromatic light: the light radiations wavelength has 0 to 1000nanometers
Precision: Precision describes the reproducibility of results. Reproducibility of results is the agreement between numerical values for two or more replicate measurements. In other words, it is the agreement between measurements that have been made in exactly the same way. The three terms which are used to describe the precision of a set of replicate data.
1.      Standard deviation: it is the root mean square deviation from mean
2.      Variance: square of standard deviation
3.      Co-efficient of variation x 100%
Sample: a generic term encompassing controls, blanks, unknowns, and processed samples.
Spectroscopy: is the measurement and interpretation of electromagnetic radiation (EMR) absorbed or emitted when the molecules or atoms or ions of a sample move from one energy state to another energy state. This change may be from ground state to excited state or excited state to ground state. At ground state, the energy of a molecule is the sum total of rotational, vibrational and electronic energies.
Sensitivity:  According to IUPAC, a method is said to be sensitive if small changes in concentration causes large changes in the response function. Sensitivity can be expressed as the slope of the linear regression calibration curve, and it is measured at the same time as the linearity tests. The sensitivity attainable with an analytical method depends on the nature of the analyte and the detection technique employed.  The sensitivity required for a specific response depends on the concentrations to be measured in the biological specimens generated in the specific study.
Internal standard: an internal standard is a substance that is added in a constant amount to all samples, blanks, and calibration standards in an analysis.
Turbidimetry: is the measurement of transmitted light as a function of concentration of suspended particles (more than 100mg / liter, high concentration)
Wave length: wavelength is the distance between two successive maxima or minima or distance between two successive troughs or peaks. Wavelength can be measured in meters, centimeters (cm or10-2), millimeters (mm or 10-3), micrometers (µ or 10-6), nanometers (nm or 10-9).
Wave number: it is the number of waves per cm.
Wave number is expressed in cm-1 or Kayser.
Unknown: a biological sample that is the subject of the analysis.
Upper limit of quantification (ULOQ): the highest amount of an analyte in a sample that can be quantitatively determined with precision and accuracy.
Qualitative analysis: a qualitative method yields information about the identity of atomic or molecular species or the functional groups in the sample.
Quantitative analysis: a quantitative method, in contrast, provides numerical information as to the relative amount of one or more of these components.





UNITS, SOLUTION PREPARATION AND DILUTIONS
For Wavelength:
1 meter = 100 centimeter (cm) or 1000 millimeter (mm)
1 millimeter = 1000 microns (µm)
1 µm = 1000 nanometer (nm)
1nm = 1000 picometer (pm)
1 mm = 10-3m
1 µm = 10-6m
1nm = 10-9m
1Å =10-10m
1pm = 10-12m
The visible Spectrum:
Wavelength region, nm
Color
Complementary Color
400 -435
Violet
Yellow - green
435-480
Blue
Yellow
480-500
Blue-green
Red
500-560
Green
Purple
560-580
Yellow-Green
Violet
580-595
Yellow
Blue
595-650
Orange
Green-blue
650-750
Red
Blue-green


For Frequency:
1 Kilo Hertz = 103 Hz
1 Mega Hertz = 106 Hz
1 Giga Hertz = 10 9 Hz
1 Tera Hertz = 10 12 Hz
For weight:
1 Kilogram = 1000 gram (gm)
1 gram = 1000 milligram (mg)
1 milligram = 1000 microgram (µg)
1 microgram = 1000 nanogram (ng)
1mg =10-3g
1 µg = 10-6g
1ng = 10-9g
Percentage solution and conversion:
1% w/v = 1gram in 100ml solvent
1% w/w = 1 gram in 100 grams of solvent
Normally %w/v solutions are used
Conversions:
1.      To convert % w/v to mg/ml : multiply 10
%w/v x 10 = mg/ml
Problem: convert 2% w/v to mg/ml
Formula: 2% w/v x 10 = 20mg/ml
Answer is 20mg/ml
(2% w/v means 2grams/ 100 ml
i.e . 2000mg/100ml
ie. 20mg/ml)
2. To convert mg/ml to %w/v: divide by 10
Problem: convert 30mg/ml to %w/v
Formula:  
Answer is 3% w/v
(Checking: 1 ml contains 30mg
100ml contains 30x 100mg = 3000mg= 3gm
i.e 3gm/100 ml = 3% w/v)
3. To convert %w/v to µg/ml: multiply by 10,000
%w/v X 10000= µg/ml
Problem: convert 2%w/v to µg/ml
Formula: 2%w/v x 10,000= 20,000 µg/ml
(2%w/v means 2grams /100ml
i.e., 2000mg in 100ml= 20mg/ml = 20000 µg/ml)
4.  What is the percent concentration of a solution that you made by taking 5.85 g of NaCl
and diluting to 100 ml with H20?
5.85 g/100 ml = 5.85% W/V solution of NaCl
5.  What is the percent concentration of a solution that you made by taking 40 g of CaCl2 and diluting to 500 ml with H20?
You set up a proportion problem.
40 g /500 ml = Xg /100 ml
X = 8g
8g/100 ml = 8% (W/V) solution
OR another way to look at this is, 40 grams solute is what percent of the 500 mL solution? The 100 is used to convert to percent.
6.  How would you make 250 ml of a 8.5% NaCl solution?
This works backwards from the others --
8.5%= 8.5 g /100 ml
Again, set up a proportion
8.5 g /100 ml= X / 250 ml
21.3 g = X
OR an alternative method is to say what 8.5% of 250 ml is?
250 x 0.085 = 21.3
Therefore you would need to weigh out 21.3 g NaCl and dilute to 250 ml with H20.
7.  How much (volume) 0.85% NaCl may be made from 2.55 g NaCl?
An 0.85% NaCl solution = 0.85 g/100 ml
Setting up a proportion again,
0.85 g /100 ml = 2.55 g/X
X = 300 ml
Therefore, 300 ml of 0.85% NaCl may be made from 2.55 g NaCl.
The definition of molar solution is a solution that contains 1 mole of solute in each liter of solution. A mole is the number of gram molecular weights (gmw). Therefore, we can also say a 1M = 1 g MW solute/liter solution.
1M NaCl solution would be
Na = MW of 23
Cl = MW of 35.5
NaCl = MW of 58.5
1M = 58.5 g of NaCl in 1 liter of solution.
It may be made by weighing out 58.5 g of NaCl and qs to 1 liter with water. The qs stand for quantity sufficient and are a term used to designate that the total volume must be 1 liter (or whatever is stated).
58.5 g NaCl qs 1 liter with H20
Examples of other solutions would be
1 M H2S04 = 98 g/L
1 M H3P04 = 98 g/L
PROBLEMS
1. How would you make a liter of 4M CaCl2?
First find molecular weights.
Ca = 40
Cl2 = 35.5 x 2 = 71
CaCl2 = 111 (MW)
Then,
1M = 111 g/L
4M = 4 (111 g/L)
= 444 g/L
Weigh out 444g CaCl2 qs 1 liter with H20
2. How would you make 300 ml of a 0.5M NaOH solution?
First find molecular weights.
Na = 23
0 = 16
H = 1
NaOH = 40 (MW)
Then,
1M = 40 g/L
0.5M = 0.5 (40 g/L)
= 20 g/L
But you only want 300 ml so . . .
20 /1000 ml = x /300 ml
6 g = x
Weigh out 6 g NaOH pellets qs. 300 ml with H20

The definition of a normal solution is a solution that contains 1 gram equivalent weight (gEW) per liter solution. An equivalent weight is equal to the molecular weight divided by the valence (replaceable H ions).
1N NaCl = 58.5 g/L
1N HCl = 36.5 g/L
1N H2S04 = 49 g/L
Problems involving normality are worked the same as those involving molarity but the valence must be considered:
1N HCL the MW= 36.5 the EW = 36.5 and 1N would be 36.5 g/L
1N H2SO4 the MW = 98 the EW = 49 and 1N would be 49 g/L
1N H3PO4 the MW = 98 the EW = 32.7 and 1N would be 32.7 g/L
PROBLEMS:
1. You weigh out 80 g of NaOH pellets and dilute to 1 liter. What is the normality?
MW of NaOH = 40
EW = 40
1N = 40 g/L
80 g/L /40 g/L = 2N
What is the molarity?
MW = 40
1M = 40 g/L
80 g/L /40 g/L = 2M
2. You weighed out 222g of CaCl2 and diluted to 1 liter. What is the normality?
EW = 111 /2 = 5.55
1N = 55.5 g/L
222 g/L /55.5 g/L = 4N
What is the molarity?
1M = 111 g/L
222 g/L /111 g/L = 2M



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