Analytical chemistry: separation, spectroscopic, electro- and thermoanalytical methods.
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1.The spectrophotometric response for the titration of a mixture of Fe3+and Cu2+ions against EDTA is given below.
The correct statment is:
(a) Volume ab≡ [Fe3+]and volume cd ≡ [Cu2+]
(b) Volume ab≡ [Cu2+]and volume cd ≡ [Fe3+]
(c) Volume ab≡ [Fe3+]and volume ab≡excess EDTA
(d) Volume ab≡ [Cu2+]and volume cd ≡excess EDTA
2.The quantitative determination of N2H4 with KIO3 proceeds in a mixture of H2O/CCl4 as follows
N2 H4 +KIO3+2HCl —-> N2 +KCl+ICl+3H2O
The end point for the titrimetric reaction is:
(a) Consumption of N2H4
(b) ICI formation
(c) Disappearance of the yellow color due to Cl2 in CCl4 layer.
(d) Displacement of the rod color due to I2 in CCl4 layer.
3.Appropriate reasons for the deviation form the Beer’s law among the following are
(a) Monochromaticity of light
(C) Very high concentration of analyte
(B) Association of analyte
(D) Dissociation of analyte.
(1) A, B and D
(b) B, C and D
(c) A, C and D
(d) A, B and C
- Beer’s law is subjected to certain real and apparent deviation.
- Real deviations are most usually encountered in relatively concentrated solutions of the absorbing compound(> 0.01 M). These deviations are due to interactions between the absorbing species and to atterations of the refractive index of the medium.
- Most common are the apparent deviations and these deviations are due to Chemical reasons arising when the absorbing compound dissociates, associates or reacts with a solvent to produce a product having a different absorption spectrum.
- Strict adherence to Beer’s law is observed only with truly monochromatic radiation.
4.On subjecting 9.5 ml solution of Pb2+of X M to polorographic measurements, Id was found to be 1 µA. When0.5 mL of 0.04 M Pb2+was added before the measurement, the Id was found to be 1.25 µA.
5.A solution of 2.0 g of brass was analysed for Cu electrogravimetrically using Pt-gauze as electrode. The weight of Pt-gauze changed from 14.5g to 16.0 g. The percentage weight of Cu in brass is
The weight of Pt-gauze as electrode increases by 16–14.5 = 1.5 g because 1.5 g of Cu is deposited on Ptgauze.
Therefore, weight of Cu in 2.0 gram brass is 1.5g.
Therefore, percentage weight of Cu in brass =1.5/2.0 x100 = 75
6.In a polarographic measurement, (aqueous KCl solution used as supporting electrolyte) anapplied potentialmore than +0.4 V, results mainly in the formation of
The major drawback of polarographic measurement is that when the applied potential exceed more than 0.4, oxidation of mercury takes place in mercurous ion (Hg22+).