ABSORPTION EXPERIMENT SPECTROPHOTOMETRY 22 OF NiSO 6H2O EQUIPMENT AND CHEMICALS

ID: 478948 • Letter: A

Question

ABSORPTION EXPERIMENT SPECTROPHOTOMETRY 22 OF NiSO 6H2O EQUIPMENT AND CHEMICALS Equipment Chemicals volumetic pipets 0.400 M Nisoa. 6Hro stock solution 5, 10. 15, 20, 25 ml) Unknown Nisos 6Hizo samples 25 mL volumetric flask Solid Nickel (1) sulfate hexahydrate 50 ml volumetric flasks (5) cuvettes (2) pec 20-D spectrophotometer INTRODUCTION The absorption of light by a solution produces the phenomenon of color, and may also be used to determine the concentration of the absorbing species in the solution. The characteristic color of a ion of light of the complementary color of the solution. The solution results fro absorption of visible light by a chemical substance occurs because the energy levels of electrons in molecules are quantized, and transitions between these levels lead to absorption of visible ultraviolet, or infrared radiation. Since the wavelength of absorption depends upon the nature of the absorbing species, absorption spectrophotometry can be used for the purposes of identification of the absorbing species. The amount of light absorbed by a substance can be described quantitatively. The percentage of light passing through a medium is called the percent transmittance (%1) of the solution. The amount of light absorbed is termed absorbance (A). A perfectly transparent medium has an absorbance of zero, and a totally opaque medium has an infinite absorbance. Most colored solutions have absorbances between 0 and 1. For chemists, the absorbance is the most useful quantity, and we shall focus only on absorbance in this experiment In general, there is a direct relationship between concentration and absorbance. This relationship is known as the Beer-Lambert Law and states that within a specified concentration concentration and absorbance increase proportionally. As such, the concentration of an unknown sample can be determined directly from the Beer-Lambert law by measurement of the absorbance of the sample. In scientific work, concentrations of unknown absorbing species are determined using the Beer- Lambert Law by measuring the absorbances of a group of solutions of known concentrations. A a calibration curve of absorbance vs. concentration should produce a straight line passing through origin. By measuring the absorbance of the unknown species, the calibration curve is used to the determine the concentration of the unknown 17 Chemistry 132 EXPERIMENT 22

Explanation / Answer

B) Absorbance and concentration data

Concentration of stock nickel sulfate hexahydrate solution = 0.400 M

Total volume of prepared standard solutions = 50.0 mL (you used 50.0 mL volumetric flasks).

Sample #

Volume of stock solution (mL)

Absorbance

Concentration of prepared standard solution (M)

0.186

0.04 (see sample calculation below)

0.367

0.08

0.526

0.12

0.674

0.16

0.796

0.20

Sample calculation:

Volume of stock solution, V1 = 5.0 mL.

Concentration of stock solution, M1 = 0.400 M.

Volume of prepared standard solution, V2 = 50.0 mL.

Use the dilution law:

M1*V1 = M2*V2

===> (0.400 M)*(5.0 mL) = M2*(50.0 mL)

===> M2 = (5.0)*(0.400 M)/(50.0) = 0.04 M

Prepare the calibration curve: plot absorbance along y-axis and concentration of standard solution along x-axis.

The plot is linear with a slope of 3.8175 and y-intercept 0.0517.

C) Determination of concentration of unknown liquid

Unknown id: 5-E

Absorbance = 0.337

1) Use the equation of the line; put y = 0.337 and obtain

0.337 = 3.8175x + 0.0517

===> 3.8175x = 0.337 – 0.0517 = 0.2853

===> x = 0.2853/3.8175 = 0.0747 0.075

The concentration of the diluted unknown species is 0.075 M (ans).

2) We obtained 10.0 mL of the unknown and diluted the same to 25.0 mL; the concentration of the diluted unknown sample is 0.075 M.

Use the dilution law: M1*V1 = M2*V2 where M1 = ?; V1 = 10.0 mL; M2 = 0.075 M and V2 = 25.0 mL.

Plug in values and obtain

M1*(10.0 mL) = (0.075 M)*(25.0 mL)

===> M1 = (0.075 M)*(25.0)/(10.0) = 0.1875 M

The concentration of the original unknown before dilution = 0.1875 M (ans).

D) Determination of concentration of unknown prepared from solid NiSO4.6H2O

1) Molar mass of NiSO4.6H2O = 262.75 g/mol.

2) Molarity to be prepared = 0.0500 M

3) Grams solid NiSO4.6H2O = 0.328 g

4) Grams solid NiSO4.6H2O actually used = 0.392 g

5) Absorbance of solution = 0.295

6) Concentration of unknown from calibration curve = 0.075 (you obtained this in part C)

Calculate the molarity of your solution prepared from solid NiSO4.6H2O as

Molarity of NiSO4.6H2O = (mass taken)/(molar mass)/(volume of solution in L) = (0.392 g/262.75 g/mol)/[(25.0 mL)*(1 L/1000 mL)] = 0.0596 M 0.06 M (ans).

6) The concentrations differ slightly (by 0.015 M); determination of the concentration by diluting the unknown solution is better. Measurement of solid can lead to errors due to loss of sample or accidentally taking more sample. Moreover, the solid sample can lose water of hydration to the atmosphere and thus register a lower mass than you are required to take.