3. 1,10-Phenanthroline forms a bright red complex (1) and (2). Da with iron(II)1

Question

3. 1,10-Phenanthroline forms a bright red complex (1) and (2). Da with iron(II)1. The complex has an analytical wave- points. length of 510 nm and has a high molar absorptivity. Hydroxylamine hydrochloride is added to keep the of iron(ll) was treated with hydroxylamine hydrochlo- and (2). Draw &DS; (4) A solution containing an unknown amount iron in a reduced +2 state, and the pH is controlled us- ride, acetate buffer, and 1,10-phenanthroline as de- iron in a reduced -2 stat, nd the is controlled us-ride, acetate buffer, and 1,10-phenanthroline as de ing an acetate buffer. A series of iron(ll)-phen scribed above. The percent transmittance of the sam- anthroline solutions were made by adding various ple at = 510 nm was 52.4 when read against a blank amounts of 4.300 x 10 M Fe to 5 mL each of containing all but the iron(ll) solution. Determine the hydroxylamine hydrochloride, acetate buffer, and molar concentration of iron(II) in the unknown. 1,10-phenanthroline. Distilled water was added to make the total volume of each solution 50.00 mL.The (5) Calculate the molar absorptivity for percent transmittance of each solution was read, us- ing a blank containing everything but the iron(II) solu- tion. Data for the determination are volume of stock volume of stock solution, mL 96T solution, mL %T 7.00 21.78 9.00 14.09 1.00 80.35 2.00 64.71 5.00 33.6510.00 11.32 (1) Calculate the molar concentration [Fe(phenanthroline)32 in each solution. [Fe(phenanthroline)3]2-. Assume a 1.00-cm path- length. of

Explanation / Answer

1) the molar concentration in each solution = Initial concentration X intiail volume / total volume

The above plot is between absrobance V/s concentration

Absrobance = 2-log%T

the absrobance is realted to concentration as

Absorbance = concentration X absroptitivity X length

4) we can consider the obtained equation from graph

y = 10996x + 0.000

y = absrobance

x = concentration

Given %T = 52.4

Absrobance = 2 -log%T = 2 - log (52.4) = 0.28

0.28 = 10996 X concentration

concentration = 0.28 / 10996 = 2.54 X 10^-5 M

5) as written above

Absrobance = molar absroptitivity X concentraiton X length of path

molar absroptitivity = absorbance / concentration X length of path = 0.28 / 2.54 X 10^-5 M X 1 = 1.10 X 10^4 mole-1 L cm-1

Volume of stock initial molarity total volume final concentration %T absorbance 1 0.00043 50 0.0000086 80.35 0.095 2 0.00043 50 0.0000172 64.71 0.189 5 0.00043 50 0.000043 33.65 0.473 7 0.00043 50 0.0000602 21.78 0.6619 9 0.00043 50 0.0000774 14.09 0.8511 10 0.00043 50 0.000086 11.32 0.9462

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