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Mrac [35]
3 years ago
8

ompare your TLC analysis with the results of the column chromatography. More polar molecules move more slowly through silica tha

n less polar molecules. Describe why silica gel chromatography is a good method of separating methyl orange from biphenyl. Based on the structure of Biphenyl and Methyl Orange, predict which compound will have a higher Rvalue on TLC and explain your choice.

Chemistry
1 answer:
Assoli18 [71]3 years ago
7 0

Biphenyl will have a higher R value than the Methyl Orange.

Explanation:

Biphenyl is a aromatic hydrocarbon and it is a nonpolar molecule.

Methyl Orange is a organic compound with a -SO₃⁻Na⁺ polar functional group which will induce a high polarity in the compound.

You may find the chemical structures of both molecules in the attached picture.

Column chromatography, which use as stationary phase silica gel, is a good technique for separation of the Methyl Orange from Biphenyl.

Being a non-polar molecule,  Biphenyl will have a higher R value than the Methyl Orange.

To separate them you use a appropriate solvent as eluent, as exemple chloroform, and Biphenyl will elute first from the column and after that, as a separate phase, Methyl Orange will elute thus separating them.

Learn more about:

chromatography

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What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c
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Answer:

[A g + ]  =  3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q )  +  C rO4^2-  ( a q )  →Ag2CrO4 (  s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12  = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f  >>  x

1.2 * 10^− 12  = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f  = 2x = 3.12 *10^-6 M

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