Solve 0.1% w/w Concentration Problem: Beer's Law & C15H6N4

[monae]

Problem:
Assume the human eye can detect light with transmittance of 0.35.
The colored form of the indicator has ε = 8.79 x 10^5. and λmax = 580 nm. How
much indicator in ml should be used if the indicator has concentration 0.1% w/w in water.

Equation Used:
Beer's Law = A = εbc
where ε is given as 8.79 x 10^5
b = 1.00cm
A= -logT = -log(0.35) = 0.456

Using Beer's Law, I solved for C and got 5.08(10^-7)mol/L
But I don't know what to do to find the amount of indicator needed.
The molecular formula is C15H6N4 if that helps.

 

[chemisttree]

Looks like you are going to need 5.08 X 10^-7 moles per liter to see it. How much of the 0.1% w/w of the indicator (you should probably convert this to mol/L using the molecular formula) will you need?

That depends on how much solution you will have to dilute this into (what is the volume of the cell). This information is missing in your problem.

 

[Blueshift5]

To solve this problem, we can use the following steps:

1. Convert the concentration of the indicator from w/w (weight/weight) to molarity. This can be done by calculating the molar mass of C15H6N4 and using the given concentration of 0.1% w/w. The molar mass of C15H6N4 is 258.23 g/mol, so the concentration in molarity is 0.0000408 M.

2. Use the Beer's Law equation A = εbc to calculate the molar absorptivity (ε) at λmax of 580 nm. Plugging in the values given, we get ε = 8.79 x 10^5 L/mol*cm.

3. Use the molarity and molar absorptivity to solve for the path length (b) of the cuvette. Rearranging the Beer's Law equation, we get b = A/(εc). Plugging in the values, we get b = 0.456/(8.79 x 10^5 * 0.0000408) = 0.0134 cm.

4. Now that we know the path length, we can calculate the volume of the indicator needed. The path length (b) is equal to the thickness of the cuvette, which is typically 1 cm. Therefore, the volume of the cuvette is 1 cm^3 or 1 mL.

5. Finally, we can calculate the amount of indicator needed by multiplying the volume of the cuvette (1 mL) by the molarity of the indicator (0.0000408 M). This gives us 4.08 x 10^-5 moles of indicator needed.

In conclusion, to prepare a solution with a concentration of 0.1% w/w of C15H6N4, we would need to use 4.08 x 10^-5 moles of the indicator in 1 mL of water. This can be converted to grams by multiplying by the molar mass of C15H6N4, which would give us 0.0105 grams of indicator needed.