Please help, I can't do this and its stressing me out ?!?! A 25.0 mL aliquot of an aqeuous quinine solution was diluted to 50.0mL and was found to have an absorbance of 0.832 at 348nm when measured in 2.00cm cell. A second 25.0mL aliquot was mixed with 10.0ml of a solution containing 23.4ppm of quinine, and after dilution to 50.0ml, had an absorbance of 1.220 (2.00 cm cell path). Calculate the concentration in parts per million of quinine in the sample.
The standard is in the second part.
A = abc
A = 1.220
a = ?
b = 2
c = 23.4 ppm
a = A/bc = ?
Then for the unknown,
A = abc.
You know A = 0.832, a(from the calculation above), and b = 2 cm. Solve for c of the unknown in ppm.
To calculate the concentration in parts per million (ppm) of quinine in the sample, we can use the Beer-Lambert Law, which relates the absorbance (A), the molar absorptivity (ε), the path length (l), and the concentration (C) of a solution.
The formula for the Beer-Lambert Law is A = εCl, where:
A represents the absorbance,
ε represents the molar absorptivity (which is specific to each compound),
C represents the concentration,
and l represents the path length.
Given:
1) A 25.0 mL aliquot of an aqueous quinine solution was diluted to 50.0 mL and had an absorbance of 0.832 at 348 nm when measured in a 2.00 cm cell.
2) A second 25.0 mL aliquot was mixed with 10.0 mL of a solution containing 23.4 ppm of quinine, and after dilution to 50.0 mL, had an absorbance of 1.220 in a 2.00 cm cell.
To find the concentration of quinine in the sample in ppm, we need to determine the molar absorptivity (ε). Since the path length (l) and cell path (2.00 cm) remain constant, we can cancel them out when setting up a ratio of the two absorbance values.
Let's calculate the molar absorptivity (ε) first using the first aliquot:
A₁ = 0.832
C₁ = concentration (to be determined)
l = 2.00 cm
Substituting these values into the Beer-Lambert Law, we have:
A₁ = εCl
0.832 = ε * C₁ * 2.00
Next, let's calculate the molar absorptivity (ε) using the second aliquot:
A₂ = 1.220
C₂ = 23.4 ppm
l = 2.00 cm
Substituting these values into the Beer-Lambert Law, we have:
A₂ = ε * C₂ * 2.00
1.220 = ε * 23.4 * 2.00
Now we have two equations with two unknowns (ε and C₁). We can solve this system of equations simultaneously to find ε and C₁.
Simplify the equations:
1) 0.832 = 2.00 * ε * C₁
2) 1.220 = 46.8 * ε
Now, we can solve these equations simultaneously to find the molar absorptivity (ε) and the concentration (C₁).
To solve these equations, divide equation 2 by equation 1:
1.220 / 0.832 = (46.8 * ε) / (2.00 * ε)
Simplifying further:
1.468 = 23.4
From this, we can see that the concentration (C₁) cancels out and we are left with a ratio of the molar absorptivity (ε).
Therefore, the molar absorptivity is equal to 23.4.
Now, we can substitute the value of ε into either of the two original equations to find the concentration (C₁). Let's use equation 1:
0.832 = 2.00 * 23.4 * C₁
Simplifying further:
C₁ = 0.832 / (2.00 * 23.4)
C₁ ≈ 0.0178 mol/L (since we are working with mL, we can equate it to 0.0178 M)
Finally, to express the concentration in parts per million (ppm), we need to convert from molarity (M) to ppm. Since 1 ppm is equivalent to 1 mg/L, we can multiply the concentration by the molar mass of quinine to get the concentration in ppm.
The molar mass of quinine is approximately 324.43 g/mol.
Concentration in ppm = 0.0178 M * 324.43 g/mol * 10^6 (to convert from g to mg)
Concentration in ppm ≈ 5,765 ppm
Therefore, the concentration of quinine in the sample is approximately 5,765 ppm.