a solution of a metal ion has a molar absorptivity (E) of 75M^-1 at 450 nm. a 10mL sample of this solution was withdrawn and diluted to 250mL. from this diluted solution, a 5 mL sample was withdrawn and diluted to 100 mL. the absorbance of the final solution was .234. find the molarity of metal ion in the original solution.
You don't give the thickness of the sample for the final absorbance so I will assume it is the same (say 1 cm) as when the standards were made. So
A = ec*1
0.234 = 75 c
c = 0.234/75 = estimated 0.003 M but that was the final solution. The original M must be
approx 0.003 x (100/5) x (250/10) = ? M for the initial solution.
To find the molarity of the metal ion in the original solution, we can use the Beer-Lambert Law, which states that the absorbance (A) is equal to the molar absorptivity (E) times the path length (l) times the concentration (c) of the absorbing species:
A = E * l * c
First, let's find the concentration of the final solution using the absorbance and molar absorptivity values:
A = 0.234
l = 1 cm (assuming a standard cuvette width)
Using the equation, we can rearrange it to solve for concentration (c):
c = A / (E * l)
c = 0.234 / (75 M^-1 * 1 cm)
c = 0.00312 M
The concentration of the final solution is 0.00312 M.
Next, let's determine the concentration of the second solution (the 5 mL sample diluted to 100 mL):
Given:
Volume of second solution (V2) = 5 mL = 0.005 L
Volume after dilution of second solution (V2') = 100 mL = 0.1 L
To calculate the concentration after dilution, we can use the formula:
c2' = (c2 * V2) / V2'
c2' = (0.00312 M * 0.005 L) / 0.1 L
c2' = 0.000156 M
The concentration of the second solution after dilution is 0.000156 M.
Finally, let's determine the concentration of the original solution (the 10 mL sample diluted to 250 mL):
Given:
Volume of original solution (V1) = 10 mL = 0.01 L
Volume after dilution of original solution (V1') = 250 mL = 0.25 L
Using the same concentration dilution formula:
c1' = (c1 * V1) / V1'
0.000156 M = (c1 * 0.01 L) / 0.25 L
Solving for c1 (the molarity of the metal ion in the original solution):
c1 = (0.000156 M * 0.25 L) / 0.01 L
c1 = 0.0039 M
The molarity of the metal ion in the original solution is 0.0039 M.
To find the molarity of the metal ion in the original solution, we can use the Beer-Lambert Law, which states that the absorbance is directly proportional to the concentration of the absorbing species and the path length.
The Beer-Lambert Law equation is:
A = εcl
Where:
A = Absorbance
ε = Molar absorptivity (constant for a specific substance at a specific wavelength)
c = Concentration of the absorbing species
l = Path length (typically the length of the cuvette, usually 1 cm)
Let's break down the given information and solve step by step:
1. Calculate the absorbance of the 5 mL sample:
A1 = 0.234
2. Calculate the concentration of the metal ion in the final solution:
Using the Beer-Lambert Law: A1 = ε1 * c1 * l1
c1 = A1 / (ε1 * l1)
We are given:
ε1 = 75 M^-1 (molar absorptivity)
l1 = 1 cm (path length)
A1 = 0.234 (absorbance)
Plugging in the values:
c1 = 0.234 / (75 M^-1 * 1 cm)
c1 = 0.00312 M
3. Calculate the concentration of the metal ion in the 10 mL sample before dilution:
c2 = c1 * (V1 / V2)
We are given:
V1 = 10 mL (initial volume)
V2 = 250 mL (final volume)
Plugging in the values:
c2 = 0.00312 M * (10 mL / 250 mL)
c2 = 0.0001248 M
4. Calculate the concentration of the metal ion in the original solution:
c3 = c2 * (V3 / V4)
We are given:
V4 = 5 mL (volume withdrawn from the diluted solution)
V3 = 100 mL (final volume)
Plugging in the values:
c3 = 0.0001248 M * (5 mL / 100 mL)
c3 = 0.00000624 M
Therefore, the molarity of the metal ion in the original solution is 0.00000624 M.