An unknown was prepared with the concentration of 0.000630 M. A Beer's Law plot was prepared using the absorbance values from standard solutions of ASA and a line drawn through all the points passed through the origin with a slope of 1550.2 M–1 . The expected absorbance and %T values for the diluted aspirin solution prepared by the student is _______ and _________, respectively.
If you plot these as A vs C, and if it obeys Beer's Law, it gives a straight line passing through the origin with the slope of Absorbance/concn = 1550.2.
Thus A = 1550.2*0.000630 = ??
and A = log(100/%T)
0.98 is correct
but the second part is not right
To determine the expected absorbance and %T values for the diluted aspirin solution, we can use Beer's Law equation:
A = ε * b * c
where:
A = absorbance
ε = molar absorptivity (also known as the molar absorptivity constant)
b = path length (usually given in cm)
c = concentration (in M)
In this case, we are given that the slope of the Beer's Law plot is 1550.2 M–1. This slope represents the product of ε * b. Since the path length is not specified, we can assume it to be 1 cm, which is common for spectrophotometric measurements.
So, ε * b = 1550.2 M–1
Now, we can rearrange the equation and solve for ε:
ε = (1550.2 M–1) / b
Next, we need to calculate the concentration of the diluted aspirin solution. The unknown was prepared with a concentration of 0.000630 M. Let's assume that the solution was diluted by a factor of 10. Therefore, the concentration of the diluted solution is:
c_diluted = (0.000630 M) / 10
= 0.000063 M
Now we can substitute the known values into the Beer's Law equation to find the expected absorbance:
A = ε * b * c
A = [(1550.2 M–1) / 1 cm] * (1 cm) * (0.000063 M)
A = 0.0977
Therefore, the expected absorbance for the diluted aspirin solution is 0.0977.
To calculate the %T (percent transmittance), we use the equation:
%T = 100 * 10^(-A)
%T = 100 * 10^(-0.0977)
%T = 79.28
Therefore, the expected %T value for the diluted aspirin solution is 79.28%.