I would like help on four and five specifically >formula is T=e^(-ln(10) ?lc) wi

ID: 1041500 • Letter: I

Question

I would like help on four and five specifically

>formula is T=e^(-ln(10)?lc)

with l being length and c being concentration

4. basically is just asking to solve for the molar attenuation coefficient

I found the slope and the y-intercept and I'm not sure how to plug it back into the formula to get the coefficient

the formula I am using is A=-ln(T)

1 Introduction A student, August, is interested in measuring the concentration of a protein in solution. He knows from his research and his good friends, Pierre and Johann, that if you shine a laser through a solution, not all of the light will be transmitted. The ratio of transmitted light (T) will depend on the length of the container holding the solution (I), the concentration of the solution (c), and the molar attenuation coefficient (e). The relation between these parameters i the following Before August is able to measure the concentration of his protein solution, he must first determine the molar attenu- ation coefficient, a parameter that is specific to the protein. It is measured in units of L mol-1 em-1 he is To do this, August designs an experiment. He can't change the mass attenuation coefficient of the protein, but able to change the length of the container. As the length increases he knows less light to get through based on Fq Lastly, he will need a sample of known concentration of his protein. Luckily he has a bottle in the back with a known concentration of 5.8pmolL-1 (Note the micro) See the figures below for illustrations of his experiment. August performs his experiment and collects the following data. length(cm) | Transmitted Light 0.99 1.99 3.00 3.93 5.08 6.00 6.93 7.91 9.10 10.14 7431 4461 3257 2416 1219 0854 0668 0434 0291 A: 2.303 E 2 Questions Question 1 Plot this data. Is it linear? estion 2 Using your knowledge of logarithms rewrite Equation 1 so that it is linear. Your result should take for form of f(T) ml +b where m and b are constants, I is the length and T is the transmitted light ratio. Define a new named the absorbance, that is equal to -n(T). Write out this new equation. This is known as Beer's Law, named after August Beer. T:2.3I Question 3 Make a new column in your table for the absorbance. Now create a plot of A as a function should be linear. Create a best fit line and determine the slope and intercept. What should the intercept be? Question 4 Using the slope and Beer's Law, determine the molar attenuation coefficient ( Question 5 Now that August knows the molar attenuation factor of his tration of his solution. He use a concentration of his solution? protein solution he can measure the concen- container of length lem and get a transmitted light ration of 0.2214. What is the

Explanation / Answer

(1) The plot of transmitted light vs the length of the container is shown in the graph below.

Plot of T vs l (cm)

The plot is definitely not linear; rather the plot is exponential.

(2) We have T = e-ln(10)?lc

Take natural logarithms on both sides as

ln T = -ln(10)?lc*ln (e)

=====> ln T = -?lc*2.303*1 = -2.303*?lc

The equation is of the form f (T) = ml + b

where f (T) = ln (T); m = -2.303*?c and b = 0.

We define the absorbance of a solution as A = -ln (T); therefore, the linear equation takes the form

A = -ln (T) = 2.303*?lc

which is Beer’s law.

(3) Prepare the following table.

l (cm)

A = -ln (T)

0.99

0.7431

0.2969

1.99

0.4461

0.8072

3.00

0.3257

1.1218

3.93

0.2416

1.4205

5.08

0.1777

1.7276

6.00

0.1219

2.1045

6.93

0.0854

2.4604

7.91

0.0668

2.7060

9.10

0.0434

3.1373

10.14

0.0291

3.5370

Plot A vs l as below.

Plot of A vs l

The plot is definitely linear. The slope of the plot is m = 0.3414 cm-1 (the left hand side is dimensionless while x has dimension cm). The intercept of the plot is b = 0.0516.

(4) We have, from Beer’s law,

A = 2.303*?lc

and y = 0.3414x + 0.0516

Therefore,

?*c = 0.3414 cm-1

We have c = 5.8 µmol.L-1; therefore,

?*(5.8 µmol.L-1) = (0.3414 cm-1)

=====> ? = (0.3414 cm-1)/(5.8 µmol.L-1) = (0.3414 cm-1)/[(5.8 µmol.L-1)*(1 mol/106 µmol)]

=====> ? = 58862.06897 L.mol-1.cm-1 ? 58862 M-1.cm-1 (ans).

(5) We have l = 1 cm and A = 0.2214; therefore,

A = 2.303*?lc = 2.303*(58862 M-1.cm-1)*(1 cm)*c

=====> 0.2214 = 135559.186*c M-1

=====> c = (0.2214)/(135559 M-1) = 1.6332*10-6 M ? 1.63*10-6 M (ans).