CHEM 2PC3 Lecture Notes - Lecture 15: Internal Energy, Intensive And Extensive Properties, Ideal Gas

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27 Sep 2016

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Chem 2r03 - week_1 homework solutions - sept 14, 2007. Week_1 solutions: (1) consider the protein (p) / ligand (l) dissociation equilibrium: pl p + l with a 1 nm dissociation constant (k d) for the complex pl. Imagine that you have a solution with a protein concentration [p] = 100 nm. What will the concentration of the protein/ligand complex ([pl]) be after adding ligand up to a total ligand concentration of 100. Choose one option and explain. a. b. c. d. The correct answer is c. here is the explanation. Since k d = [p][l]/[pl], we know that [p]/[pl] = kd/[l]. We also know that [l], the equilibrium concentration of the free ligand, is about the same as the initial concentration of the ligand (100 m) because the initial concentration of protein is only 100 nm (or 0. 1 m). So [p]/[pl] ~ 10-5 or in other words essentially all the protein is used up to form the pl complex.

Related Questions

Dr Vlad is testing the effects of two drugs named D1 and D2 on a protein P found in the blood. He does several tests to characterize how the drugs interact with protein P. He tests them separately and determined that one protein P molecule has only one binding site for D1. Similarly, one protein P interacts only with one molecule of D2.

a) (2pts) First, Dr Vlad, tests the interaction of D1 with P in his lab. At equilibrium: the concentration of free D1 is 10 mM, and the concentration of free P is 50 mM

Given that the affinity constant is for this interaction is Kd=40 mM under these experimental conditions (pH salt, temperature, etc), what is the concentration of complex PD1 at equilibrium?

b (2pts) Then Dr Vlad adds more D1 drug is added to the mixture. When the equilibrium is established again, the concentration of free D1 is now 40 mM. What percentage of P remains free at this new equilibrium? Explain. Note: you need to know that the experimental conditions (pH, salt, temperature, etc) in part b are the same as in part a

c (2pts) The experimental conditions used by Dr Vlad in the lab reflect very closely the conditions in the blood, he can therefore use his data to calculate what happens in the blood. Using his available data, what is the concentration of free drug D1 in the blood when 50% of the protein P in the blood is bound to the drug? Explain your reasoning.

d (2pts) Now Dr Vlad considers the same protein P binding to drug D2. The experimental conditions are the same as those used for D1 above (points a and b).

P and D2 are mixed together: at equilibrium, the concentration of free P is 100 mM and the concentration of free D2 is 100 nM, the concentration of PD2 complex is 10 mM.

Which drug D1 or D2 has the highest affinity for protein P? ________

Explain how you get to that answer

e) (2pts) Dr Vlad accumulated lots of data on how D1 and D2 bind to P. In addition to Kd values (defined or determined above), he also found that the ka value for D1 binding to P is 10^-6 M^-1s^-1 and the ka value for D2 binding to P is 10^-5 M^-1s^-1. Which drug has the fastest association rate for P? And how much faster is that rate compared to the other drug? Explain briefly without calculations.

Lewis Acids and Bases

This is related to a post-lab of inorganic chemirsty.

The stability of Lewis acid-base complexes can be describedthrough the stability constant which is analogous to the Ka that describes BrÃ¸nsted-Lowryacidities. Many metal ions can coordinate (bind) many Lewis base ligands in a step-wisefashion according to:

M + L = ML K1 = [ML]/[M][L] Eq. 1--Please note that the L in the denomnator is supposed to be the alpha value.

ML + L = ML2 K2 = [ML2]/[ML][L] Eq. 2

ML2 + L = ML3 K3 = [ML3]/[ML2][L] Eq. 3

MLn-1 + L = MLn Kn = [ML4]/[ML3][L] Eq. 4

and for which the free energy for this dissociation reaction is given by:

âG = -RTLnKn Eq. 5.

where R is the gas constant and T is the temperature in Kelvin.

The ethylene diamine tetraacetic acid ligand (EDTA, a Lewis base) provides amechanism to probe the Lewis acid strength of various metal complexes. This multi-dentate ligand (multiple Lewis base sites on a single ligand) forms complexes withmetals of the form:

Mn+ + Y4- = MYn-4 Eq. 6

K = [MYn-4]/[Mn+][Yn-4] Eq. 7

where Y4- is the EDTA ligand.In this experiment, the Lewis acid strength for two different metals will beexamined by measuring the stability constants for the metal-EDTA complex. Of specificinterest is the how the Lewis acid strength of the metal ion correlates with:

1) atomic/ionic radii, 2) oxidation state, 3) electron configuration, and 4) polarizability.

You will determine the stability constants of both a Cu2+- and Zn2+-EDTA complexes as well as the corresponding free energy. Using these data, you will comparethe relative Lewis acidities of both metals and describe differences in terms of the metalion properties described above.

Experimental Procedure

You and your group will determine the concentration and ultimately the free energy of Cu(II) and Zn(II) using the hexadentate chelating ligand, EDTA.

You will be given a ~100 mM standardized solution of EDTA (check the label forexact concentration prior to beginning). You will need to weigh ~0.1 g per metal saltsample. Dissolve the metal salt into 50 mL of DI water. Add 2-3 drops of murexide toyour metal. After properly preparing your buret with EDTA, titrate your metal with EDTA. Repeat for a total of 3 titrations per metal.

Results:

For Cu(II), see below:

Values for each titration (we had three tries for each salt sample).

.1114 g Cu(II)------4.8 mL EDTA solution-------pH 3.0 found after titration

.1054 g Cu(II)-------4.1 mL EDTA solution ------pH 2.8 found after titration

.1182 g Cu(II)------4.8 mL EDTA solution------pH 2.7 found after titration

-------------------------------------------------------------------------------------------------------------------------

For Zn(II), see below:

.099 g Zn(II)-----3.5 mL EDTA solution----pH 2.5

.0965 g Zn(II)----4.4 mL EDTA solution----pH 2.7

.0952 g Zn(II)-----3.6 mL EDTA solution---pH 2.01

Questions:

1) What trends would you predict for EDTA-metal complex Kn and ÎG values when:

a) going down a row of transition metals of the same oxidation state?

b) of the same transition metal with increasing oxidation state? (For example,Iron can exist as Fe(II), Fe(III), Fe(IV), and higher oxidation states)

2. That are the differences between the concepts of Lewis and BrÃ¸nsted-Lowry acids and bases?

Hello, I am doing homework and I am not sure if my answers are correct, can someone help me verify if I am doing this right or wrong? My responses are underneath the questions.

You are given a protein solution with a concentration of 0.25 mg/ml. Answer these questions.

A.) We need 20 Âµg for an experiment. What volume of the protein solution do we need?

Convert 20 Î¼g and you will get 0.02 mg C = M/V rearrange for Volume V= M/C , V= (0.02 mg)/(0.25 mg/ ml) = 0.08 ml or 80 Î¼l

B.) Suppose we want to prepare a solution containing 150 Âµg of the protein at a concentration of 0.50 mg/ml. To do this we will first dry down enough protein solution to get 150 Âµg. How much solution do we need to start with? How much H2O do we add to get the desired concentration?

You will convert 150 Î¼g to 0.15 mg and divide by 0.25 mg/ ml to get 0.60 ml that needs to be dried down to obtain 0.5 mg/ml, then you will add enough water so that the total volume is 0.5 ml.

C.) If the protein has a M.W. = 15,000, express its initial concentration in moles/liter, in Âµmoles/ml and in Âµmoles/Âµl. If we want 2 Âµmoles for a reaction, what volume do we need?

Convert 0.25 mg/mL = 0.00025 g/mL = 0.00025 g/0.001 L = 0.25 g/L If M.W. of the protein is 15,000 g/mol then 0.25 g = 0.25/15,000 moles = 1.67 x 10^-5 moles. This many moles is contained in a liter (L) therefore concentration is 1.67 x 10^-5 moles/L = 1.67Î¼moles/L = 1.67 Î¼moles/1000 mL = 1.67 x 10^-3 Î¼moles/mL = 1.67 x 10^-3 Î¼moles/1000 Î¼L = 1.67 x 10^-6 Î¼moles/Î¼L. The concentration of the solution is 1.67 x 10^-6 Î¼moles/Î¼L, 1 Î¼mole is present in 1/(1.67 x 10^-6) Î¼L and if 2 Î¼moles are present in 2/(1.67 x 10^-5) Î¼L = 1.20 x 10^-5 Î¼L = 1.20 L

D.) Suppose we want 1 ml of a 10 Âµg/ml solution. How much H2O and protein stock must we add to get this?

C1V1 = C2V2 (250 Î¼g/ Î¼l)(V1) = (10 Âµg/ml)(1ml) V1= 0.04 ml or 40 Î¼l (Volume of protein stock for solution) Volume of water 1.0 ml- 0.04ml = 0.96 ml

E.) Suppose we want 100 Âµl of a 0.1 Âµg/Âµl solution. How much H2O and protein stock must we add to get this?

C1V1 = C2V2 (0.25 Î¼g/ Î¼l)(V1)= (0.1 Î¼g/ Î¼l)(100 Î¼l) V1=40.0 Î¼l (Volume of protein stock for solution) Volume of water 100 Î¼l- 40.0 Î¼l = 60.0 Î¼l

CHEM 2PC3 Lecture Notes - Lecture 15: Internal Energy, Intensive And Extensive Properties, Ideal Gas

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