PROBLEM SET 1

TAs: Badal Rana, Hassan Allahrakha

QUESTION 1

a) What is the pH of 5 mM HCl?

Answer:

HCl is a strong acid thus fully dissociates

pH= -log[ H+] = -log [HCl] = ?

b) If concentration of HCl is increased 10 fold to 50mM, what is the pH?

Solve the same Same approach as part a.

QUESTION 1 c) What is the pH of 5 mM acetic acid? ( Acetic acid pKa =

4.76 )

Answer:

Acetic acid is a weak acid, therefore it only partially dissociates;

CH3COOH ⇌ CH3COO- + H+

The amount of dissociation at equilibrium is reflected in the Keq, i.e., Ka,

which is 10-pKa

Therefore use the Ka equality and an ICE table to solve for [H+]

𝐾𝑎 = [𝐻 +][𝐶𝐻3𝐶𝑂𝑂−][𝐶𝐻3𝐶𝑂𝑂𝐻] 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

CH3COOH ⇌

CH3COO- + H+

Inital 6mM 0mM 0mM

Change -x +x +x

Equilibrium 6mM-x +x +x

QUESTION 1 PART C CONT. c)

And since X = H+

To simplify the equation, assume minimal dissociation so that you can remove the [H+] from the denominator as insignificant. Then solve for H+.

Remember, you can only remove [H+] from the denominator if there is <5% dissociation. So check if there is <5% dissociation:

Divide the calculated [H+] by the starting acetic acid concentration and express as a percent.. If > 5%, you cannot simplify the equation, but must instead solve the complete quadratic equation.

For this, rearrange the Ka equality to the form Ax2 + Bx + C = 0; where

][105

][3

2

XMx

XKa

][105

][3

2

HMx

HKa

For final answer, convert [H+] to pH

QUESTION 1

d) If the concentration of acetic is increased 10 fold to 50 mM, what is the pH?

Answer: Do the same thing as part c ….the pH should decrease.

QUESTION 1

e) What is the pH of 6mM of phosphoric acid? (For H3PO4, pKa1= 2.12, pkA2= 7.21, pKa3 = 12.32). Use Ka equation to solve for pH using pKa1 and compare it to pKa2 to see if the dissociation of second proton is significant or not.

Answer:

Ka1 = 10-pka1

Let’s assume minimal dissociation; thus remove the [H+] from the denominator

QUESTION 1 PART E CONT.

Usually the second dissociation of the proton can beIgnored if the pH of thepKa1 is ideally 2 units away from pKa2.

e)

QUESTION 1 f) pH is concentration of phophoric acid increased 10 fold to 60

mM

Answer: Same thing as part e except use 60mM instead of 6 mM

QUESTION 1

g) With each of the above acids, a 10-fold change in concentration resulted in a different amount of change in pH. Calculate the differential in pH for each acid and list with the acid in order of increasing pKa. What well-known principle is at play? Explain.

Answer:

For each acid subtract the pH at the higher acid concentration from that at the higher acid concentration.

How is the differential changing as a function of the pKa.

The principle at play is a fundamental one from gen chem. Hint: Think about the balance between the two sides of the chemical equation for acid dissociation.

Explain what’s going on.

QUESTION 2

a) What is pH of the PBS?

Answer: Since this a mixture/buffer.. Use the Henderson-Hasselbach equation.

Na2HPO4 is HPO42- in solution while KH2PO4 is H2PO4

1- in solution; this is a buffer using the pKa2

pH= pKa + log( [ conjugate base ] /[conjugate acid ])

There are three pKas given, try to figure out which one to plug in. Hint: look up disassociation of H3PO4

QUESTION 2

b) Why does PBS contain NaCl and KCl, and why is their total amount 140 mM?

Answer: Hint: think of the normal concentrations of serum. What do these concentrations do for the antibodies and potential cells present?

QUESTION 2

c) What are the concentrations of acetic acid and acetate in the 0.06 M Na-acetate, pH 3.0 used for the elution of IgG?

Answer: This is a buffer system; use H-H equation

pH= pKa + log( [ conjugate base ] /[conjugate acid ])

Hint: 0.06M is not the final concentration of acetate. Remember that putting Na-acetate in solution would cause acetate to establish an equilibrium thus the total concentration at equilibrium of acetate and acetic acid would equal 0.06M

QUESTION 2

d) What is the final pH of 2ml fraction containing purified IgG? Assume pH change by IgG is insignificant.

Answer:

Acetic acid will be titrated by Tris base to form more acetate. Remember an acid plus a base will neutralize each other. When acetic acid is neutralized, it will form acetate

Acetic Acid Acetate Tris Base

Initial Found in part c found in part c (70 x 10-6 L) x(1M)= 70 x 10-6 mol

Change

End

Plug into H-H equation : pH= pKa + log( [ conjugate base ] /[conjugate acid ]) pH= 4.76 + log(72 x 10-6 mol/48 x 10-6 mol) = 4.93

QUESTION 2

e) Why does 0.05 M Na-acetate, pH 3.0, elute IgG and why is aliquot of 1.0M Tris Base put in the collection tubes?

Answer: What do low pHs do to stable structures? And what would raising the pH back up using a base do to the same structure?

QUESTION 2

f) What type of chromatography is this? Explain.

Answer: Hint: IgG and G protein are ATTRACTED to each other.

QUESTION 3

Understand that in order to donate the methyl group, methionine must converted to SAM (an activated form that donates methyl more readily). The the delta G standard for this conversion is given.

a)What is the Keq’ for the overall reaction?

Answer:

QUESTION 3

b) What is the ΔG’ for the overall reaction at the concentrations of metabolites in E. coli?

Answer:

Hint: what is the difference between the delta G naught and delta G? The delta G naught is founded experimentally using SET conditions such as temperature, concentrations etc. Delta G on the other can be found at any temperature, concentration, usually before equilibrium is established. Consider Delta G as the “right here, right now, real world” measurement of Gibbs Free Energy

QUESTION 3

c) Calculate the change a 1000 fold decrease of PPi have to ΔG’?

Answer:

Approach this problem as the previous part but remember there was a 1000 fold decrease to PPi.

QUESTION 3

d)Assume that the ΔGo’ PPPi hydrolysis ( step 2) is the same as PPi hydrolysis (found in a table). This is value is given as -33.6kJ/mol . What is the ΔGo’ of step 1?

Answer: Understand how the overall relates to step 1 and step 2. Hint: You are given the overall and and step 2 delta G naught values.

QUESTON 3

e) What drives this reaction?

Both delta G values are negative therefore spontaneous thus driving the reaction forward.

QUESTION 4

a) Calculate the ΔGo for myoglobin for FOLDING?

Answer: Carefully read the paragraph. Are the values given for unfolding or folding? Even though the magnitudes won’t change, the sign of the delta H naught and delta S naught will depending on if use them for FOLDING or UNFOLDING.

QUESTION 4

b) Is unfolding driven by entropy or enthalpy?

Answer:. Remember to plug values that correspond to unfolding since this question is asking about unfolding. From the answer to previous part, you’ll be able to determine if folding is favorable or favorable. A negative delta G corresponds to a favorable reaction while a positive delta G would correspond to a unfavorable reaction. If folding is favorable, then unfolding would be unfavorable and vice versa. You would then see which term would contribute the most unfolding using the correct sign values

QUESTION 4

c) What's the Keq for the FOLDING of myoglobin?

Not much to this question, just remember it’s asking for folding…the delta G naught was solved for in part a, granted you want to make sure you do part a correctly in order to get the right answer for this question

ΔGo = -RTlnKeq

QUESTION 4

d) What does the Keq tell you about unfolded vs. folded molecules?

Hint: try to think of Keq as equal to a ratio. Ex: if Keq= 40…then that means the ratio of products to reactants is 40 to 1

Hypothetically, if unfolding was unfavorable and folding was favorable and a large Keq is solved for…What will this tell you about the number of folded molecules compared to unfolded?

Use this same logic and apply it to your numbers.

QUESTION 5

The pentapeptide Val-Arg-Gln-Trp-Cys has N-terminal and C-terminal pKas of 9.5 and 4.5 respectively. The side chain Pka’s are Arg 12.5 and Cys 8.0.

a) Write this peptide’s sequence using single letter abbreviations. Ex: Val-Trp-Phe-Gln-Arg

Answer: VWFQR

QUESTION 5

b)