TestMagic

Intrepidjss · 2009 November 5th, 02:28 AM

Hey Guys, I am a college grad but taking a standardized exam, hence I do not have professors to go to for help. Could some one please help me with few questions?

Question 1. Which of the following is closest to the pH of a solution that contains 5 millimolar per liter of H+ ions.

A. 1.2
B. 2.3
C. 3.7
D. 6.5
E. 7.5

The answer is B, but I don't know how to calculate.

Question 2. The ionic strength of MgCl2 is

A. 0.25
B. 0.5
C. 1.0
D. 1.5
E. 5.0

The answer is D, I don't know how to do it though.

Question 3. What is the pKa of trimethylammonium in water if the base ionization constant Kb for trimethylamine is 7.4 * 19^-5? (log 7.4 * 10^-5 = -4.13)

A. -4.13
B. 2.87
C. 4.13
D. 9.87
E. 11.13

The answer is D. I don't know how to get the answer though.

Question 4. In an intact cell, the free energy change associated with an enzyme catalyzed reaction is frequently different from the standard free energy change of the same reaction because in the intact cell the

A. activation energy is different
B. reaction is always near equilibrium
C. enzyme may be regulated allosterically
D. Reactants are not at 1 M concentrations
E. Reaction may be catalyzed by more than one enzyme.

The answer is D, but can some one explain a little more?

Question 5. The pH of a 10^-4 molar HCl solution prepared with pure water at pH 7.0 is closest to.

A. 6.0
B. 6.9
C. 7.1
D. 7.9
E. 8.0

The answer is B.

Question 6. The nucleoside adenosine exists in a pronated form with a pKa of 3.8. The percentage of pronated form at pH 4.8 is closest to

A. 1
B. 9
C. 50
D. 91
E. 99

The answer is B.

milkvincent · 2009 November 5th, 07:21 PM

Q1: pH = -log[H⁺]
pH = -log(5x10^-3)
= -(log5 + log10^-3)
= -(log5 + -3) <- log5 = something between 0 and 1
= -(0.X + -3)
= -(-2.X)
=2.X

Q2: I don't know the definition of ionic strength, but I believe it's in wikipedia. Try to search for ionic strength in wikipedia.

Q3: Let HA = the protonated form of trimethylammonium and A^- be the conjugated base (trimethylammonia)

The equation of K_a: HA <-> H⁺ + A^-
The equation of K_b: A^- + H₂O <-> HA + OH^-

K_a = [A^-][H⁺] / [HA]
K_b = [HA][OH^-]/A^-

So, K_a * K_b = [H⁺][OH^-] <<Just do the simplification by multiplying Ka and Kb will give you this.

since [H⁺][OH^-] = K_w, K_a*K_b = K_w = 10^-14

or in other words, pK_a + pK_b = pK_w = 14

4.13 + pK_b = 14
pK_b = 9.87

Q4 Standard free energy change (chemical definition) is the free energy change when all the species in the equation is at 1M at 25 ^oC and 1 atm

However, in cells, the concentrations of almost all species are not at 1M. Hence, the free energy change is different from the standard.

Remember that enzymes/catalysts will NEVER alter the free energy change of an reaction. It does not matter how many enzymes are involved or how enzymes are regulated, they DO NOT affect the free energy change. Therefore, C and E are wrong.

The statement A in itself is correct. However, activation energy has nothing to do with the free energy change of a reaction so it is wrong.

B is wrong because if reactions are at equilibrium, the cells will die. The cells maintain the concentrations of substrate such that reactions do not reach equilibrium to ensure they always proceed forward.

Q5 Are you sure it's 10^-4 and not 10^-8? I think this question is in the 1994 exam and it's 10^-8 in the question.

Q6: K_a = [A^-][H⁺] / [HA]

pK_a = -log[H+] - log([A^-]/[HA])
pK_a = pH - log([A^-]/[HA])
log([A^-]/[HA]) = pH - pK_a
log([A^-]/[HA]) = 1
[A^-]/[HA] = 10 / 1

therefore, percentage of protonated form = [HA] / ([HA]+A^-) = 1/(10+1) = 1/11 < 1/11 is slightly less than 10%, so B is the answer.

No offense, but your concepts about acid-base equilibrium seems quite weak. You may want to do some problems from your general chemistry text about acid-base equilibrium if you have time.

Intrepidjss · 2009 November 5th, 11:53 PM

Thank you so much for your help. These really helped me. It has been a long time since I have looked at acid base reactions.

Intrepidjss · 2009 November 6th, 01:09 AM

-4.13 + pKb = 14
pKb = 9.87

For question 3, how does that add up? Would it be 14 + 4.13?

milkvincent · 2009 November 6th, 01:23 AM

Quote:

Originally Posted by Intrepidjss

-4.13 + pKb = 14
pKb = 9.87

For question 3, how does that add up? Would it be 14 + 4.13?

Sorry, It should be 4.13 + pKb = 14, no minus. I edited it now it should be correct now.

2009 November 5th, 02:28 AM	#1 (permalink)
Intrepidjss I JUST got here. Join Date: Oct 2005 Posts: 15	Few Questions. Could you please help? Hey Guys, I am a college grad but taking a standardized exam, hence I do not have professors to go to for help. Could some one please help me with few questions? Question 1. Which of the following is closest to the pH of a solution that contains 5 millimolar per liter of H+ ions. A. 1.2 B. 2.3 C. 3.7 D. 6.5 E. 7.5 The answer is B, but I don't know how to calculate. Question 2. The ionic strength of MgCl2 is A. 0.25 B. 0.5 C. 1.0 D. 1.5 E. 5.0 The answer is D, I don't know how to do it though. Question 3. What is the pKa of trimethylammonium in water if the base ionization constant Kb for trimethylamine is 7.4 * 19^-5? (log 7.4 * 10^-5 = -4.13) A. -4.13 B. 2.87 C. 4.13 D. 9.87 E. 11.13 The answer is D. I don't know how to get the answer though. Question 4. In an intact cell, the free energy change associated with an enzyme catalyzed reaction is frequently different from the standard free energy change of the same reaction because in the intact cell the A. activation energy is different B. reaction is always near equilibrium C. enzyme may be regulated allosterically D. Reactants are not at 1 M concentrations E. Reaction may be catalyzed by more than one enzyme. The answer is D, but can some one explain a little more? Question 5. The pH of a 10^-4 molar HCl solution prepared with pure water at pH 7.0 is closest to. A. 6.0 B. 6.9 C. 7.1 D. 7.9 E. 8.0 The answer is B. Question 6. The nucleoside adenosine exists in a pronated form with a pKa of 3.8. The percentage of pronated form at pH 4.8 is closest to A. 1 B. 9 C. 50 D. 91 E. 99 The answer is B.

2009 November 5th, 07:21 PM	#2 (permalink)
milkvincent Eager! Join Date: Feb 2008 Posts: 56	Q1: pH = -log[H⁺] pH = -log(5x10^-3) = -(log5 + log10^-3) = -(log5 + -3) <- log5 = something between 0 and 1 = -(0.X + -3) = -(-2.X) =2.X Q2: I don't know the definition of ionic strength, but I believe it's in wikipedia. Try to search for ionic strength in wikipedia. Q3: Let HA = the protonated form of trimethylammonium and A^- be the conjugated base (trimethylammonia) The equation of K_a: HA <-> H⁺ + A^- The equation of K_b: A^- + H₂O <-> HA + OH^- K_a = [A^-][H⁺] / [HA] K_b = [HA][OH^-]/A^- So, K_a * K_b = [H⁺][OH^-] <<Just do the simplification by multiplying Ka and Kb will give you this. since [H⁺][OH^-] = K_w, K_aK_b = K_w = 10^-14 or in other words, pK_a + pK_b = pK_w = 14 4.13 + pK_b = 14 pK_b = 9.87 Q4 Standard free energy change (chemical definition) is the free energy change when all the species in the equation is at 1M at 25 ^oC and 1 atm However, in cells, the concentrations of almost all species are not at 1M. Hence, the free energy change is different from the standard. Remember that enzymes/catalysts will NEVER alter the free energy change of an reaction. It does not matter how many enzymes are involved or how enzymes are regulated, they DO NOT affect the free energy change. Therefore, C and E are wrong. The statement A in itself is correct. However, activation energy has nothing to do with the free energy change of a reaction so it is wrong. B is wrong because if reactions are at equilibrium, the cells will die. The cells maintain the concentrations of substrate such that reactions do not reach equilibrium to ensure they always proceed forward. Q5 Are you sure it's 10^-4 and not 10^-8? I think this question is in the 1994 exam and it's 10^-8 in the question. Q6: K_a = [A^-][H⁺] / [HA] pK_a = -log[H+] - log([A^-]/[HA]) pK_a = pH - log([A^-]/[HA]) log([A^-]/[HA]) = pH - pK_a log([A^-]/[HA]) = 1 [A^-]/[HA] = 10 / 1 therefore, percentage of protonated form = [HA] / ([HA]+A^-) = 1/(10+1) = 1/11 < 1/11 is slightly less than 10%, so B is the answer. No offense, but your concepts about acid-base equilibrium seems quite weak. You may want to do some problems from your general chemistry text about acid-base equilibrium if you have time. Last edited by milkvincent : 2009 November 6th at 01:25 AM.*

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

2009 November 5th, 11:53 PM	#3 (permalink)
Intrepidjss I JUST got here. Join Date: Oct 2005 Posts: 15	Thank you so much for your help. These really helped me. It has been a long time since I have looked at acid base reactions.

2009 November 6th, 01:09 AM	#4 (permalink)
Intrepidjss I JUST got here. Join Date: Oct 2005 Posts: 15	-4.13 + pKb = 14 pKb = 9.87 For question 3, how does that add up? Would it be 14 + 4.13?

TestMagic

Free GMAT, GRE, SAT Test, TOEFL practice tests