Download Exam 2 with Questions and Answers | Principles of Chemistry II | CH 302 and more Exams Chemistry in PDF only on Docsity!
This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
labrake - ch
— Unit 6 Exam —
Spring 2013 - TTh 11am (51535) 7-9pm March 6, 2013 001-175 in WCH 1. 176-325 in PAI 3. 326 and up in MEZ 1.
001 3.5 points Which one of the following is a strong acid?
- H 2 SO 4 correct
- (COOH) 2
- HSO− 3
- H 3 PO 4
- HClO
Explanation: Sulfuric acid (H 2 SO 4 ) is a strong acid since it ionizes completely in water. All of the others only partially ionize.
002 3.5 points The Ka of lactic acid is 1.4 x 10−^4. If you mix 100 mL of a 0.1 M solution of lactic acid with 4 mL of 1.0M solution of NaOH, what is the pH of the resulting solution?
- 3.67 correct
Explanation: A 100 mL solution of 0.1 M lactic acid will contain 0.01 moles of lactic acid. 4mL of a 1.0M solution of NaOH will contain. moles of OH−^ ions. HA(aq) + OH−(aq) ↔ A−(aq) + H 2 O pH = pKa + log
[lactate] [lacticacid] pKa = -log(1.4 x 10−^3 ) = 3. pH = 3.85 + log
003 3.5 points Given that the Ka for HF is 7.2 x 10−^4 , what is the equilibrium constant for the following reaction?
F−(aq) + H 3 O+(aq) ⇀↽ HF(aq) + H 2 O(l)
- 1
- 7.2 x 10−^4
- 1 x 10−^14
- 1.4 x 10^3 correct
- 1.4 x 10−^11
Explanation: The reaction is simply the reverse of the reaction of HF with water. Thus the equi- librium constant for this reaction is simply 1/Ka. K =
[HF]
[F−][H 3 O+]
Ka K = 1/(7.2 x 10−^4 ) = 1.4 x 10^3
004 3.5 points What is the pH of a 0.2 M solution of pyri- dinium chloride (C 5 H 5 NHCl)? Kb for pyri- dine (C 5 H 5 N) is 1. 5 × 10 −^9.
2.938 correct
Explanation: MC 5 H 5 NHCl = 0.2 M Kb = 1. 5 × 10 −^9 It’s a salt of a weak base (BHX). This means you need a Ka for the weak acid BH+:
Ka =
Kw Kb
=
1. 0 × 10 −^14
1. 5 × 10 −^9
= 6. 66667 × 10 −^6
You CAN use the approximation for the equilibrium which means that
[H+] =
Ka · CBH+
=
(6. 66667 × 10 −^6 ) (0.2)
= 0.0011547 M
pH = − log(0.0011547) = 2. 93753
005 3.5 points Nitrous acid, HNO 2 has a Ka of 4.6 x 10−^4. If 100 mL of a 0.2 M solution of nitrous acid is mixed with a 200 mL of a 0.1 M solution of NaOH which of the following is NOT true of the resulting solution?
- [Na+] > [H 3 O+]
- [NO− 2 ] = [OH−] correct
- they are all true
- [NO− 2 ] > [HNO 2 ]
- [OH−] > [H 3 O+] Explanation: Mixing these two solution will result in a solution that is basic as essentially all the HNO 2 will be converted into NO− 2. This will result in a basic solution with a concentration of NO− 2 that is .2/3 = .067 M. The false statement is that the [NO− 2 ] = [OH−]. They will be very different.
006 3.5 points Write the equilibrium expression for the reac- tion 2 NO(aq) + 4 H 2 O(ℓ) ←→ 3 H 2 (aq) + 2 H+(aq) + 2 NO− 3 (aq)
1. K =
[NO]^2
[H+]^2 · [NO− 3 ]^2
2. K =
[H 2 ]^3 · [H+]^2 · [NO− 3 ]^2
[NO]^2 · [H 2 O]^4
3. K =
[H 2 ]^3 · [H+]^2 · [NO− 3 ]^2
[NO]^2
correct
4. K =
[NO]^2
[H 2 ]^3 · [H+]^2 · [NO− 3 ]^2
5. K =
[H+]^2 · [NO− 3 ]^2
[NO]^2
Explanation: Set up K, products in the numerator, reac- tants in the denominator, all raised to respec- tive stoichiometric coefficients.
007 3.5 points A 0.00001 M solution of KOH has a pH of
- 9.0 correct
Explanation:
RATIO =
[base] [acid]
= 10^0.^22
1.66 best matches 5/3 or 5 : 3
010 3.5 points What is the conjugate acid of O^2 −?
- O−
- H 2 O
- HOOH
- OH^2 −
- OH−^ correct
Explanation: The conjugate acid of a species is that species with an added proton.
011 3.5 points If 100 mL of each of the following solutions is mixed, which one produces a buffer?
- 1.0 M NH 4 Cl(aq) + 0.4 M HCl(aq)
- 1.0 M NH 4 Cl(aq) + 1.0 M KOH(aq)
- 1.0 M NH 3 (aq) + 0.4 M HCl(aq) correct
- 1.0 M NH 3 (aq) + 1.0 M HCl(aq)
- 1.0 M NH 3 (aq) + 0.6 M KOH(aq)
Explanation:
012 3.5 points Which of the following produces the STRONGEST conjugate base?
- CH 3 COOH (pKa = 4.75)
- HF (pKa = 3.45)
- HClO (pKa = 7.53)
- HIO (pKa = 10.64) correct
- HCOOH (pKa = 3.75)
Explanation:
013 3.5 points A group who investigated the formation of ammonia from its elements determined that Kc is 90 at 773 K and 3 at 873 K. Is this reaction endothermic or exothermic?
- endothermic
- exothermic correct
- Neither; the reaction is at equilibrium.
- Cannot tell; insufficient information is given.
Explanation: The K is observed to decrease with increas- ing temperature of reaction, which is charac- teristic of exothermic reactions.
014 3.5 points The equilibrium N 2 (g) + O 2 (g) ⇀↽ 2 NO(g) ∆H = 200 kJ/mol exists in a closed container. Which perturba- tion would favor the formation of NO(g)?
- Increasing the temperature correct
- Removing some O 2 (g) from the reaction
- Decreasing the pressure
- Decreasing the temperature
- Increasing the pressure
Explanation:
015 3.5 points Benzoic acid reacts with water to form the benzoate ion by the following reaction
C 6 H 5 COOH(aq) + H 2 O(l) ⇀↽ C 6 H 5 COO−(aq) + H 3 O+(aq)
The equilibrium constant for this reaction is 6.4 x 10−^5. In a 0.1 M solution of benzoic acid, what is the concentration of the benzoate ion at equilibrium?
- 2.5 x 10−^3 M correct
- 0.1 M
- 6.4 x 10−^5 M
- 8.0 x 10−^3 M
- 0.0975 M
Explanation: The concentration of the benzoate ion and the hydronium ion will be the same at equi- librium. Thus this problem is the same as solving for the hydronium ion concentration.
Ka =
[benzoate][H 3 O+] [benzoic acid] Assuming a change of ”x” in the concentra- tion of benzoic acid gives
Ka =
(x)(x) (0. 1 − x)
x^2
. 1 x =
(Ka)(0.1) = 2.5x10−^3
016 3.5 points At 25◦C, ∆G◦ r for the reaction
2 SO 2 (g) + O 2 (g) → 2 SO 3 (g)
is − 141 .74 kJ · mol−^1. What is the value of Kp for this reaction?
- 01 × 1024 correct
- 74 × 1026
- 56 × 1013
- 65 × 1012
Explanation: When the relationship
∆G◦ r = −RT lnK
is used with all gaseous reactants and prod- ucts, K represents Kp.
017 3.5 points You are investigating the reaction
N 2 (g) + 3H 2 (g) ⇀↽ 2NH 3 (g)
At a particular temperature you find an equilibrium mixture for which the partial pressure of N 2 is 1 atm, the partial pressure of H 2 is 0.1 atm, and the partial pressure of NH 3 is 0.1 atm.
You find a second container with a gas mix- ture that has a partial pressure of N 2 of 0. atm, partial pressure of H 2 of 1 atm, and a partial pressure of NH 3 of 1 atm. What can you say about this second system?
- Reaching equilibrium for the second sys- tem requires the reaction generating more re- actants
- Reaching equilibrium for the second sys- tem requires the reaction forming more prod- ucts
- The second system is at equilibrium cor- rect
Explanation: For this reaction K =
P^2 NH 3
P^3 H 2 PN 2
(0.1)^2
(0.1)^3 (1)
For the second system you find
Q =
P^2 NH 3
P^3 H 2 PN 2
(1)^2
(1)^3 (.1)
Q = K so this system is also at equilibrium.
018 3.5 points Which of the following is TRUE?
- When the value of Q is large, the equilib- rium lies on the product side of the equilib- rium reaction.
- A small value of K means that the equilib- rium concentrations of the reactants are small compared to the equilibrium concentrations of the products.
021 3.5 points This question is merely a placeholder for the points in the hand-graded portion of the exam.?
Explanation:
This question is merely a placeholder for the points in the hand-graded portion of the exam.
