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Chemistry 1252

Exam #4 KEY (90 minutes) August 6, 2012

First Name Last Name

Signature

Page

Points

2

20

3

26

4

27

5

27

Total

100

SID# 800‐

Score

Grader

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You may assume the 5% rule on all problems without checking the rule. Please read each question carefully and answer it completely and clearly. Most of the factual information you need is contained in the text of the problem. Answer the questions you find easy first; questions which appear difficult at first usually require a revision of your thought process. Always keep in mind the intended examination topics if a problem appears too hard. The quiz should have 5 pages for a total of 10 questions. Individual point values for free response are given in the corner of each answer space. You must show your work to receive any credit for problems involving calculations. All calculated answers and work must include units, and should be rounded to the appropriate number of significant figures!

88.91

1

A
Free Response Instructions: Solve the following problems in the boxes provided below each question.
Be sure to show all your work and units, and round your answers to the appropriate significant figures at the end of each part. FOR ALL PARTS: 1 for MISSING UNITS 1. A 100.0 mL sample of 0.100 M methylamine (CH3NH2, Kb = 3.710‐4) is titrated with 0.250 M HNO3. (a) Calculate the pH after the addition of 15.0 mL of acid. CH3NH2(aq) + H3O+(aq)  CH3NH3+(aq) + H2O(l) +1 correct reaction 0.0100 mol 0.00375 mol 0 mol |
+1 correct BCA B
C
‐0.00375 mol ‐0.00375 mol + 0.00375 mol |

+1 Kb  Ka ‐14
‐4
0.00625 mol 0 mol 0.00375 mol Ka = 1.010 /3.710 = 2.710‐11 A
+1 H‐H equation setup with base/acid (consistent)
+1 answer ‐14 / 3.710‐4) + log (0.00625 mol / 0.00375 mol) = 10.79 pH = log(1.010 5 (b) What is the pH at the equivalence point? 0.1000 L(0.100 M CH3NH2) = 0.0100 mol CH3NH2  0.0100 mol H3O+ = 0.250 M HNO3(V)  V = 0.0400 L For incorrect reactions, +3 maximum.
CH3NH2(aq) + H3O+(aq)  CH3NH3+(aq) + H2O(l) 0.0100 mol 0.0100 mol 0 mol | +1 correct reaction + After B
C
‐0.0100 mol ‐0.0100 mol + 0.0100 mol | +1 Vtot, +1 mol  M (consistent) +
NH
A
0 mol 0 mol 0.0100 mol |
[CH
3
3 ] = 0.0100 mol / 0.140 L

= 0.071429 M CH3NH3+ +1 correct reaction/ICE table +
+
+1 Ka setup with correct Ka CH
3NH3 (aq) + H2O(l) CH3NH2(aq) + H3O (aq) ‐11 = x2 / 0.071429 M ‐ x 0 M 0 M 2.70310
I
0.071429 M | ‐x | + x
+x

x = 1.38910‐6 +1 answer C x x pH = ‐log(x) = 5.86
0.071429 M – x | E
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(c) What major post‐equivalence point species are present? (List species present AFTER the strong reaction but BEFORE equilibrium is established.)
H2O
H3O+
CH3NH3+

+1/2 each
NO3
2

(d) In part (c), what effect allows you to ignore the weak acid in solution when calculating the pH? Explain. If “weak versus strong,” +1 description. Common‐ion effect: +1 Although CH3NH3+ is present, H3O+ is a common‐ion (product) in the reaction +2 common‐ion explained CH3NH3+(aq) + H2O(l) CH3NH2(aq) + H3O+(aq), so this reaction will be suppressed.
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2. (a) Describe how to make 1.00 L of 1:1 HCO3/CO32 buffer solution from solid Na2CO3 and 1.00 M HCl.
(Assume that the solid does not displace the volume of the solution.) Be specific with numerical amounts! +4 ratio of 2 (weak base): 1(strong