Cliffs AP Chemistry, 3rd Edition

Section II (Free-Response Questions)

Scoring Guidelines

One point deduction for mathematical error (maximum once per question)

One point deduction for error in significant figures* (maximum once per question)

*number of significant figures must be correct within +/– one digit

Part A: Question 1

1. Given: () () () ()
   .

C H NH aq H O C H NH aq OH aq

K for C H NHb 5 6 10

25 2 2 25 3

23 2

4

D

#

++,

=

+ -

(a) Given: 65.987 mL of 0.250 M C 2 H 5 NH 2

Restatement: Find [OH–]

Step 1: Rewrite the balanced equation for the ionization of ethylamine.

C H NH 25 2 ++H O 2 DC N NH 25 3 + OH-

Step 2:Write the expression for the base-dissociation constant.

K CHNH.

C H NH OH

b 56 10

2

3

25

(^254)
==#

• 6

7 7

@

A A

Step 3: Create a chart showing initial and final concentrations (at equilibrium) of the in-

volved species. Let xbe the amount of C 2 H 5 NH 3 +that forms from C 2 H 5 NH 2. Because

C 2 H 5 NH 3 +is in a 1:1 molar ratio with OH–, [OH–] also equals x.

Species Initial Concentration Final Concentration (at equilibrium)

C 2 H 5 NH 2 0.250 M 0.250 −x

C 2 H 5 NH 3 + 0 M x

OH– 0 M x

Step 4: Substitute the equilibrium concentrations from the chart into the equilibrium ex-

pression and solve for x.

..
()( )
K CHNH

C H NH OH

x

xx

b 2 56 10 0 250

3

25

(^254)
===# -

• 
  
  
  
  • 7 A 7 A -
    Answers and Explanations for the Practice Test