Cliffs AP Chemistry, 3rd Edition

(singke) #1
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



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