You have a choice in solving for x. The first method would require the quadratic equa-

tion—not a good idea because compared to the magnitude of 0.250, the value of xis negli-

gible. If you used the quadratic, you would be wasting time. The second method would

assume that [C 2 H 5 NH 2 ] remains constant at 0.250 M; 5.6 × 10 –4= x^2 / 0.250.

x = [OH–] = 0.012 M

By the way, the 65.987 mL is not needed because concentration is independent of the

amount of solution measured.

(b) Restatement: Find pOH of solution.

pOH = −log [OH–]

pOH = −log [0.012] = 1.92

(c) Restatement: Find % ionization of ethylamine

%%whole .. %.%

part

===## (^100) 0 2500 012 100 4 8
(d) Given: 15.000 g C 2 H 5 NH 3 Br + 250.00 mL 0.100 M C 2 H 5 NH 2
Restatement: Find pH of solution.
Step 1:Note that when C 2 H 5 NH 3 Br dissolves in water, it dissociates into C 2 H 5 NH 3 +
and Br–. Furthermore, C 2 H 5 NH 3 +is a weak acid.
Step 2:Rewrite the balanced equation at equilibrium for the reaction.
C H NH 25 3 DC H NH 25 2 +H

• 
  
  
  
  • Step 3: Write the equilibrium expression.
    //.
    .
    KKK
    CHNH
    C H NH H
    10 5 6 10
    18 10
    awb
    25 3
    (^252144)
    11
    
    
    
    

    == =

    
    
  
  
  
  


• 
  

  --

  
  

7 +

(^67)
A
@ A
Step 4:Calculate the initial concentrations of the species of interest.
[C 2 H 5 NH 2 ] = 0.100 M (given)
.
.

..

CHNH g

gC H NH

CHNH

mole C H NH

liter M

1 126 05

15 000 1

0 250

(^1) 0 476
Br
(^3) Br
3
3
25 25 3
25
#^25

=

• 
  
  
  
  • 7 A
    [H+] = 0
    Step 5:
    Species Initial Concentration Final Concentration (at equilibrium)
    C 2 H 5 NH 2 0.100 M 0.100 + x
    C 2 H 5 NH 3 + 0.476 M 0.476 −x
    H+ 0 M x
    Part IV: AP Chemistry Practice Test