General Chemistry Principles, Patterns, and Applications, 2011

B To calculate the pH, we need to determine the H + concentration. Unfortunately, H + does not appear in
either the chemical equation or the equilibrium constant expression. However, [H + ] and [OH − ] in an
aqueous solution are related by K w = [H + ][OH − ]. Hence if we can determine [OH − ], we can calculate [H + ] and
then the pH. The initial concentration of CH 3CH 2NH 2 is 0.225 M, and the initial [OH − ] is 1.00 × 10 −7 M.
Because ethylamine is a weak base, the extent of the reaction will be small, and it makes sense to
let x equal the amount of CH 3CH 2NH 2 that reacts with water. The change in [CH 3CH 2NH 2] is therefore −x,
and the change in both [CH 3CH 2NH 3+ ] and [OH − ] is +x. To see whether the autoionization of water can safely
be ignored, we substitute K b and C B into Equation 16.46:
K b C B = (6.5 × 10 −4 )(0.225) = 1.5 × 10 −4 > 1.0 × 10 −6
Thus the simplifying assumption is valid, and we will not include [OH − ] due to the autoionization of water
in our calculations.
H2O 1
( ) + CH 3CH 2NH 2( aq) CH 3CH 2NH 3+ ( aq) +OH - ( aq)
[CH3CH2NH2] [CH3CH2NH3 + ] [OH − ]
initial 0.225 0 1.00 × 10 −7
change −x +x +x
final (0.225 − x) x x
Substituting the quantities from the last line of the table into the equilibrium constant expression,
Kb = [CH 3CH 2NH 3+][OH-][CH 3CH 2NH 2] = (x)(x)0.225 - x = 6.5 ´10 - 4
As before, we assume the amount of CH 3CH 2NH 2 that ionizes is small compared with the initial
concentration, so [CH 3CH 2NH 2] f = 0.225 − x ≈ 0.225. With this assumption, we can simplify the equilibrium
equation and solve for x:
Kb = x20.225x = 6.5 ´10 - 4 = 0.012 = [CH 3CH 2NH 3+]f = [OH-]f
The percent ionization is therefore
percent ionization = [OH-]CB ´100 = 0.012 M0.225 M ´100 = 5.4%
which is at the upper limit of the approximately 5% range that can be ignored. The final hydroxide
concentration is thus 0.012 M.
C We can now determine the [H + ] using the expression for K w:
Kw1.01´10 -148.4 ´10 -13 M = [OH-][H+] = (0.012 M)[H + =
] [ H+]
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