Homework Answers
a)0.260 M alanine hydrochloride
H2A ---------------> H+ + HA-
0.260 0 0
0.260 - x x x
Ka1 = x^2 / 0.260 - x
4.53 x 10^-3 = x^2 / 0.260 – x
1.18 x 10^-3 -4.53 x 10^-3 x = x^2
Solve as quadratic equation.
x = 0.032160 M
[H+] = 0.032160 M
pH= -log [H+] = -log 0.032160
pH = 1.50
B)
HA -------------> H+ + A-
0.260 0 0
0.260- x x x
Ka2 = x^2 / 0.260- x
1.36 x 10^-10 = x^2 / 0.260 - x
x = 5.95 x 10^-6
[H+] = 5.95 x 10^-6M
pH = -log[H+] = -log (5.95 x 10^-6)
pH = 5.23
c)
A- + H2O ------------> HA + OH-
0.260 0 0
0.260 - x x x
Kb1 = x^2 / 0.260 - x
10^-14 / 1.36 x 10^-10 = x^2 / 0.260 - x
7.35 x 10^-5 = x^2 / 0.260 - x
x = 4.37 x 10^-3
[OH-] = 4.37 x 10^-3 M
pOH= -log [OH-] = -log (4.37 x 10^-3 )
= 2.36
We know that,
pH + pOH = 14
then pOH = 14- pH
=1 4- 2.36
pH = 11.64
Add Answer to:
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