A. If the Kb of a weak acid is 5.9 x 10^-6 what is the ph of a 0.24 M solution of this base?
ph=
B. A certain weak base has a Kb of 7.70x10^-7. What concentration of this base will produce a ph of 10.03?
answer in M
A)
This is a base in water so, let the base be "B" and + = HB+ the protonated base "HB+"
there are free OH- ions so, expect a basic pH
B + H2O <-> HB+ + OH-
The equilibrium Kb:
Kb = [HB+][OH-]/[B]
initially:
[HB+] = 0
[OH-] = 0
[B] = M
the change
[HB+] = x
[OH-] = x
[B] = - x
in equilibrium
[HB+] = 0 + x
[OH-] = 0 + x
[B] = M - x
Now substitute in Kb
Kb = [HB+][OH-]/[B]
Kb = x*x/(M-x)
x^2 + Kbx - M*Kb = 0
Kb = Kw/Ka = (10^-14)/(5.9*10^-6) = 1.6949*10^-9
x^2 + ( 1.6949*10^-9 )x - (0.24)( 1.6949*10^-9 ) = 0
solve for x
x = 2.0*10^-5
substitute:
[HB+] = 0 + x = 2.0*10^-5M
[OH-] = 0 + x = 2.0*10^-5 M
pH = 14 + pOH = 14 + log(2.0*10^-5) = 9.30
B)
for
Kb = [BH+][OH-]/[B]
pH = 10.03
pOH = 14-10.03 = 3.97
[OH-] = 10^-pOH = 10^-3.97
then, HB+ = OH
7.7*10^-7 = (10^-3.97)(10^-3.97) / (M-10^-3.97)
M = (10^-3.97)^2 /(7.7*10^-7 ) + 10^-3.97
M = 0.01501 mol per liter
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