Question

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

Answer 1

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