Question

Use the information provided in appendix to answer the following questions:

a) Calculate the K_b of Acetate ion

b) Calculate the pK_a of the anilinium ion (C₆H₅NH₃⁺)

c) Calculate the K_b of NO₂^-

d) Calculate the pK_b of the benzoate ion

Answer 1

First thing first: any aqueous solution that contains acetate ions WILL contain some CH3COOH, thanks to the following reversible reaction:

CH3COO- + H2O <=> CH3COOH + OH-

But to answer your question, it depends on the concentration of the acetate ion.

For any conjugate acid-base pair, Ka x Kb = Kw

Kw = 1.0e-14

So if we know Ka for acetic acid (1.8e-5), then we can easily determine the value of Kb for the acetate ion:

(1.8e-5) x Kb = 1.0e-14

Kb = 5.6e-10

2.

The anilinium ion (C6H5NH3+) is the conjugate acid of the weak base, aniline. We need to calculate Ka for this weak acid.

Ka = Kw / Kb = (1 x 10^-14) / (4.2 x 10^-10) = 2.4 x 10^-5

When anilinium ion hydrolyzes in water, it donates a proton to water to form H3O+ and C6H5NH2.
Set up an ICE chart.

Molarity . . . . . .C6H5NH3+ + H2O <==> H3O+ + C6H5NH2
Initial .. .. . . . . . . . .0.25 . . . . . . . . . . . .. . . .0 . . . . . . . . 0
Change . . . . . . . . . .-x . . . . . . . . . . . . . . . . x . . . . . . . . x
At Equil. . . . . . . . .0.25 - x . . . . . . . . . . . . . .x . . . . . . . .x

Ka = [H3O+][C6H5NH2] / [C6H5NH3+] = (x)(x) / (0.25 - x) = 2.4 x 10^-5

The value of x is going to be small compared to 0.25, so I drop it from that term to make the problem much easier. (How do you know if x is negligible? If M / Ka is > 100, x is negligible. In our case,
0.25 / 2.4 x 10^-5 = 10,000).

x^2 / 0.25 = 2.4 x 10^-5
x^2 = 6.0 x 10^-6
x = 2.4 x 10^-3 = [H3O+]

3.

The product of the acid-dissociation constant for an acid and the base-dissociation constant for its conjugate base equals the ion-product constant for water: Ka x Kb = Kw

The ion-product constant for water is 1.0 x 10^-14.

So, rearranging the equation, Kb = Kw/Ka = (1.0 x 10^-14)/(4.5 x 10^-4) =

2.2 x 10^-11.

4.

You need:
pKa propionic acid = 4.87
Molarity of CH3CH2COOH = 0.085M
Molarity of CH3CH2COOK = 0.06M
Equation:
pH = pKa + log ([salt]/[acid] )
pH = 4.87 + log (0.06/0.085)
pH = 4.87 + log 0.706
pH = 4.87 + (-0.15)
pH = 4.72

4.
The Kb value for (CH3)3N is 6.5 x 10-5.
pKb = -log (6.5*10^-5) = 4.19