Question

1. What is the pH of 0.025M aqueous solution of sodium propionate, NaC₃H₅O₂? what is the concentraion of propionic acid in the solution?

2. Obtain

a) the K_b value for NO₂^-

b) the K_a value for C₅H₅NH⁺

Answer 1

1.

Sodium propionate NaC₃H₅O₂ = C₂H₅COONa

[C₂H₅COONa] = 0.025 M

C₂H₅COONa     +     H2O     <----->     C₂H₅COOH     +     OH^-

Ka of propanoic acid (C2H5COOH) = 1.34 x 10^-5

Kb = Kw / Ka
     = (1.0 x 10^-14) / (1.34 x 10^-5)
    = 7.46 x 10^-10

Now,

Kb = [C₂H₅COOH] [OH^-] / [C₂H₅COONa]

or, 7.46 x 10^-10 = (x) (x) / (0.025 - x)

or, 7.46 x 10^-10 = x² / (0.025)            ; (Kb is very small)

x² = (7.46 x 10^-10) (0.025)

x² = 1.87 x 10^-11

x = 4.32 x 10^-6

So, x = 4.32 x 10^-6 M = [C₂H₅COOH] = [OH^-]

2.

(a)

NO₂^-     +     H₂O     <----->     HNO₂     +     OH^-

Nitrous acid (HNO2) is a weak acid with Ka= 4.5 x 10^-4.

Kb = Kw / Ka
     = (1.0 x 10^-14) / (4.5 x 10^-4)
     = 2.22 x 10^-11

(b)

C₅H₅NH⁺     +     H₂O     <----->     H₃O⁺     +     C₅H₅N

Pyridine (C5H5N) is a weak base with Kb = 1.7 x 10^-9.

Ka = Kw / Kb
     = (1.0 x 10^-14) / (1.7 x 10^-9)
     = 5.88 x 10^-6