Question

Question #1)

a) What is the pH of a 0.273 M aqueous solution of potassium hypochlorite, KClO at 25 °C? (K_a for HClO = 3.5×10^-8)

b) The value of K_a for acetic acid is 1.80×10^-5. What is the value of K_b, for its conjugate base, CH₃COO^-?

c) The hydronium ion concentration of an aqueous solution of 0.456 M ammonia is ...

[H₃O⁺] = ____ M.

d) In the laboratory, a general chemistry student measured the pH of a 0.456 M aqueous solution of ammonia to be 11.439.

Use the information she obtained to determine the K_b for this base.

K_b(experiment) = _____________

Answer 1

a)

Ka HClO = 3.5 x 10^-8

we know that kw = ka * kb (kw is constant of water 1 x 10^-14)

Kb = Kw / Ka = 1x 10^-14 / 3.5 x 10^-8 = 2.86 x 10^-7

OCl + H₂O === HClO + OH^-

write the Kb expression (ratio of products to reactants):

Kb = [OH] [HClO] / [OCl]

2.86 x 10^-7 = x² / 0.273

x² = 2.86 x 10^-7 * 0.273 = 7.807 x 10^-8

x = 0.00027941 M, this is the concentration of OH

calculate POH with -log OH

POH = - log ( 0.00027941) = 3.55

remember that PH + POH = 14

PH = 14 - 3.55 = 10.446

b)

ka of acetic acid is 1.8 x 10^-5

kw = ka * kb = 1 x 10^-14

kb = kw / ka = 1 x 10^-14 /  1.8 x 10^-5 = 5.56 x 10^-10