Question

Determine the degree of ionization and ph of 0.05 m ammonia solution give that kb for the ammonia is 1.77 and 225

Answer 1

Answer:

Explanation:

Ka × Kb = Kw

Ka × 1.77×10^-5 = 10^-14

Ka = 10^-14 / (1.77×10^-5)

Ka = 5.64×10^-10

Degree of ionization is calculated by-

α = √(Ka / C)

α = √(5.64×10^-10 / 0.05)

α = 1.044×10^-4

For the given base -

pOH = -log(α)

pOH = -log(1.044×10^-4)

pOH = 3.981

pH is calculated by-

pH = 14 - pOH

pH = 14 - 3.981

pH = 10.01

Hope it helps..

Answer 2

Answer:

poh = 3.981

pH = 10.01

Explanation:

hope it help uh⛅✔