Question

An aqueous ammonium bromide solution has a pH of 5.56. Calculate the equilibrium concentration of ammonium ion in the solution:

NH₄⁺(aq) + H₂O(ℓ) ⇌ NH₃(aq) + H₃O⁺(aq)

K_a of ammonium bromide = 5.5 × 10^-10

[NH₄⁺] = _____ M

Answer 1

we know that

pH = -log [H30+]

so

5.56 = -log [H30+]

[H30+] = 2.75423 x 10-6

consider the given equation

NH4+ + H20 ---> NH3 + H30+

Ka = [NH3] [H30+] / [NH4+]

Let the initial concentration of [NH4+] be C

then

at equilibrium

[NH3] = y

[H30+] = y

[NH4+] = C - y

so

[NH3] = [H30+]= y = 2.75423 x 10-6

so

Ka = [y][y]/ [NH4+]

5.5 x 10-10 = y2 / [NH4+]

[NH4+] = y2 / 5.6 x 10-10

[NH4+] = ( 2.75423 x 10-6)^2 / 5.6 x 10-10

[NH4+] = 0.013546

so

at equilibrium , [NH4+] = 0.013546 M