Question

What is the pH of a 0.608 M solution of ammonium bromide? The Kb value for ammonia is 1.8 x 10–5.

Answer 1

pH of the 0.608 M solution of ammonium bromide = 4.74

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NH4Br exists in solution as NH4+ and Br- ions,

In aqueous solutions the NH4+ ions will exists in the following equilibrium,

NH4+ + H2O <--> NH3 + H3O+

Ka = [NH3][H3O+]/[NH4+]

Kb of NH3 is given as 1.8 x 10^-5

Since Ka x Kb = 1 x 10^-14

Ka = 1 x 10^-14 / 1.8 x 10^-5 = 5.556 x 10^-10

Concentration of NH4Br = 0.608 M

i.e. [NH4+] = 0.608 M, during dissociation [NH3] = [H3O+]

If y moles of H3O+ is formed the [NH4+] = [NH4+] -y

So, Ka = y x y /(0.608 - y)

Since y is very small quantity, Ka can be rewritten as Ka = y^2/0.608 = 5.556 x 10^-10

y^2 = 5.556 x 10^-10 x 0.608 = 3.378 x 10^-10

y = sqrt [3.378 x 10^-10 ] = 1.838 x 10^-5 M

y = [H3O+] = 1.838 x 10^-5 M

pH = -log [H3O+] = - log (1.838 x 10^-5) = 4.7356