[dgt20 (OP)]

What is the pH of a 0.15 M solution of weak acid ammonium bromide? The Kb value for ammonia is 1.8 × 10^–5

I keep getting:

x=√(1.8*10^-5)(0.15)
x= 5.1*10^-3
-log(5.1*10^-3)
=pOH = 2.79
14-2.79
pH=11.71

[chiralSPO]

Ammonium bromide is an acid, so solutions of it should have a pH less than 7.

The K_b value provided refers to the equilibrium:

NH₃ + H₂O ↔ NH₄⁺ + OH^–

K_b = [NH₄⁺]*[OH^–]/[NH₃]

However, because you are adding NH₄⁺, you need to figure out how much of it will dissociate back to neutral NH₃ (protonating OH^–), which means you also need to take into account K_w to find the K_a of NH₄⁺

K_b*K_a = K_w

so:

K_a = K_w/K_b
= [H⁺]*[OH^–]*[NH₃]/([NH₄⁺]*[OH^–])

the [OH^–] terms cancel, leaving:

K_a = [H⁺]*[NH₃]/[NH₄⁺]

and because we know the numerical values of  K_w and K_b, we can calculate the numerical value of  K_a, and then solve the expression below for [H⁺]:

K_a = [H⁺]*[NH₃]/[NH₄⁺]
=(x)*(x)/(0.15–x)

and then pH is –log₁₀(x)

good luck!

[Constantinut]

Thanks for the way it is calculated, it is very beneficial to me. It helped me a lot.

[aricjoshua]

Thanks for the clear explanation of this chemistry exercise. However, because you're adding NH4+, you'll need to calculate how much of it will dissociate back to neutral NH3 (protonating OH–), which means you'll need to include Kw when calculating NH4+ Ka.