how do you find the pH of 0.020 ammonium chloride?

Question

okay it dissociates by NH4CL ---> NH4+ +CL-

so the concentration of NH4+ is .20M so do I just take the -log of .20 to get the pH or am I doing it wrong?

Answer 1

Let me think this through:

You can ignore the chloride.

Kb of NH3 is 1.8*10^-5

Ka = Kw/Kb = 5.6*10^-10

NH3 + H+ < --------> NH4+

Ka = [NH3][H+]/[NH4+] = x^2/0.020, solve for x, x = 3.35*10^-6

therefore [H+] = 3.35*10^-6 moles, the negative log (pH) is 5.47
round pH to 5.5

please check for errors.

Answer 2

Perhaps you mistyped the concentration: 0.02 M or 0.2 M is the right data? Anyway, the NH4+ is a weak acid with Ka = 5*10^(-10) and reacts with water:

NH4+ + H2O <==> NH3 + H3O+

So if in the solution [NH3] = [H3O+] = x then [NH4+] = 0.2 - x = 0.2 approx.

Now, Ka = [NH3]*[H3O+]/[NH4+] so 5*10^(-10) = x^2/0.2

x = [H3O+] = 10^(-5), pH = 5

(If 0.02 is the right concentration then pH = 5.5)