[Juliana]

How would I calculate the pH of a 0.50 M solution of NH3.  The Kb of NH3 is 1.8x10^-5

Thanks

[Borek]

Please read forum rules.

Ammonia is a weak base. Start with the reaction equation, set up ICE table, calculate pOH, convert to pH.

[Juliana]

Please read forum rules.

Ammonia is a weak base. Start with the reaction equation, set up ICE table, calculate pOH, convert to pH.

OK I read them sorry.

Here is what I did:

Kb = 1.8 x 10-5 = [NH4+][OH-]/[NH3] = x2/0.50

Then

(1.8 x 10-5)(0.50) = x2

So using quadratics I got 3.0 x 10^-3

pH = -log 3.0 x 10^-3

OR

pH = log x 1/3.0 x 10^-3

pH = 2.52

This is how I was taught how to do these problems...however someone else said that 2.52 = the pOH and that I need to find that first and then sub in the formula pOH + pH = 14

However my notes say that pH = -log (value)

Thanks

[Astrokel]

No!, the value u obtained is a actually the concentration of OH-, right?

so in this case, when u take -lg(value), you are actually calculating pOH. Yes then you can use pH + pOH = 14

your note is correct or probably unclear, pH = -log (value)

See the pH, the H tells you the value you use should be [H+] so if you use [OH-] it doesn't work.

kelvin

[Borek]

Kb = 1.8 x 10-5 = [NH4+][OH-]/[NH3] = x2/0.50

That's not correct (although in this case error is luckilly small enough it can be neglected). You are forgetting [NH₃] is consumed in the reaction, so its final concentration can be lower than 0.50.

However my notes say that pH = -log (value)

pH = -log([H⁺])
pOH = -log([OH^-])

[Astrokel]

Kb = 1.8 x 10-5 = [NH4+][OH-]/[NH3] = x2/0.50

As Borek stated, it would be good if you state your assumptions before your calculations.