[THC]

20 mL 1 M NaOH is mixed with 10 mL 4 M NH₃.
Calculate pH for this solution.


My suggestion:
OH^- + NH₃  + H₂O <-> 2 OH^- + NH₄⁺

So I find [OH^-] from the NH₃ which is 10^(-(½*(pKb - logc))). Then I find n(OH-), divide it by the total volume and use pOH = -log[OH-] and pH =14-pOH.

Is this correct? Or do I have to take the buffer solution into account as well?

[Borek]

No buffer here - there is ammonia, but no NH₄⁺. Start with fully dissociated NaOH - will ammonia change anything?

[THC]

No buffer here - there is ammonia, but no NH₄⁺. Start with fully dissociated NaOH - will ammonia change anything?

Why is that? Ammonia is a weak base, so
 NH3 + H₂O <--> NH₄⁺ + OH^-
But does the NaOH disrupt the equilibrium? Does this have anything to do with [OH^-]*[H₃O⁺] = 1*10^-14?

Sorry, but I'm not sure I understand. Could you elaborate a bit?
Thanks.

[Borek]

Well, there are traces of NH₄⁺ - which can be neglected. Let's take a closer look at the dissociation constant definition:

K_b = [NH₄⁺][OH^-]/[NH₃]

it can be rearranged to

[NH₄⁺]/[NH₃] = 10^pOH-pK_b

What for? Well, let's assume OH^- comes from NaOH only. We know pK_b to be 9.25. Now you should be able to calculate NH₄⁺ concentration at this pOH - if it is small enough when compared to NaOH concentration, OH_- from ammonia can be neglected.

If not - that's completely different problem. But we will deal with it ONLY if it'll be necessary

[THC]

Ok, so pH = 14 + log[0,02], right?