Buffered solutions...?

Question

A buffered solution is made by adding 50.0g NH4Cl to 1.00L of a 0.75M solution of NH3. Calculate the pH of the final solution (assume no volume change)

what do i do?!?!
thanks!

Answer 1

Run around and scream.

Honestly, you think thru what's going on.
NH4+ + H2O -> [NH4OH] + H+
NH4OH is hydrated ammonia= NH3
The basic dissociation of ammonia is:
[NH4+][OH-]/[NH3] = Kb.
We can use this and then convert our results to [H+] and then pH. The mol wt of NH4Cl is 54, so the solution is (50/54) or appx 0.9 M in [NH4]+. The [NH3] molality is 0.75 M. So we substitute:
0.9 [OH-]/0.75 = 1.8x10-5
[OH-] = 1.5 x10-5 M appx
Since [H+][OH-] = 10x10-15 in aqueous solution, the [H+] = 6.7x 10-10 M
To find pH, pH= - log [H+] = 10 - log 6.7 = 9.2 appx

Answer 2

We can calculate moles of NH4Cl:

n= m/MM

so n= 50/53.45=0.93 moles.

NH3 has 0.75 moles per L

so Ka= [H+]*[NH3]/[NH4+]
ka=5.8*10^-10
and[H+]=Ka*[NH4+]/NH3]

so we can calculate [H+]=5.8*10^-10*0.93/0.75

and pH=9.14