[pumpkin]

This is the question:
How many millilitres of 0.200 M HCl must be added to 50.0 mL of a 0.200 M
solution of NH3 in order to prepare a buffer that has a pH of 8.60? 
Kb of NH3 = 1.8 x 10^-5

We haven't done any examples like this in class, and I couldn't find anything similar in the textbook I own.
This is how far I've gotten:

moles NH3=0.05L*(0.200mol/1L)=0.01 moles NH3
Kb(NH3)=1.8 x 10^-5   Therefore pKb=-log(1.8 x 10^-5)=4.74
So to have a pH of 8.60, the pOH must be 14-8.6= 5.40

--> pOH=pKb+log([salt]/[base])
Therefore the log([salt]/[base])=5.40-4.74=0.66  -> ([salt]/[base])=10^0.66 = 4.571
So i must have 4.571 parts salt per 1 part base, correct?

I'm not sure where to go from here. I've been working on this for the past hour or two and just can seem to figure it out. Thanks for your help,
-pumpkin

[UG]

You have got the number of moles of base as 0.01 mol, when you add HCl to the solution, the amount of base is going to decrease down to say 0.01 - x while the increase in the salt NH₄⁺ is going to be x mol. Substitute this back into the original equation to find the amount of salt, this is going to be the same as the amount of HCl you have added, finding the volume of HCl then should be easy.

[pumpkin]

I don't really understand what to do based on your explanation

[UG]

Ok, let me try again...
So when you added HCl to the ammonia solution, the reaction that occurs is HCl + NH₃  NH₄⁺ + Cl^-
You have used the equation pOH=pKb+log([salt]/[base]) to find that ([salt]/[base]) = 4.571
The equation here is working in concentration units of base and salt but working with mol is also ok.
The initial amount of NH₃ calculated = 0.01 mol, this is going to decrease when you add the acid correct? The addition of 'x' moles of HCl causes 'x' moles of the salt to form, therefore the amount of NH₃ remaining is going to be (0.01 - x) mol
So then, you substitute this back into the equation to get ([salt]/[base]) = 4.571 = (x)/(0.01 - x)
You can then figure out 'x' which is the amount (in mol) of salt formed but is also the amount of HCl added (we are assuming the reaction of HCl with ammonia goes to completion). The volume of HCl (in litres) then is just mol divided by the molarity.

[Borek]

In short: assume when you add acid you stoichiometrically convert NH₃ to NH₄⁺. Calculate how much acid you have to add to reach the NH₄⁺/NH₃ ratio of 4.751.

[pumpkin]

ok, so if 4.571= (x/(0.01-x)), then x=0.082 moles HCl added.
So Volume(HCl)=(moles)/(Molarity)=0.082mol/0.200M=0.41 litres=410mL HCl.

Is this correct? It seems like A LOT of HCl to add when I only have 50mL of NH₃
If it is correct, why does it take so much HCl?

[UG]

HCl is 0.0082 mol rather

[pumpkin]

ahh I see. thanks for all of the help. This forum is awesome

[pumpkin]

The initial amount of NH₃ calculated = 0.01 mol, this is going to decrease when you add the acid correct? The addition of 'x' moles of HCl causes 'x' moles of the salt to form, therefore the amount of NH₃ remaining is going to be (0.01 - x) mol

Another question actually: What would the ICE table for this particular part look like for this?
I tried to figure it out but couldn't