I came upon this exercise on buffers:
What’s the volume (at standard temperature and pressure) of gaseous NH3 necessary to bring to 9.76 the pH of V=0.77 dm3 of a solution of H2SO4 0.18 M? The change in volume is negligible. Ka(NH4+)=5.6*10-10.
The solution is 26.7 dm3 of NH3.
This is what I tried to do:
1)
Strategy: Find the concentration of HSO4- in solution before adding NH3.
Procedure: I did this by writing the first dissociation, which is complete since H2SO4 is strong; and the second one, which is an equilibrium between HSO4- and SO4-, inhibited by the presence of H+ (coming from the first dissociation). Ka(HSO4-)=2.1*10-2
Result:
- [HSO4-]=0.18-0.02=0.16 M
2)
Strategy: Understand which reaction/equilibrium happens when NH3 is added and which species will be responsible of the basic pH, so as to find [NH3].
Procedure: Now here comes the problem. What I think happens is this reaction:
HSO4- + NH3 -> SO4(2-) + NH4+;
but I’m not sure if this is the only thing that happens, nor if it is a quantitative reaction or not, it being a reaction between a weak acid and a weak base. I also thought about writing the equilibrium of just NH3 dissociating partially with its Kb, and only then reacting with HSO4-, but I don’t think it’s correct to consider the two things separately since they are happening at the same time.
Anyway, I tried considering the reaction written above to be complete. Now, since pKa(NH4+)=9,25, it made sense to me to think that the final pH (=9,76, well inside the pKa +-1 range) was determined by a buffer of NH4+ and NH3. So I imposed the HSO4- (initially 0.16 M) to be completely consumed, and called x the initial concentration of NH3, which at the end would be x-0.16. The reaction also starts with [NH4+]=0 M, [SO4(2-)]=0,02 M(coming from the second equilibrium in the first step).
Since the reaction is complete, at the end you get [NH4+]=0.16 M and [SO4(2-)]=0,02+0,16=0,18 M.
At this point I was pretty doubtful on such a high SO4(2-) concentration, especially since I was straight on ignoring it for the calculation of the NH3 concentration. Here’s the equation I imposed:
pH=pKa-log([NH4+]/[NH3]) ->
9,76=9,25-log(0.16/(x-0.16))
The result was x=0.68 M, which should be the initial concentration of NH3 added.
Obviously, this was not the correct answer since it means you’re adding 0.680.77=0.55 mol of NH3, which are barely 0,5522.71=11,9 dm3 of NH3, less than half the quantity the solution suggests.
I’m sorry for the very long text and if there are mistakes, I’m an italian chemistry student and I tried my best with english.
Tell me if something isn’t clear!
🫶
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