[L3ul]

I'm being asked to find the volume of a [itex]\ce{NH3}[/itex] solution ([itex]0.1\ \mathrm{M}[/itex]) to be added to [itex]10\ \mathrm{mL}[/itex] of a [itex]\ce{AgNO3}[/itex] solution ([itex]0.1\ \mathrm{M}[/itex]) to assure the 'complete formation of [itex]\ce{[Ag(NH3)2]+}[/itex]'  . Both constants of formation are given: [itex]\beta_1=10^{3.3}; \ \beta_2=10^{7.2}[/itex]

As far as I know, a complete complexation would mean that [itex][\ce{Ag+}]<10^{-5}[/itex]. I tried working it out with the common method of solving those kind of complex equilibria, but the equations I get seem pretty much impossible to solve because of the volume that I have to find, which gets incorporated in them.

Their solution is really weird, but much simpler; they invoke this mathematical condition, which makes the problem easily solvable:

[itex]\frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+]}}>10^2[/itex]

Starting from this, I can find [itex][\ce{NH3}][/itex] present in the solution, therefore the volume.

Where did it came from?

[Borek]

No idea what they did, all I can say is how I would try to approach this kind of a problem. I would start with the overall stability constant, easy to calculate from the stepwise constants. Then it should be much easier to solve.

Also, I don't like the condition of [Ag⁺] < 10^-5. If anything better condition would be one that says something about a fraction of the Ag⁺ that is not complexed. 10^-5 M Ag⁺ when the total is 10^-4 M is way worse as "complete" than 10^-3 M left out of 1 M.

[L3ul]

That would be:

[tex]\beta_2=\frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}}[/tex]
[tex]\frac{\beta_2}{\beta_1}=\frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}[/tex]
[tex]\ce{[Ag+]_0=[Ag+]}+\ce{[Ag(NH3)2+]}+\ce{[Ag(NH3)+]}[/tex]
[tex]\ce{[NH3]_0=[NH3]}+2\ce{[Ag(NH3)2+]}+\ce{[Ag(NH3)+]}[/tex]

[itex]\ce{[Ag+]_0}[/itex] and [itex]\ce{[NH3]_0}[/itex] can be expressed with respect to the volume of the [itex]\ce{NH3}[/itex] solution, but the equations I get are really complicated...
What approximations should I go for besides [itex]\ce{[Ag+]} << (\ce{[Ag(NH3)2+]}+\ce{[Ag(NH3)+]}) [/itex]?

[AWK]

Adding an excess of ammonia over the need of complex you do not know an exact concentration of complex. I would be solve the problem in a few iteration. At the beginning; enough ammonia needed from stoichiometry of complete complex formation and calculation starting concentration of complex.
1 iteration - calculation of ammonia needed for equilibrium from β₂. Recalculation of complex concentration (a new volume of ammonia added)
2 iteration -calculation a new ammonia ... as in 1.
End of iteration if ammonia volume changes less than eg 0.1 cm³ (your choose)

Without any calculations I expect volume of ammonia as 25-27 cm³ and 3-4 iterations.

[Borek]

Using overall stability constant is equivalent to ignoring presence of Ag(NH₃)⁺. It doesn't have to work, so once you finish the simplified calculations you should check if the assumption works.

[L3ul]

@AWK, someone proposed a spreadsheet to bruteforce the behavior of the solution if we were to add any volume of [itex]\ce{NH3}[/itex] in the interval 20mL-24.5mL.
The problem's that not a single volume (nor the answer the problem has, 24.3mL) gives us a clue that the reaction is 'complete'. Here's the discussion we had: http://chemistry.stackexchange.com/questions/48497/the-mathematical-condition-for-a-complete-complexation-reaction

@Borek, this gives us ~20mL, which doesn't coincide with their answer .

[AWK]

If you would like solve the problem in such precise manner the ammonia mass balance should also include NH₄⁺ from dissociation of ammonia.  My suggestion is completely sufficient for working without ionic strength.
Neither Borek, nor I do not know the criterion 10² comes from. Such value may come from β₂/β₁ at equilibrium ammonia concentration ~0.0125 M.
Iterative calculations give the final ammonia concentration less than 0.02 M.
So it may be a kind of tentative criterion, in my opinion rather useless in this crude calculations.

[L3ul]

It seems that with your method, after 8 iterations, [itex]\ce{[Ag(NH3)2+]}[/itex] reaches a constant value throughout the next iterations, therefore the complexation is complete.
The answer is: 24.25ml (24.3mL rounded to 3 sig figs).

Slight note: the problems specifies that the dissociation of [itex]\ce{NH3}[/itex] is negligible

It seems like your method is the best for now (and it's the most sound one), AWK, but it's almost impossible to use it in under 15mins. This problem was given in the contest for selecting the IChO team of my country and it shouldn't take that long to solve...

I'll try and search the author of the problem for an explanation; any information I find will be posted here.
Thank you

[AWK]

The first iteration gives 4.3 cm³ added + 20 = 24.3.

[L3ul]

Really? What am I doing wrongly, then?
I considered 20mL of [itex]\ce{NH3}[/itex], calculated [itex]\ce{[Ag+]}=5.01\cdot 10^{-4} \ \mathrm{M}[/itex], and from that the stoichiometric volume required to be added for the next iteration: [itex]V=\dfrac{\ce{[Ag+]}\cdot 30\cdot 2}{0.1}=\mathrm{0.3mL}[/itex].

I might've misunderstood your algorithm...

[AWK]

0.1 M NH₃ and 10 cm³ 0.1 M AgNO₃.
From stoichiometry of complex ==> 30 cm³ 0.0333 M Ag(NH₃)⁺
β₂=[Ag(NH₃)₂⁺]/([Ag⁺]·[NH₃]²)
After rearangement
[NH₃]=SQRT([Ag(NH₃)₂⁺]/[Ag⁺]·β₂)=SQRT(0.0333/10^7.2-5)=0.0145 M NH₃ in excess
c₁V₁=c₂V₂
V2=30·0.0145/0.1=4.35 cm³
V(NH₃)=20+4.35=24.35; V(complex)=34.35 c(complex)=0.02911
next iteration ==> I expect it will be sufficient