The mathematical condition for a complete complexation reaction

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


\begin{align\*}
\beta\_1 &= \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}} =10^{3.3} \\
\beta\_2 &= \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}} =10^{7.2}
\end{align\*}


As far as I know, a complete complexation would mean that $[\ce{Ag+}]<10^{-5}$. 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.


\begin{align\*}
\dfrac{\beta\_2}{\beta\_1} &= \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}} \\
\ce{[Ag+]\_0} &= \ce{[Ag+]} + \ce{[Ag(NH3)2+]} + \ce{[Ag(NH3)+]} \\
\ce{[NH3]\_0} &= \ce{[NH3]} +2\ce{[Ag(NH3)2+]} + \ce{[Ag(NH3)+]}
\end{align\*}


$\ce{[Ag+]\_0}$ and $\ce{[NH3]\_0}$ can be expressed with respect to the volume of the $\ce{NH3}$ solution, but the equations I get are really complicated...


What approximations should I go for besides $\ce{[Ag+]} << (\ce{[Ag(NH3)2+]}+\ce{[Ag(NH3)+]})$?




---


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


$$\dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+]}}>10^2$$


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


Where did it come from?




**Ok, to start let's just check the given answer of 24.3 mL.** 


The total solution is 34.3 ml (10 mL Ag + 24.3 ml NH3).


We'll assume that all 0.001 moles of silver are as species $\ce{[Ag(NH3)2^+]}$ so:


$\ce{[Ag(NH3)2^+]} \approx (\dfrac{10}{34.3})0.100 = 0.0291 $ 


$\ce{[NH3]\_{initial} =}(\dfrac{24.3}{34.3})0.100 = 0.0708$


We'll assume $\ce{NH4^+}$ and $\ce{[Ag(NH3)^+]}$ are both negligible so:


$\ce{[NH3] =} 0.0708 - 2(0.0291) = 0.0126$


Now from the equation 


$\dfrac{\beta\_2}{\beta\_1}= 10^{3.9} = 7.94 \times 10^3 = \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$


$\ce{[Ag(NH3)^+] =} \dfrac{[\ce{Ag(NH3)2^+}]}{(7.94 \times 10^3)\ce{[NH3]}} = \dfrac{0.0291}{(7.94 \times 10^3)(0.0126)} = 2.91 \times 10^{-4}$


$\ce{[Ag^+] =} \dfrac{[\ce{Ag(NH3)^+}]}{(2.00E3)\ce{[NH3]}} = \dfrac{2.91 \times 10^{-4}}{(2.00E3)(0.0126)} = 1.15 \times 10^{-5}$


fraction of total Ag as $\ce{Ag(NH3)^+}$


$\dfrac{2.91 \times 10^{-4}}{0.0291} = 0.01 $


fraction of total Ag as $\ce{Ag^+}$


$\dfrac{1.15 \times 10^{-5}}{0.0291} = 3.97 \times 10^{-4} $


Limits on ratios
----------------


Given the two equations:


$\mathrm{K}\_1= 10^{3.3} = \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}}$ 


$\mathrm{K}\_2 = 10^{7.2} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}}$ 


then we can derive a third


$\mathrm{K}\_3 = 10^{7.2 - 3.3} = 10^{3.9} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$


It is possible to solve for some lower limits on the ratios assuming that the $\ce{[NH3]}$ can at most be 0.1 M. So:


$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)+]}} \geqslant \frac{1}{\mathrm{K}\_1\ce{[NH3]}} = 5.01 \times 10^{-3}$


$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}\_2\ce{[NH3]^2}} = 6.31 \times 10^{-6}$


$\frac{\ce{[Ag(NH3)+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}\_3\ce{[NH3]}} = 1.26 \times 10^{-3}$


So the absolute upper limit on the % of the silver as $\ce{[Ag(NH3)2^+]}$ is 99.874%. 


Quick and dirty calculations...
-------------------------------


To check this I brute forced some calcs in a spreadsheet. 


The calculations below ignores the reactions below (alkali solutions do not produce appreciable amounts of silver hydroxide due to the favorable energetics for the reaction to the oxide): 
$$ \ce{NH3 + H2O <-> NH4^+ + OH^{-}}$$
$$\ce{Ag^+ + OH^{-} <-> AgOH(s)}$$
$$\ce{AgOH(s) -> Ag2O(s) + H2O} \text{ (pK = 2.875)}$$



```
 NH3    [Ag(NH3)2+] [NH3]tot    [NH3]       [Ag(NH3)+]  [Ag+]       mol(Ag)     F*
 20.1   3.32E-02    6.68E-02    3.32E-04    1.26E-02    1.90E-02    1.950E-03   4.87E-01
 20.2   3.31E-02    6.69E-02    6.62E-04    6.30E-03    4.75E-03    1.334E-03   2.50E-01
 20.3   3.30E-02    6.70E-02    9.90E-04    4.20E-03    2.12E-03    1.191E-03   1.61E-01
 20.4   3.29E-02    6.71E-02    1.32E-03    3.15E-03    1.20E-03    1.132E-03   1.17E-01
 20.5   3.28E-02    6.72E-02    1.64E-03    2.52E-03    7.68E-04    1.100E-03   9.11E-02
 20.6   3.27E-02    6.73E-02    1.96E-03    2.10E-03    5.35E-04    1.081E-03   7.46E-02
 20.7   3.26E-02    6.74E-02    2.28E-03    1.80E-03    3.95E-04    1.067E-03   6.31E-02
 20.8   3.25E-02    6.75E-02    2.60E-03    1.57E-03    3.03E-04    1.058E-03   5.47E-02
 20.9   3.24E-02    6.76E-02    2.91E-03    1.40E-03    2.40E-04    1.051E-03   4.82E-02
 21.0   3.23E-02    6.77E-02    3.23E-03    1.26E-03    1.95E-04    1.045E-03   4.31E-02
 21.1   3.22E-02    6.78E-02    3.54E-03    1.14E-03    1.62E-04    1.041E-03   3.91E-02
 21.2   3.21E-02    6.79E-02    3.85E-03    1.05E-03    1.36E-04    1.037E-03   3.57E-02
 21.3   3.19E-02    6.81E-02    4.15E-03    9.69E-04    1.17E-04    1.034E-03   3.29E-02
 21.4   3.18E-02    6.82E-02    4.46E-03    9.00E-04    1.01E-04    1.031E-03   3.05E-02
 21.5   3.17E-02    6.83E-02    4.76E-03    8.40E-04    8.82E-05    1.029E-03   2.84E-02
 21.6   3.16E-02    6.84E-02    5.06E-03    7.87E-04    7.77E-05    1.027E-03   2.66E-02
 21.7   3.15E-02    6.85E-02    5.36E-03    7.41E-04    6.91E-05    1.026E-03   2.50E-02
 21.8   3.14E-02    6.86E-02    5.66E-03    7.00E-04    6.18E-05    1.024E-03   2.36E-02
 21.9   3.13E-02    6.87E-02    5.96E-03    6.63E-04    5.56E-05    1.023E-03   2.24E-02
 22.0   3.13E-02    6.88E-02    6.25E-03    6.30E-04    5.04E-05    1.022E-03   2.13E-02
 22.1   3.12E-02    6.88E-02    6.54E-03    6.00E-04    4.58E-05    1.021E-03   2.03E-02
 22.2   3.11E-02    6.89E-02    6.83E-03    5.72E-04    4.19E-05    1.020E-03   1.94E-02
 22.3   3.10E-02    6.90E-02    7.12E-03    5.48E-04    3.85E-05    1.019E-03   1.86E-02
 22.4   3.09E-02    6.91E-02    7.41E-03    5.25E-04    3.54E-05    1.018E-03   1.78E-02
 22.5   3.08E-02    6.92E-02    7.69E-03    5.04E-04    3.27E-05    1.017E-03   1.71E-02
 22.6   3.07E-02    6.93E-02    7.98E-03    4.84E-04    3.04E-05    1.017E-03   1.65E-02
 22.7   3.06E-02    6.94E-02    8.26E-03    4.66E-04    2.82E-05    1.016E-03   1.59E-02
 22.8   3.05E-02    6.95E-02    8.54E-03    4.50E-04    2.63E-05    1.016E-03   1.54E-02
 22.9   3.04E-02    6.96E-02    8.81E-03    4.34E-04    2.46E-05    1.015E-03   1.49E-02
 23.0   3.03E-02    6.97E-02    9.09E-03    4.20E-04    2.31E-05    1.015E-03   1.44E-02
 23.1   3.02E-02    6.98E-02    9.37E-03    4.06E-04    2.17E-05    1.014E-03   1.40E-02
 23.2   3.01E-02    6.99E-02    9.64E-03    3.94E-04    2.04E-05    1.014E-03   1.36E-02
 23.3   3.00E-02    7.00E-02    9.91E-03    3.82E-04    1.93E-05    1.013E-03   1.32E-02
 23.4   2.99E-02    7.01E-02    1.02E-02    3.70E-04    1.82E-05    1.013E-03   1.28E-02
 23.5   2.99E-02    7.01E-02    1.04E-02    3.60E-04    1.72E-05    1.013E-03   1.25E-02
 23.6   2.98E-02    7.02E-02    1.07E-02    3.50E-04    1.63E-05    1.012E-03   1.22E-02
 23.7   2.97E-02    7.03E-02    1.10E-02    3.40E-04    1.55E-05    1.012E-03   1.19E-02
 23.8   2.96E-02    7.04E-02    1.12E-02    3.31E-04    1.47E-05    1.012E-03   1.16E-02
 23.9   2.95E-02    7.05E-02    1.15E-02    3.23E-04    1.40E-05    1.011E-03   1.13E-02
 24.0   2.94E-02    7.06E-02    1.18E-02    3.15E-04    1.34E-05    1.011E-03   1.10E-02
 24.1   2.93E-02    7.07E-02    1.20E-02    3.07E-04    1.28E-05    1.011E-03   1.08E-02
 24.2   2.92E-02    7.08E-02    1.23E-02    3.00E-04    1.22E-05    1.011E-03   1.06E-02
 24.3   2.92E-02    7.08E-02    1.25E-02    2.93E-04    1.17E-05    1.010E-03   1.03E-02
 24.4   2.91E-02    7.09E-02    1.28E-02    2.86E-04    1.12E-05    1.010E-03   1.01E-02
 24.5   2.90E-02    7.10E-02    1.30E-02    2.80E-04    1.07E-05    1.010E-03   9.93E-03
 24.6   2.89E-02    7.11E-02    1.33E-02    2.74E-04    1.03E-05    1.010E-03   9.73E-03
 24.7   2.88E-02    7.12E-02    1.35E-02    2.68E-04    9.89E-06    1.010E-03   9.55E-03
 24.8   2.87E-02    7.13E-02    1.38E-02    2.62E-04    9.51E-06    1.009E-03   9.37E-03
 24.9   2.87E-02    7.13E-02    1.40E-02    2.57E-04    9.15E-06    1.009E-03   9.20E-03
 25.0   2.86E-02    7.14E-02    1.43E-02    2.52E-04    8.82E-06    1.009E-03   9.04E-03

```

(1) In the table the column **[Ag(NH3)2+]** assumes that the 1 millimole of silver is all $\ce{Ag(NH3)\_2^+}$. The value is expressed as a concentration so it falls due to the dilution of the extra $\ce{NH3}$. In the column **mol(Ag)** all the silver species are totaled. The value should be 1.00E-03 moles or 1 mM Ag. Values significantly above that indicate that the calculations are in error. (Note that the way the calculations are done the mM Ag will always be too high.) 


(2) The column **[NH3]tot** is the concentration of all NH3 (mMol NH3)/(total volume)


(3) The column **[NH3]** is the "free" ammonia. It assumes ammonia is either free $\ce{NH3}$ or $\ce{Ag(NH3)\_2^+}$ and thus it ignores the single amine complex $\ce{Ag(NH3)^+}$. It calculates free ammonia as [NH3]tot - 2\*$[\ce{Ag(NH3)\_2^+}]$


(4) The column labeled F\* is the fraction of silver as $\ce{Ag(NH3)^+}$ and $\ce{Ag^+}$. Thus the real decision is what fraction is low enough since the fraction continues to fall as excess ammonia is added. 


Conclusion
----------


The choice of the "end point" is arbitrary...



```
mL        end point
----      -----------------------------------
21.4      97% [Ag(NH3)2^+]
21.6      [Ag^+] 10% of [Ag(NH3)^+]
22.1      98% [Ag(NH3)2^+]
24.3      [Ag(NH3)^+] 1% of [Ag(NH3)2^+]
24.4      99% [Ag(NH3)2^+]
24.6      [Ag^+] = 1*10^-5

```

 4

Let $V\_0 = 10~\mathrm{ml}$ be the initial volume of $\ce{AgNO3}$, $c\_{\ce{AgNO3}} = 0.1~\mathrm{mol\, l^{-1}}$ its concentration, and $c\_{\ce{NH3}} = 0.1~\mathrm{mol\, l^{-1}}$ the concentration of the $\ce{NH3}$.


If complete complexation means that $\ce{[Ag+]}<10^{-5}~\mathrm{mol\, l^{-1}}$ we could use it as a target value, i.e.
$$\gamma\_1=\frac{\ce{[Ag(NH3)+]}}{\ce{[NH3]}}=\beta\_1 \ce{[Ag+]} = 10^{3.3}\cdot10^{-5} = 10^{-1.7} \tag{1}$$
and
$$\gamma\_2=\frac{\ce{[Ag(NH3)\_2+]}}{\ce{[NH3]^2}}=\beta\_2 \ce{[Ag+]} = 10^{7.2}\cdot10^{-5} \,\mathrm{mol^{-1}\, l} = 10^{2.2}\,\mathrm{mol^{-1}\, l}\tag{2}$$


Let $x$ be the number of moles of $\ce{Ag(NH3)+}$, $y$ the number of moles of $\ce{Ag(NH3)2+}$, and $z$ the number of moles of free $\ce{NH3}$ in the solution.


With this we get for the number of moles of silver (neglecting free $\ce{Ag+}$)


$$c\_{\ce{AgNO3}} \cdot V\_0 = 10^{-3}~\mathrm{mol} \approx x + y \tag{3}$$


and for the number of moles of free and bound $\ce{NH3}$


$$x + 2y + z. \tag{4}$$


The complete volume of the solution is given as


$$V = V\_0 + \frac{x + 2y + z}{c\_{NH\_3}} \tag{5}$$


With this we can write equations $(1)$ and $(2)$ as


$$\gamma\_1 = \frac{x}{z} = 10^{-1.7} \tag{6}$$
and
$$\gamma\_2 = \frac{y}{z^2} \cdot V = \frac{y}{z^2} \cdot (V\_0 + \frac{x + 2y + z}{c\_{NH\_3}}) = 10^{2.2}\,\mathrm{mol^{-1}\, l} \tag{7}$$


$(3)$, $(6)$ and $(7)$ are tree equations with three unknowns which we can solve.


 1