# 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?

• You seem to be overthinking the problem. The goal would be to have $$\ce{[Ag^{+}] +[Ag(NH3)^{+}]} << [\ce{Ag(NH3)2^{+}}]$$ The problem is sloppy with significant figures. Think of having 0.100 M $\ce{AgNO3}$ which would mean that $$\ce{[Ag+] +[Ag(NH3)^{+}] < 0.001}$$ would be insignificant. – MaxW Mar 25 '16 at 20:42
• Considering that the concentrations of silver nitrate and ammonia, as well as both constants, are only given to 1 significant figure, I think the "correct" answer is 2E1 mL of NH3 without doing any calculating. ;-) – MaxW Mar 25 '16 at 20:52
• @MaxW, In my country, students and teachers haven't really got into the habit of using significant figures; what we do is we consider everything to 3 decimal places. Therefore, a 0.1M solution should be treated as a 0.100M one. ( just noting, their answer is 24.3mL; your answer rounded to one sig fig ;) ) As for your approximation, I applied it, but I still can't get a solvable system of equations; all I'm left with is 2 unknowns ( the volume and $\ce{[NH3]}$ ) in a single equation – L3ul Mar 26 '16 at 19:48
• I'll work it out later tonight if someone doesn't beat me too it. Sorry my suggestion didn't help. I was just shooting from the hip without really working the problem out. // You may have to consider the fraction of silver in each of the three forms. It'll be messy but doable... – MaxW Mar 26 '16 at 19:49

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

• Now that I fixed the formulas the numbers make more sense. I had thought with the old calculations that the diamine complex was forming too fast and I didn't stop and try to figure out why my gut was telling me that the calculations were wrong. – MaxW Mar 27 '16 at 16:26
• Why would they choose such an arbitrary condition without specifying it? I think we're missing something... – L3ul Mar 28 '16 at 4:57
• @MaW, I'm writing this in case it might be of interest to you or someone else: after exploring other problems and asking some professors of Analytical Chemistry, it seems that there's a quite common arbitrary condition hanging around in our country's analytical problems, which states that: 'a reaction is considered complete when 99.9% of the reactants go into products.' What's interesting is that the author of our problem had decided to change this condition without any reason whatsoever. (I doubt there's even a practical reason for these conditions) – L3ul May 7 '16 at 5:22

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.

• You made some mistakes. Let $\ce{V_{Ag}}$ be the volume of silver solution used which is 10 mL. Let $\ce{V_{NH3}}$ be the volume of ammonia. In equation (3) the millimoles of $\ce{Ag^+}$ is $\ce{(10 + V_{NH3})(1E-5)}$. // The OP's problem is that you're pulling the assumption that $[\ce{Ag^+}] = 1 \times 10^{-5}$ out of thin air. Why shouldn't it be $2 \times 10^{-5}$? // Doing the calc you should get 24.6 mL from my table... – MaxW Mar 27 '16 at 23:55
• @MaxW, That's the assumption that I've noticed to be tackled in most problems with precipitation reactions; which says that if the concentration of the cation or anion becomes $10^5$ smaller, the reaction is complete. – L3ul Mar 28 '16 at 5:01
• Their argument for such value was that most analysis apparatus wouldn't be able to detect a lower concentration, which seems reasonable (Unless we're dealing with concentrations lower than 0.01 M) – L3ul Mar 28 '16 at 5:04
• @MaxW: $\ce{[Ag+]} < 10^{-5}$ is the given condition for completeness. Equation (3) gives the total amount of silver which is constant summed over all silver species. If you do the calculation the result is indeed 24.6 ml. – aventurin Mar 28 '16 at 9:02
• That's true. Corrected it. – aventurin Mar 28 '16 at 17:13