A lone pair can do more than sit on an atom: it can form a bond to something that has room for it. That broader idea of acids and bases explains why metal ions grab ammonia or cyanide, and why that makes some insoluble salts dissolve.
Key idea
A Lewis base donates an electron pair; a Lewis acid accepts one. Metal ions are Lewis acids. A metal ion bonded to several Lewis bases (ligands) is a complex ion, such as $\mathrm{[Ag(NH_3)_2]^+}$. Its formation constant $K_f$ measures how tightly it holds together, and forming it can dissolve a salt: $K = K_{sp} \times K_f$.
Why it works
$\mathrm{BF_3}$ has only six electrons around boron and an empty orbital, so it accepts the lone pair of $\mathrm{NH_3}$ with no proton involved. A metal ion's empty orbitals accept lone pairs the same way. When ammonia ties up $\mathrm{Ag^+}$ as $\mathrm{[Ag(NH_3)_2]^+}$, free $\mathrm{Ag^+}$ drops, $Q$ falls below $K_{sp}$ for AgCl, and more AgCl dissolves. Adding the two reactions multiplies their constants.
Worked examples
Worked example: Dissolving AgCl in ammonia
$K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10}$ and $K_f(\mathrm{[Ag(NH_3)_2]^+}) = 1.7 \times 10^{7}$. Find the molar solubility of AgCl in $1.0\ \text{M}$ $\mathrm{NH_3}$.
- Add: $$\mathrm{AgCl(s)} + \mathrm{2NH_3} \rightleftharpoons \mathrm{[Ag(NH_3)_2]^+} + \mathrm{Cl^-}$$ $K = (1.8 \times 10^{-10})(1.7 \times 10^{7}) = 3.1 \times 10^{-3}$.
- ICE: $[\mathrm{[Ag(NH_3)_2]^+}] = [\mathrm{Cl^-}] = s$, $[\mathrm{NH_3}] = 1.0 - 2s$. So $\dfrac{s^2}{(1.0 - 2s)^2} = 3.1 \times 10^{-3}$.
- Square root: $\dfrac{s}{1.0 - 2s} = 0.0553$, so $s = \dfrac{0.0553}{1.111} = 0.050\ \text{M}$.
Answer: $0.050\ \text{M}$, nearly 4000 times its solubility in water.
Worked example: How little free metal ion is left
A solution has $0.010\ \text{M}$ silver, nearly all as $\mathrm{[Ag(NH_3)_2]^+}$, with $0.98\ \text{M}$ free $\mathrm{NH_3}$. Estimate $[\mathrm{Ag^+}]$.
- $K_f = \dfrac{[\mathrm{[Ag(NH_3)_2]^+}]}{[\mathrm{Ag^+}][\mathrm{NH_3}]^2}$, so $[\mathrm{Ag^+}] = \dfrac{0.010}{(1.7 \times 10^{7})(0.98)^2}$.
- $[\mathrm{Ag^+}] = 6.1 \times 10^{-10}\ \text{M}$: almost every silver ion is tied up.
Answer: $6.1 \times 10^{-10}\ \text{M}$.
| Complex ion | $K_f$ |
|---|---|
| $\mathrm{[Ag(NH_3)_2]^+}$ | $1.7 \times 10^{7}$ |
| $\mathrm{[Zn(NH_3)_4]^{2+}}$ | $2.9 \times 10^{9}$ |
| $\mathrm{[Cu(NH_3)_4]^{2+}}$ | $1.1 \times 10^{13}$ |
| $\mathrm{[Ag(S_2O_3)_2]^{3-}}$ | $4.7 \times 10^{13}$ |
| $\mathrm{[Ag(CN)_2]^-}$ | $1.0 \times 10^{21}$ |
Common mistake
Thinking a Lewis acid must contain hydrogen, or that forming a complex ion makes a salt less soluble. $\mathrm{BF_3}$ and $\mathrm{Ag^+}$ are Lewis acids with no hydrogen at all, and complex formation pulls metal ions out of the solubility equilibrium, so more solid dissolves.
Photographic "fixer" (sodium thiosulfate) works this way: it dissolves the unexposed silver bromide as $\mathrm{[Ag(S_2O_3)_2]^{3-}}$.