According to my knowledge, I know that $\ce{AgCl}$ dissolves in dilute ammonia, $\ce{AgBr}$ dissolves in concentrated ammonia and $\ce{AgI}$ does not dissolve even with concentrated ammonia.

What is the reason for this?

  • $\begingroup$ Multiply the Ksp of the salt by the Kf of the complex ion to get K for the reaction between ligand and salt. The solubility trend of the salts is AgCl>AgBr>AgI so with the same ligand that's the solubility trend of the salts and ligand together. $\endgroup$ – Brinn Belyea Aug 3 '14 at 19:26

Silver chloride is very thermodynamically stable in ammonia solution. The $K$ for the dissolution of silver chloride in ammonia solution is only $\pu{2.9E-3}.$ suggesting that the product — silver ammonia complex ion — is much less thermodynamically stable than the reactant, silver chloride.

$$\ce{AgCl + 2 NH3 <=> [Ag(NH3)2]+ + Cl-}$$

$$ \begin{align} K &= K_\mathrm{sp}(\ce{AgCl})\times K_\mathrm{f}(\ce{[Ag(NH3)2]+})\\ & = \pu{1.8E-10}\times\pu{1.6E7}\\ & = \pu{2.9E-3} \end{align} $$

However, silver chloride will dissolve in an excess of concentrated ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride has a minuscule $K_\mathrm{sp}$ value, some free silver ions are present in solution. Addition of concentrated ammonia solution effectively consumes this free silver ion as ammonia, a good σ-electron pair donor ligand, combines with the silver ion.

Ions themselves are not precipitates, as all bulk matter must be approximately electrically neutral, and so the silver chloride dissolves when hit with excess, concentrated ammonia.

Also, note that the cyanide ion is also a good complexing agent with respect to silver. The cyanide ion has a surfeit of electrons on the carbon, and carbon is even less able to stabilize negative charge than nitrogen, making the cyanide ion reactive as a complexing agent. Hitting silver chloride and other silver salts with a source of cyanide ion will similarly cause the silver salt to dissolve into ions suspended in solution.

| improve this answer | |
  • 6
    $\begingroup$ In fact you can predict exactly how much AgCl you can dissolve: so long as $\ce{[[Ag(NH_3)_2]^+]\cdot[ {Cl^{-}}]}< 0.0029\cdot\ce{ [NH_3]^2 }$ all the silver will be dissolved. That's why people talk of "excess ammonia"; you need way more than a 1:1 molar ratio. Because AgBr and AgI have even smaller solubility products than AgCl, the factor on the right is much less and the amount that you can dissolve even in pure ammonia is similarly much less. $\endgroup$ – Silvio Levy Aug 4 '14 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.