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Halides are group 7 elements. They generally have a low melting and boiling point. They are highly reactive. Silver chloride and silver bromide are not soluble in water. So they will be precipitates. However, I do know that they have varying solubilities. So, they will form different amounts of precipitates. I want to know why they have different solubilities and which is the most soluble halide.

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The solubilities of silver halides decreases down the periodic table:

  • $\ce{AgF}\ : K_{sp} = 205$
  • $\ce{AgCl}: K_{sp} = 1.8\times10^{-10}$
  • $\ce{AgBr}: K_{sp} =5.2\times10^{-13}$
  • $\ce{AgI}\ \ \ : K_{sp} =8.3\times10^{-17}$

The rationale for this trend is typically described using the concept of hard/soft acids and bases. (See J. Chem. Educ., 1968, 45, page 581 and also page 643 for early discussions on HSAB theory.) Briefly, ions with similar size and high polarizability will lend themselves to more covalent interactions, and therefore will have decreased solubility in water. HSAB classifies ions as either hard or soft, with soft being large and polarizable and hard being small and highly charged. H-H and S-S interactions are strong while H-S or S-H interactions are weaker. Ag is classified as soft and the halides go from hard (F-) to soft (I-).

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