Question #13e47

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Question #13e47

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Answer 1 · 2015-09-14T01:43:07

Because you actually have one mercury cation and two bromide anions.

Explanation:

The thing to remember about mercury(I) bromide, ${Hg}_{2} {Br}_{2}$ $"Hg"_2"Br"_2$ , is that it contains two ${Hg}^{+}$ $"Hg"^(+)$ cations bonded into one ${Hg}_{2}^{2 +}$ $"Hg"_2^(2+)$ cation.

Mercury is still in its +1 oxidation state, it's just that two such cations form a metal-metal bond and exist as ${Hg}_{2}^{2 +}$ $"Hg"_2^(2+)$ .

This means that the compound is formed using two bromide anions, ${Br}^{-}$ $"Br"^(-)$ , since you'd need two of those to balance the +2 positive charge of the cation.

These mercury cations are actually called polycations because they feature two mercury(I) cations bonded together.

This means that when mercury(I) bromide is placed in aqueous solution, it's dissociation equilibrium will be

${Hg}_{2} {Br}_{2(s]} ⇌ {Hg}_{2(aq]}^{2 +} + 2 {Br}_{(aq]}^{-}$ $"Hg"_2"Br"_text(2(s]) rightleftharpoons "Hg"_text(2(aq])^(2+) + color(red)(2)"Br"_text((aq])^(-)$

By definition, the solubility product constant, $K_{s p}$ $K_(sp)$ , will be

$K_{s p} = [{Hg}_{2}^{2 +}] \cdot {[{Br}^{-}]}^{2}$ $K_(sp) = ["Hg"_2^(2+)] * ["Br"^(-)]^color(red)(2)$

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Question #13e47

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