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Question

Write the dissociation equilibrium equation for lead (II) bromide, PbBr2(s)

Answer 1

For equilibrium equations you write K = [products]/[reactants] and in this case the chemical equation for the dissolution of lead (II) bromide is PbBr_{2(s) -->} Pb²⁺_(aq) + 2Br^-_(aq) ; since only aqueous products/reactants are included in equilibrium expressions (hence the brackets for concentrations), the solid PbBr₂ is not included, only the two products are. Therefore the expression should simply read K = [Pb²⁺][Br^-]²

The Br^- is squared in the equation because coefficients in the chemical equations become exponents in the equilibrium expressions. See bolded PbBr_{2(s) -->} Pb²⁺_(aq) + 2Br^-_(aq) K = [Pb²⁺][Br^-]²

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