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Solubility Equilibria K sp

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Solubility & Solubility Product Even insoluble salts dissolve somewhat in water –insoluble = less than 0.1 g per 100 g H 2 O The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced AB (s) a A + (aq) + b B - (aq) Equilibrium constant called solubility product K sp = [A + ] a [B - ] b If undissolved solid is in equilibrium with the solution, the solution is saturated Larger K = More Soluble –for salts that produce same the number of ions

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K sp is the solubility product equilibrium constant at a given temperature. The solubility of a substance represents an equilibrium position for the system. First, write the dissociation equation of the ionic compound in water, including states. Then, set up the expression for K (you will only have the ions in this expression since the salt is a solid). Solve for a numerical value of K sp using given data. Remember that for every one formula unit that breaks up, you will get equal molar concentrations for ions that are in a 1:1 ratio in the compound. If the ratio is not 1:1, adjust accordingly.

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Example Determine the balanced equation for the dissociation of the salt AgI (s) Ag + (aq) + I - (aq) Determine the expression for the solubility product –Same as the Equilibrium Constant Expression K sp = [Ag + ] [I - ] Calculate the solubility of AgI in water at 25°C if the value of K sp = 1.5 x

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Define the concentrations of dissolved ions in terms of x AgI (s) Ag + (aq) + I - (aq) Stoichiometry tells us that we get 1 mole of Ag + and 1 mol I - for each mole of AgI dissolved Let x = [Ag + ], then [I - ] = x Example Calculate the solubility of AgI in water at 25°C if the value of K sp = 1.5 x

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Plug the ion concentrations into the expression for the solubility product and solve for K sp [Ag + ] = [I - ] = x [Ag + ] = 1.2 x mol/L = [AgI] The solubility of AgI (at 25°C) = 1.2 x M Example Calculate the solubility of AgI in water at 25°C if the value of K sp = 1.5 x

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