For any slightly soluble salt placed in water, an equilibrium is established bet

Question

For any slightly soluble salt placed in water, an equilibrium is established between the solid and ions in solution. Consider the following reaction: Ca(IO3)2 (s) <----> Ca 2+ (aq) + 2IO3- a) Write the equilibrium constant expression for this reaction (Ksp). b) When calcium Iodate is added to water at 25 degrees celsius , the equilibrium concentration of calcium ion was found to be 0.011M. Calculate the value of Ksp for this salt.

Explanation / Answer

a) Equilibrium constant, K = [Ca 2+] * [IO3-]^2 ----> Note: concentration of solid is assumed to remain constant and is taken as 1 M (unity). In this case equilibrium constant is Ksp or solubility product. b) Considering Ca(IO3)2 as a strong electrolyte that is it will completely dissociate into its constituent ions. From the balanced chemical equation mentioned in the problem we can see that for each mole of Ca2+ there are 2 mole of IO3-, thus concentration of IO3- will be twice the concentration of Ca2+. [Ca2+] = 0.011 M and [IO3-] = 0.011 * 2 M = 0.022 M. Now calculate the Ksp, Ksp = [Ca 2+] * [IO3-]^2 = 0.011 M (0.022 M)^2 = 4.84 * 10^-4 M^3

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