Question

Suppose the solubility product constant of calcium iodate is 6.47 x 10^-6. If the concentration of calcium iodate is 0.45 M. What would be the concentrations of Ca²⁺ and IO₃^- at equilibrium? Would this solution form a precipitate? Is the solution unsaturated, saturated, or supersaturated?

Note* Ksp = [Ca²⁺][IO₃^-]²

Answer 1

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Calcuin iodate = Cal I02), Ca (IO3), ² Cara + 2103 (64) 1s 2s. Here s is the molar solubility in moy for [ca2+]=s . [10₂] = 25 so ksp = [Ca2+] [103] 6-47x10-6 = Sx (25)? - 483 SE5-47810-6 = 0.0117m (coztleq = S = 0.01170 7 At equilibrium I Concentration [103] es = 25 = 2*0.0117 = 0-02341 12340) of Ca 2+ & IO3 As our concentration is 0.45M 8 we have required only 0.0117 m for precipitation, so our solution is super saturated & we obtain with precipitate.