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A 5.000 mL sample was taken of a saturated solution of barium iodate at 10 ºC. The sample was mixed with 5.000 mL of 0.4000 M KI and 5.000 mL of 1.500 M HCl. The transmittance of the mixture was read and compared to a calibration curve which indicated a concentration of I3- as 20.55 x 10-4 Calculate Ksp of barium iodate? Ba(IO3)2 . 2H2O --> Ba2+ + 2IO3- + 2H2O g

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Answer: Ksp (Ba(IO3)2)= 4.34*10^-9

Explanation:

Total solution Volume : 15 mL = 0.015 L

moles of I3- = 20.55*10^-4 *0.015 L = 3.0825 *10^-5 moles

1 mol of IO3- gives 3 moles of I3- ===> mol of IO3- = 3.0825 *10^-5 moles/3 = 1.0275 *10^-5 moles

concentration of IO3- in original sample : 1.0275 *10^-5 mole/ 0.005 L = 0.002055 M

concentration of Ba2+ in original sample : 0.002055 M/2 = 0.0010275 M

Ksp (Ba(IO3)2) = [Ba2+] [IO3- ]2 = 4.34*10^-9

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