 # Question & Answer: In an experiment to calculate the K_sp of Calcium Iodate [Ca (IO_3)_2], a 10 mL solution of a….. In an experiment to calculate the K_sp of Calcium Iodate [Ca (IO_3)_2], a 10 mL solution of a saturated solution of Calcium Iodate was titrated against 0.0250 M solution of Sodium thiosulfate (Na_2S_2, O_3) using starch indicator. The end point of the titration was reached at 37.8 mL of Sodium thiosulfate. The reaction is, IO^-_3 + 6 S_2O^2_3 + 6 H^+ rightarrow I^- + 3 S_ 4O^2-_6 + 6 H_2O Calculate the Solubility Product of Calcium Iodate.

Ans. Part 1: Calculate molarity of [Ca(IO3)2] solution

Total moles of S2O32- consumed = Molarity of Na2S2O3 x Volume in liters

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= 0.0250 M x 0.0378 L

= (0.0250 mol/L) x 0.0378 L                         ; [1 M = 1 mol/L]

0.0009450 mol

In balanced reaction, 1 mol IO3 is neutralized by 6 moles S2O32-.

So,

Moles of IO3 neutralized = (1 mol IO3 / 6 moles S2O32-) x 0.0009450 mol S2O32-

0.0001575 mol

Therefore, the 10.0 mL saturated solution consists of 0.0001575 mol [Ca(IO3)2].

Now,

Molarity of saturated [Ca(IO3)2] solution = Moles / Volume of solution in liters

= 0.0001575 mol/ 0.010 L

= 0.000001575 mol/ L

1.575 x 10-6 M

Part 2: Calculate Ksp

[Ca(IO3)2] ———–> Ca2+(aq) + 2 IO3(aq)

Stoichiometry: 1 mol [Ca(IO3)2] dissociates into 1 mol Ca2+ and 2 moles IO3.

So,

[Ca2+] = [Ca(IO3)2] = 1.575 x 10-6 M

[IO3] = 2 x [Ca(IO3)2] = 2 x 1.575 x 10-6 M = 3.150 x 10-6 M

Now,

Ksp = [Ca2+] [IO3]2 = (1.575 x 10-6) (3.150 x 10-6) = 1.563 x 10-17

Therefore, Ksp of [Ca(IO3)2] = 1.563 x 10-17