Question & Answer: In an experiment to calculate the K_sp of Calcium Iodate [Ca (IO_3)_2], a 10 mL solution of a…..

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

Total moles of S₂O₃^2- consumed = Molarity of Na₂S₂O₃ x Volume in liters

= 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 S₂O₃^2-.

So,

Moles of IO₃^– neutralized = (1 mol IO₃^– / 6 moles S₂O₃^2-) x 0.0009450 mol S₂O₃^2-

= 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