## Expert Answer

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

Total moles of S_{2}O_{3}^{2-} consumed = Molarity of Na_{2}S_{2}O_{3} 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 IO_{3}^{–} is neutralized by 6 moles S_{2}O_{3}^{2-}.

So,

Moles of IO_{3}^{–} neutralized = (1 mol IO_{3}^{–} / 6 moles S_{2}O_{3}^{2-}) x 0.0009450 mol S_{2}O_{3}^{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}] ———–> Ca^{2+}(aq) + 2 IO_{3}^{–}(aq)

Stoichiometry: 1 mol [Ca(IO3)_{2}] dissociates into 1 mol Ca^{2+} and 2 moles IO_{3}^{–}.

So,

[Ca^{2+}] = [Ca(IO3)_{2}] = 1.575 x 10^{-6} M

[IO_{3}^{–}] = 2 x [Ca(IO3)_{2}] = 2 x 1.575 x 10^{-6} M = 3.150 x 10^{-6} M

Now,

Ksp = [Ca^{2+}] [IO_{3}^{–}]^{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}**