Question

Complete the table using the following data:

IO₃^- (aq) + 6S₂O₃^2- (aq) + 6H₃O⁺ (aq) --> I^- (aq) + 3S₄O₆^2- (aq) + 9H₂O (l)

Sodium Thiosulfate Solution Concentration (M)	0.051 M
Volume Calcium Iodate Solution Added (L)	0.01 L
Thiosulfate Solution Volume	0.01814 L
Moles Thiosulfate Used	0.00092514
Moles Iodate
Equilibrium Concentration of Iodate Ion (M)
Equilibrium Concentration of Calcium Ion (M)
Molar Solubility of Calcium Iodate (M)
Ksp, Calcium Iodate

Answer 1

Sodium thiosulfate concentration(M)	0.051M
Volume calcium iodate used(L )	0.01L
Thiosulfate solution volume	0.01814L
moles of thiosulfate used	0.00092514
moles of Iodate	0.00015419
Equillibrium concentration of Iodate ion(M)	0.005479M
Equillibeium conecentration of calcium ion(M)	0.0027395M
Molar solubility of Calcium Iodate	0.0027395 mol/L
Ksp of calcium Iodate	8.22×10-8

Explanation

i) IO3^- (aq) + 6S2O3^2-(aq) + 6H3O⁺(aq) ------> I^-(aq) + 3S4O6^2-(aq) + 9H2O(l)

Stoichiometrically , 6 moles of S2O3^2-reacts with 1mole of IO3^-

Therefore,

Number of moles of IO3^- = Number of moles of S2O3^2-/6

ii) Total volume = 0.01L + 0.01814L = 0.02814L

Equillibrium concentration of IO3^- = 0.00015419mol/0.02814m = 0.005479M

iii)Equillibrium concentration of Ca⁺ = 0.005479M/2 = 0.0027395M

iv) Molar solubility of Ca(IO3)2 = Concentration of Ca²⁺

v) Ksp = [Ca²⁺][IO3^-]²

= (0.0027395M)×(0.005479M)²

= 8.22×10^-8 M