Question

Solubility product constant and the common ion effect.

Sodium Thiosulfate (M)

Volume Calcium Iodate(mL)

Initial Volume of Sodium Thiosulfate Test 1(mL)

Final Volume of Sodium Thiosulfate Test 1(mL)

0.045145

5

19

26.9

Sodium Thiosulfate (M)

Volume Calcium Iodate w/ added ions(mL)

Initial Volume of Sodium Thiosulfate Test 1(mL)

Final Volume of Sodium Thiosulfate Test 1(mL)

0.045145

5

17

23

*****(0.1M) = [Ca2+]*****

EQN: IO3 -+ 6S2O3 2- + 6H3O+ à I- + 3S4O6 2- + 9H2O

Calculate

Part A:(Without added ions)

-- Moles Thiosulfate Used

– Moles Iodate

– Equilibrium Concentration of Iodate Ion

– Equilibrium Concentration of Calcium

– Molar Solubility of Calcium Iodate

– Ksp

– % Error of Ksp

Part B:(With added Ca2+ ions)

-- Moles Thiosulfate Used

– Moles Iodate

– Equilibrium Concentration of Iodate Ion

– Equilibrium Concentration of Calcium

– Molar Solubility of Calcium Iodate

– Average Molar Solubility of Calcium Iodate

Answer 1

PartA

IO3- + 6S2O3- + 6H3O+ -------> I- + 3S4O62- + 9H2O

i) moles of thiosulphate used = (0.045145mol/1000ml)×7.9ml = 0.0003566

¡¡) moles of iodate = 0.0003566/3= 0.0000594

iii) Equilibrium concentration of Iodate ion = (0.0000594/5)×100= 0.01188M

iv) Equilibrium concentration of CalCium ion = 0.01188M/2=0.00594M

v) Molar solubility of CalCium Iodate = 0.00594 M

vi) Ksp = [ Ca2+] [ IO3-]^2= 0.00594M × (0.01188 M)^2

= 8.38×10^-7

vii) Accepted Ksp = 7.1×10^-7

%error of Ksp = (1.28×10^-7/7.1×10^-7)×100=18%

Part B

¡) Moles of thiosulphate used = (0.045145/1000)×6=0.000271

ii) Moles of iodate = 0.000271/6= 0.0000451

iii) Equilibrium concentration of Iodate ion = (0.0000451/5)×1000= 0.00902M

¡v) Equilibrium Concentration of CalCium ion = 0.00451M

v) Molar solubility of CalCium Iodate = 0.00451M