Question

A 0.1358-g portion of potassium iodate (MW 214.00), about 2 g of potassium iodide, and 2 mL of 6 M hydrochloric acid were dissolved in 25 mL of distilled water. The triiodide formed during the ensuring reaction was titrated to the starch endpoint with 31.94 mL of a thiosulfate solution. A 25.00-mL triiodide sample solution was titrated to the endpoint with 21.33 mL of the standardized thiosulfate solution. Calculate the concentrations of the thiosulfate solution and the triiodide solution.

Answer 1

Since, KIO₃ contain 59.326% I^-.

In, 2g of KIO₃ contain 1.18652 g of I^-.

Since, KI contain 76.49% I^-.

In, 0.1358g of KI contain 0.1039 g of I^-.

Initial I^- concentration = 1.18652 + 0.1039 = 1.29042 g of I₃^-

1.29042g at I₃^- can be titrated using 31.94 mL thiosulfate solution

25 mL distilled water and 2 mL HCl = 27 mL = 0.027 L

[I₃^-] =1.29042 g / (380.7 g/mole x 0.027 L)

[I₃^-] = 0.12554 mole/L = 0.12554 M

27 mL I₃^- 0.12554 M solution can be titrated using 31.94 mL Na₂S₂O₃ (X) M solution.

V_Na2S2O3 = 21.33 mL

V_I3^- = 25 mL

C_Na2S2O3 x V_Na2S2O3 = C_I3- x V_I3-

C_Na2S2O3 x 21.33 mL = 0.12554 M x 25 mL

C_Na2S2O3= (0.12554 M x 25 mL) / 21.33 mL

C_Na2S2O3 = 0.14714 M

ANSWERS:

[Na₂S₂O₃] = 0.14714 mole/L

[I₃^-] = 0.12554 mole/L