Answer to Question #160647 in General Chemistry for Dexther Villanueva

Question

A sample of impure KI weighing 0.6000g is dissolved in water, the solution is acidified, and 25.00ml of 0.0400M KIO3 is added. The iodate is reduced to I2 and the iodide is oxidized to I2. The I2 is boiled off, the solution cooled and an excess of pure KI is added to react with the unused KIO3. The iodine produced is titrated with 40.28ml of 0.1053M Na2S2O3. Calculate the percentage of KI in the sample. Molar masses: KI = 166.0, KIO3 = 214.0

A. 5.88%

B. 20.75%

C. none of the given answers

D. 55.33%

E. 48.65%

Answer 1

Solution:

KIO₃(aq) + 5 KI(aq) + 3 H₂O(l) = 3 I₂(s) + 6 KOH(aq)  .....1

2 Na₂S2O₃ + I₂ = Na₂S₄O₆ + 2 NaI  ....2

Moles of thiosulfate in the back titration  =  0.1053 $\frac{moles}{L}$ * 0.04028 L  = 0.004241484 moles 

moles of excess iodine these reacted with according to our equation 2 = $\frac{0.004241484}{2}$ = 0.002120742 moles

moles of KIO3 that produced this I2 according to eq 1 = $\frac{0.002120742}{3}$

= 0.000706914 moles 

Initial amount of KIO₃ added  = 0.0400 $\frac{moles}{L}$ * 0.025L = 0.001 moles 

moles of KIO₃ that reacted with the KI = 0.001 moles - 0.000706914 moles 

= 0.000293086 moles of KIO₃ 

these reacted with 0.000293086 * 5 moles of KI = 0.00146543 moles of KI

mass of KI in the initial reaction = 0.00146543 moles * 166.00 $\frac{g}{mole}$ = 0.24326138 g

% KI in the sample = $\dfrac{0.24326138}{0.6}$ = 40.54%