Question

12. Black titration. Impure KS.Os (FM 270.32, 0.507 3 g) was analyzed by treatment with excess standard NazC O4 in H2504 by Reaction 6-4. After reaction with 50.00 mL of 0.050 06 M Na2C204, the excess oxalate required 16.52 mL of 0.020 13 M KnO in Reaction 6-1. Find the weight percent of K.S,Osin the impure reagent.

Answer 1

The difference in amount of Na,C20, added to K,S,O and the amount of oxalte that reacts with KMnO4 is the amount ofNa,C,o that reacts with K,S,Og The reaction between K,S,03 andNa,C,0, in acidic meadium is, So, the mole ratio between K^S,0g and Na,C20 is 1:1 The reaction between C,o and KMno, (MnO;) in acidic meadium is, So, the mole ratio between CO and KMnO, is 5:2. Total moles of Na,C20, added-Molarity х volume- ( 0.0 50 06 M) x( 50.00 mL )-2.503 mmol Moles of KMnO, added(0.02013 M)x (16.52 mL)0.3325 mmol Moles of Na,C,0, react with KMno 0.3325 mmol KMnO,x22 5molNa,co,-0.83 13 mmol Na.С,04 2 mol KMnO Moles of Na2C204 that reacts with K^S2O, 2.503 mmo 0.8313 mmol1.672 mmol Moles of K,S,O 1.672 mmol Na,C,0, 1 mol Na,c,O, 1 mol K,S,o 1.672 mmol K S,O. - 1.672x103 mol K,s,o Mass ofK,S,03 moles x formula mas.672x10 mol)x(270.32 g/mol) 0.4520 g Weight percentage is, 90K,s,o, -massof sample mass of KS,O3 100% 0.45208 x 100% 0.5073 g - 89.10%