Question

A 1.2 g. of a mixture containing H2C2O4.2H2O and KHC2O4.H2O and impurities of a

neutral salt, consumed 18.9 ml of 0.5 N NaOH for complete neutralization. On

titiration with KMnO4 solution 0.4 g of the same substance needed 21.55 ml of 0.25

N KMnO4. Calculate the percentage composition of the substance.

Please answer with explanation.​

Answer 1

Answer:

Explanation:

Solution :

Let the percentage of H2C2O4.2H2O be a% and the percentage of KHC2O4.H2O be b%

(a). 1.2100(a63+b146)×1000=37.80×25

(b). 0.40100(a63+b73)×1000=43.10×0.125

a=H2C2O4.2H2O=14.3%

(b). =KHC2O4.H2O=81.7%

Answer 2

Answer:

something is wrong in your question

i think it is not complete neutralisation on titrating with naoh i think

Explanation:

equivalent of h2c2o4.2h2o + equivalent of khc2o4.h2o = equivalent of naoh

let mass of each be x and y

 x/126 + y/146 = 0.00945

146x+126y=173.84

1.2 mixture contains x grams and y grams

0.4 mixture contains x/3 grams and y/3 grams

equivalent of 1 + equivalent of 2 = 0.005387

x/378 + y/ 438 = 0.005387

438x+378y=890.40