To determine the chromium content of the stainless steel Weigh 5.199 g of steel

Question

To determine the chromium content of the stainless steel Weigh 5.199 g of steel chips, which is dissolved in a strong oxidizing acid, and diluted to 1000 mL in a volumetric flask. The chromium is oxidised hereby oxidation state +6. From the volumetric flask is taken 20.00 mL with 19.55 mL of 0.02034 M KMnO4. In a blind determination titrated 25.00 mL of the same about 0.1 M FeSO4 with 24.89 mL of 0.02034 M KMnO4.
The following chemical processes take place:

Calculate w / w% chromium in special steel.

Explanation / Answer

The above titration is called back titration.

In this

For blind determination; Consumed volume of 0.02034 M KMnO4 is : 24.89 mL

For sample determination; Consumed volume of 0.02034 M KMnO4 is : 19.55 mL

Excess volume of 0.02034 M KMnO4 = 24.89 - 19.55 = 5.34 mL

5.34 mL of 0.02034 M KMnO4 is unreacted which is equal to 5.36 mL of 0.1 M FeSO4

Hence the consumed volume of 0.1 M FeSO4 for the titration of 20 mL of Chromium sample solution = 5.36 mL

Normality of Chromium = 5.36 x 0.1 / 20 = 0.0268 N, Molarity M = 0.0268/3 = 0.0089 M

Wt of Chromium in 1000 mL sample solution = M x Mol wt of Chromium x volume / 1000 = 0.0089 x 52 x 1000 /1000

= 0.4628 g

% Chromium content (%w/w) = 0.4628 x 100 / 5.199 = 8.90 % w/w

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