ALTERNATIVE TO PRACTICAL A

A  is a solution of KMnO\(_4\) containing 0.020 moldm\(^{-3}\). B is a solution of Fe\(^{2+}\) obtained by dissolving 3.8 g of iron granules in 250 cm\(^3\) of solution.

(a)Put A in the burette. Pipette 20.0 cm\(^3\) or 25 cm\(^3\) of B into a conical flask and add 10 cm\(^3\) of H\(_2\)SO\(_4\). Titrate it with A.

     Repeat the titration to obtain concordant titre values. Tabulate your result and calculate the average volume of A used. 

     MnO\(_4\)\(^-\)\(_{(aq)}\) +  5Fe\(^{2+}\)\(_{(aq)}\)  +  8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\)  + Fe\(^{3+}\)\(_{(aq)}\) +  4H\(_2\)O\(_{(l)}\)                 

[ KMnO\(_4\) = 158.0, Fe = 56.0 ]

(b) From your results and information provided, calculate the:

(i) concentration of B in moldm\(^{-3}\)

(ii)concentration of B in gdm\(^{-3}\)

(iii) mass of Fe\(^{2+}\) in 250 cm\(^3\) of B.

(iv) percentage of Fe\(^{2+}\) in the granules.

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Discussions (2)

Ufot1
7 months ago

(ii) Conc. of B in g/dm3 cannot be calculated from 3.8g of Fe because of the presence of impurities.
So, it is calculated thus;
mol/dm3 = g/dm3/molar mass
so g/dm3 = mol/dm3 × molar mass
g/dm3 = 0.07268 × 56
= 4.07 g/dm3 Ans.
To confirm: if 4.07g = 1000cm3
then mass of Fe in 250cm3 = 1.02g

Nwachukwu_Oscar
1 month ago

correct!

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