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.
Explanation
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Discussions (2)

(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


