Question

Can anyone give readings for titration process of M/10 standard oxalic acid solution to determine concentration and strength of KMnO4 solution?

Answer 1

Answer:

here it is. Reading for the experiment determination of conc. in molarity of kmno4 by titrating it against a std soln of oxalic acid.

Answer 2

Answer:

A  $\frac{M}{10}$  standard oxalic acid solution will react with 0.04 moles of KMnO₄.

Explanation:

The general reaction between an oxalic acid solution with KMnO₄ is given as,

$5C_2O_4^-^2+2MnO^-_4+16H^+\rightarrow 2Mn^+^2+8H_2O+10CO_2$        (1)

Where,

C₂O₄⁻²=Oxalate ions

MnO₄⁻=Permagnate ions

H⁺=hydrogen ions

Mn⁺²=Manganese(II) ions

H₂O=water

CO₂=carbon dioxide

From reaction (1) we can see that 5 mole(M) of oxalic acid reacts with 2 moles of permanganate ions.

$5M \ C_2O^-^2_4= 2 \ moles\ MnO^-_4$

$1M \ C_2O^-^2_4= \frac{2}{5} \ moles\ MnO^-_4$     (2)

So,$\frac{M}{10}$ C₂O₄⁻² reaction can be derived from equation (2) which is given as,

$\frac{M}{10} \ C_2O^-^2_4= \frac{2}{5\times 10} \ moles\ MnO^-_4$

$\frac{M}{10} \ C_2O^-^2_4= 0.04 \ moles\ MnO^-_4$

Hence,a $\frac{M}{10}$ standard oxalic acid solution will react with 0.04 moles of KMnO₄.