Question

A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 22 mL of the KMnO4 solution?

Answer 1

Solution.

$Mn^{7+} + 3e = Mn^{4+}$

$O^{-1} - 1e = O^0$

$Cn(KMnO4) = \frac{C(KMnO4)}{3}$

$m(H2O2) = \frac{Cn(KMnO4) \times V(KMnO4) \times Meq(H2O2)}{1000}$

$m(H2O2) = \frac{0.56 \times 22 \times 34.01}{1000} = 0.42 \ g$

Answer:

m(H2O2) = 0.42 g