Question

Over time, hydrogen peroxide, H₂O₂, degrades into water and oxygen gas.  A bottle of hydrogen peroxide is expired and you need to determine the concentration.   A titration is performed using the following equation:

3 H₂O_2(aq) + 2 NaMnO_4(aq) → 3 O_2(g) + 2 MnO_2(aq) + 2 NaOH_(aq) + 2 H₂O_(l)

Note that permanganate ion, MnO₄^-, is purple, while manganese (IV) ion, Mn⁴⁺, is colorless.

A quantity of 683 mL of 3.44 M NaMnO₄ was measured and placed in a beaker. The solution was dark purple. The unknown concentration of H₂O₂ was dripped into the beaker drop by drop until the solution was colorless. A total of 39.48 mL was added before the purple color disappeared.  

What is the molarity of the original hydrogen peroxide solution?

Answer 1

3 H₂O_2(aq) + 2 NaMnO_4(aq) → 3 O_2(g) + 2 MnO_2(aq) + 2 NaOH_(aq) + 2 H₂O_(l)

3 moles        2 moles

NaMnO4                                                      H2O2

M1   = 3.44M                                              M2 =

V1   = 683ml                                               V2 = 39.48ml

n1   = 2                                                      n2 = 3

                  M1V1/n1    =    M2V2/n2

                          M2      = M1V1n2/n1V2

                                    = 3.44*683*3/(2*39.48)    = 89.3M >>>>answer

the molarity of the original hydrogen peroxide solution   = 89.3M >>>>answer