In the basic solution MnO4^- oxidizes NO2^- to NO3^- and is reduced to MnO2.

Question

Calculate the volume of 0.10 M KMnO4 solution that would be required to oxidize 30 mL of 0.10 M NaNO2

Answer 1

N(KMnO4) = 3*M(KMnO4)

MnO4-  +  3e-  =  MnO2  (+7 +4)

N(KMnO4) = 0.1M*3 =0.3N

N(NaNO2)= 2*M(NaNO2)

NO2-  -------->NO3- +2e- (+3 +5)

N(NaNO2 )= 2*0.1M= 0.2N

nEq(KMnO4)=nEq(NaNO2)

Vx*N(KMnO4) = V(NaNO2) * N(NaNO2)

Vx =  V(NaNO2)*N(NaNO2)/N(KMnO4) = 0.030l*0.2N/0.3N= 0,02l=20ml

Answer 2

MnO4-  +  3e-   + 2H2O  -->  MnO2   + 4OH-

NO2-  +  2OH-  -->  NO3- + 2e-  +  H2O

mequ(MnO4-) = mequ(NO2-)

=

mmoles(MnO4-)*3 = mmoles(NO2-)*2

=

XmL * 0.10M * 3 = 30mL * 0.10M * 2

then

X = 30mL * 0.10M * 2 / (0.10M * 3) =

= 20mL