25ml of a solution of sodium nitrite was acidified then titrated with 0.1mol per litre potassium permanganate. If 12.5ml were required for reaction and the half-equation for nitrite oxidation is:
N02¯ + H20 ? N03¯ + 2H(one positive charge) + 2e¯
What is the concentration of sodium nitrate, in mol per litre to 3 decimal places.
NO2- + H2O ——> NO3- +2H+ +2e-
MnO4- -+ 5e———-> Mn2+
n°eqKMnO4 = n°eqNaNO2
N=z*M z=5 N(KMnO4) = 5*0.1M = 0.5N
neqKMnO4 = 0.5eq/l*0.0125l=0.00625eq=n°eqNaNO2
N(NaNO2) = n°eqNaNO2/V= 0.00625eq/0.025l=0.25N
N=z*M M=N/z z=2 M=0.25/2=0.125M