Answer to Question #100549 in Physical Chemistry for Jamilu Ibrahim

Question

Manganate ion MnO4- reacts with nitrogen IV oxide gas in an acid medium to give Mn2+ and NO3-. In the determination of NO2 (a pollutant) in a sample of air collected from an agrochemical industry, 10.5cm3 of 4.5 x 10-4 (or 0.00045M) of KMnO4 solution (in acid) was required for equivalence.
(a) Write down a balanced ionic equation for each half reaction above, hence the net ionic equation.
(b) calculate the number of moles of NO2 gas in the air sample
(c) calculate the number of NO2 molecules in the air sample
(d) What percentage of NO2 gas is present in 110.05g air sample collected
(e) Express the quantity of NO2 gas present in the air sample collected in
(i) ppm
(ii) microgram per metre cube

Answer 1

(a) MnO₄^- + 8H⁺ + 5e^- = Mn²⁺ + 4H₂O

NO₂ + H₂O = NO₃^- + 2H⁺ + e^-

MnO₄^- + H₂O + 5NO₂ = Mn²⁺ + 5NO₃^- + 2H⁺

(b) n(KMnO₄) = C x V = 0.00045 mol/L x 0.0105‬ L = 4.725x10^-6 mol

1 mol MnO₄^- - 5 mol NO₂

4.725x10^-6 mol MnO₄^- - x mol NO₂

x = 2.4x10^-5 mol NO₂

(c) N(NO₂) = n(NO₂) x N_a = 2.4x10^-5 x 6.02x10²³ = 14.4x10¹⁸

(d) m(NO₂) = n(NO₂) x Mr = 2.4x10^-5 mol x 46 g/mol = 1.1x10^-3 g

w(NO₂) = 0.0011 / 110.05 x 100% = 0.001 %

(i) w(NO₂) = 0.001 % = 10 ppm

(ii) m(NO₂) = 1.1x10^-3 g = 1100 μg

V(air sample) = m(air sample) / p(air sample) = 110.05 g / 1.293 g/cm³ = 85 cm³

1100 μg / 85 cm³ = 12.9 μg/cm³