Question

A gaseous mixture was passed at the rate of 2.5 L/min through a solution of NaOH for a total of 1 hour. The SO₂ in the mixture was retained as sulphite ion:

\[\ce{SO2(g) + 2OH- -> SO3^2 - (g) + H2O(l)}\]

After acidification with HCl the sulphite was titrated with 5 ml of 0.003 M KIO₃

\[\ce{IO_3^- + 2H2SO3 + 2Cl- -> 2SO^{2-}_4 + ICl^-_2 + 2H+ + H2O}\]

The concentration of SO₂ will be ______ ppm if density of gaseous mixture is 1.6 gm/L.

विकल्प

• 2

• 4

• 8

• 6

Answer 1

The concentration of SO₂ will be 8 ppm if density of gaseous mixture is 1.6 gm/L.

Explanation:

\[\ce{SO2(g) + 2OH- -> SO3^{2-} (g) + H2O(l)}\]

\[\ce{IO_3^- + 2H2SO3 + 2Cl- -> 2SO^{2-}_4 + ICl^-_2 + 2H+ + H2O}\]

Millimoles of \[\ce{IO^-3}\] consumed = 5 × 0.003 ⇒ 0.015

Millimoles of \[\ce{SO^{2–}_3}\] will be = 0.015 × 2 = 0.03

Hence millimoles of SO₂ = 0.03

ppm of SO₂ = `[(0.03 xx 10^-3 xx 64)/(2.5 xx 60 xx 1.6) xx 10^6]`

= 8 ppm