Question

1.104 of impure sample of NaNO2 was acidified and analysed using excess iodide ion. Then I3- was titrated with 0.300 M Na2S2O3
If the titre volume is 29.25cm cube, calculate the % of NaNO2 in the sample

Answer 1

Task:

1.104 of impure sample of NaNO₂ was acidified and analysed using excess iodide ion. Then I₃^- was titrated with 0.300 M Na₂S₂O₃. If the titre volume is 29.25 cm³, calculate the % of NaNO₂ in the sample

Solution:

The amount of NaNO₂ in a sample can be determined by acidifying the solution to form nitrous acid, HNO₂, allowing the nitrous acid to react with an excess of iodide ion, and then titrating the triiodide ion (I₃^–) in the resulting solution with thiosulfate (S₂O₃^2–) solution.

NaNO₂ + H⁺ = HNO₂ + Na⁺

The balanced redox equations are:

(1) 2H⁺ + 2HNO₂ + 3I^- = 2NO + I₃^- + H₂O

(2) I₃^- + 2S₂O₃^2- = 3I^- + S₄O₆^2-

According to the reaction equations:

n(HNO₂) / 2 = n( I₃^-) = n(S₂O₃^2-) / 2

n(HNO₂) = n(S₂O₃^2-) = C(S₂O₃^2-)*V(S₂O₃^2-);

n(HNO₂) = C(S₂O₃^2-)*V(S₂O₃^2-) = 0.300 M * 0.02925 L = 0.008775 mol;

n(NaNO₂) = n(HNO₂) = 0.008775 mol;

n(NaNO₂) = m(NaNO₂) / M(NaNO₂)

M(NaNO₂) = Ar(Na) + Ar(N) + 2*Ar(O) = 23 + 14 + 2+16 = 69 g/mol;

m(NaNO₂) = n(NaNO₂) * M(NaNO₂) = 0.008775 mol * 69 g/mol = 0.6055 g

w(NaNO₂) = [m(NaNO₂) / m(sample)]*100% = [0.6055 / 1.104] * 100% = 54.85%

w(NaNO₂) = 54.85%

Answer: 54.85% NaNO₂ in the sample.