Question

A sample of wastewater is analyzed for IO3 ions. To a 10 mL sample of the water, 10 mL of 0.02 M NaHSO3 and a drop of starch are added. It takes 532 seconds for the indicator to change color. What is the Concentration of IO3 in the wastewater? K= -40.9

Answer 1

The balanced chemical equations for the reactions taking place are

IO₃^- + 3 HSO₃^- ---------> I^- + 3 SO₄^2- + 3 H⁺ ……(1)

IO₃^- + 5 I^- + 6 H⁺ --------> 3 I₂ + 3 H₂O …….(2)

I₂ + HSO₃^- + H₂O -------> 2 I^- + HSO₄^- + 2 H⁺ ……(3)

HSO₃^- reacts with IO₃^- to generate I₂ that reacts with HSO₃^- to re-generate I^-. HSO₃^- is completely consumed in the reaction.

As per stoichiometric equation (1),

1 mole IO₃^- = 3 moles HSO₃^-.

Since HSO₃^- is completely consumed in the reaction, hence, moles HSO₃^- used in the reaction = (initial moles of HSO₃^-) – (final moles of HSO₃^-)

The initial number of millimole(s) of HSO₃^- = (10 mL)*(0.02 M) = 0.2 mmole.

Millimole(s) of IO₃^- = (millimoles of HSO₃^-)*(1 mole IO₃^-)/(3 moles HSO₃^-) = 0.067 mmole.

Concentration of IO₃^- = (0.067 mmole)/(10 mL) = 0.0067 mmol/mL = 0.0067 M (ans).