Chapter 5, Problem 100RQ

One way to analyze a sample for nitrite ion is to acidify a solution containing ${\text{NO}}_{\text{2}}{}^{\text{-}}$ and then allow the ${\text{HNO}}_{\text{2}}$ that is formed to react with iodide ion in the presence of excess ${\text{I}}^{\text{-}}$. The reaction is

${\text{2HNO}}_{\text{2}}+2{\text{H}}^{\text{+}}+3{\text{I}}^{\text{-}}\to 2{\text{NO + 2H}}_{\text{2}}{\text{O + I}}_{\text{3}}{}^{\text{-}}$

Then the ${\text{I}}_{\text{3}}{}^{\text{-}}$ is titrated with ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ solution using starch as an indicator.

${\text{I}}_{\text{3}}{}^{\text{-}}{\text{ + 2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\to 3{\text{I}}^{\text{-}}+{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}$

In a typical analysis, a 1.104 g sample that was known to contain ${\text{NaNO}}_{\text{2}}$ was treated as described above. The titration required 29.25 mL of 0.3000 M ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ solution to reach the end point.

(a) How many moles of ${\text{I}}_{\text{3}}{}^{\text{-}}$ had been produced in the first reaction?

(b) How many moles of ${\text{NO}}_{\text{2}}{}^{\text{-}}$ had been in the original 1.104 g sample?

(c) What was the percentage by mass of ${\text{NaNO}}_{\text{2}}$ in the original sample?