Question

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

Short answer

For the first reaction, 15.3 g of NaHSO3 are needed. An additional 1.00 L of the initial NaIO3 solution is needed for the second reaction.

Step-by-step solution

Balance the given equations

The balanced equations are:\n\(\mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}+2\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{5I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) + 6 H^{+} \longrightarrow 3 \mathrm{I}_{2}(\mathrm{s})+3\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\).\n

Calculate molar mass of NaIO3 and NaHSO3

The molecular mass of NaIO3 is \(85.0+15.999 \times 3+15.999 \times 3=197.9 \, g/mol\).\nThe molecular mass of NaHSO3 is \(22.99+1.008+32.06+3 \times 15.999 = 104.06 \, g/mol\).

Find the moles of NaIO3

The number of moles of NaIO3 in the 5.00 L solution can be calculated as \(5.00 L \times \frac{5.80g}{L} \times \frac{1 mol}{197.9 g} = 0.147 mol\).

Calculate the mass of NaHSO3 needed

From stoichiometry of reaction (1), for every 1 mole of IO3- we need 1 mole of HSO3-. Therefore, we need 0.147 moles of NaHSO3. This corresponds to a mass of \(0.147 mol \times 104.06 g/mol = 15.3 g\).

Calculate Additional volume of starting solution needed

For the second reaction, we need 1 mole of IO3- for every 5 moles of I-. In the first step, we produced 0.147 moles of I-, therefore, we need \(0.147 mol/5 = 0.0294 mol\) of IO3-. The volume of the starting solution containing this quantity is given by \(0.0294 mol \times \frac{197.9 g}{mol} \times \frac{L}{5.80 g} = 1.00 L\).

Key concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. In these expressions, reactants and products interact to form a balanced equation. Balance is maintained when the total number of each type of atom on one side mirrors the total on the other. This illustrates the law of conservation of mass, meaning no atoms are lost or gained during the reaction.
Balancing a chemical equation often entails changing the coefficients (the numbers in front of molecules) to achieve an equal number of each type of atom on both sides. For example, in the two-step iodine production process, the equations provided initially were not balanced.

• For the first reaction, the balanced form is \[\mathrm{IO}_3^{−}(aq)+\mathrm{HSO}_3^{−}(aq) \rightarrow \mathrm{I}^{−}(aq) +\mathrm{SO}_4^{2−}(aq)+2\mathrm{H}_2O(l).\]
• For the second reaction, the balanced equation is \[5\mathrm{I}^{−}(aq)+\mathrm{IO}_3^{−}(aq) + 6\ H^{+} \rightarrow 3 \mathrm{I}_2(s)+3\mathrm{H}_2O(l).\]

Balancing these equations precisely is crucial to accurately reflect the stoichiometry—a concept that dictates the relationship between reactants and products in a reaction. This balanced form is employed to calculate the exact amount of each reactant needed or product formed.

Iodine Production

Iodine production from sodium iodate involves a two-step chemical reaction under acidic conditions.
The first step converts iodate ions (\(\mathrm{IO}_3^{-}\)) into iodide ions (\(\mathrm{I}^{-}\)). In this conversion, sodium bisulfite (\(\mathrm{NaHSO}_3\)) is used.

• This reaction forms sulfate ions (\(\mathrm{SO}_4^{2-}\)) as a by-product.
• As each iodate ion converts, one iodide ion and sulfate ion are produced.

Once the iodide ions are generated, the second reaction can commence, which converts iodide ions and leftover iodate ions into elemental iodine (\(\mathrm{I}_2\)). The first reaction is crucial as it provides the substrate for the second, directly impacting the amount of iodine produced.
This process emphasizes the importance of understanding chemical reaction pathways and the interplay between steps to obtain the desired end product.

Molar Mass Calculations

Molar mass calculations play a significant role in stoichiometry. They help determine how much of each substance is involved in a chemical reaction. The molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol).
For sodium iodate (\(\mathrm{NaIO}_3\)), the molar mass is calculated as follows:

• Sodium (Na): 22.99 g/mol
• Iodine (I): 126.90 g/mol
• Oxygen (O): 15.999 g/mol, with three atoms totaling 47.997 g/mol

The total molar mass of \(\mathrm{NaIO}_3\) is thus:\[22.99\ g/mol + 126.90\ g/mol + 47.997\ g/mol = 197.89\ g/mol.\]
Similarly, for sodium bisulfite (\(\mathrm{NaHSO}_3\)), the constituents' molar masses include:

• Sodium (Na): 22.99 g/mol
• Hydrogen (H): 1.008 g/mol
• Sulfur (S): 32.06 g/mol
• Oxygen (O): 15.999 g/mol, with three atoms adding to 47.997 g/mol

This gives \(\mathrm{NaHSO}_3\) a total molar mass of:\[22.99\ g/mol + 1.008\ g/mol + 32.06\ g/mol + 47.997\ g/mol = 104.055\ g/mol.\]
Knowing these values allows for accurate calculations of moles in reactions and the quantities of reactants or products.