Question

Iodine reacts with thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), to give iodide ion and the tetrathionate ion, \(\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}\). (a) Write a balanced net ionic equation for the reaction. (b) If \(25.0 \mathrm{~g}\) of iodine is dissolved in enough water to make \(1.50 \mathrm{~L}\) of solution, what volume of \(0.244 M\) sodium thiosulfate will be needed for complete reaction?

Short answer

Answer: The balanced net ionic equation for the reaction is: 2 I⁻(aq) + S₄O₆²⁻(aq) → I₂(s) + 2 S₂O₃²⁻(aq). To react completely with 25.0 g of iodine, 0.807 L of a 0.244 M sodium thiosulfate solution is required.

Step-by-step solution

Write the balanced net ionic equation

The reaction between iodine and thiosulfate ion is as follows: \(\displaystyle \textrm{I}_{2}( s) +\textrm{ S}_{2}\textrm{O}_{3}^{2-}( aq) \rightarrow \textrm{I}^{-}( aq) +\textrm{ S}_{4}\textrm{O}_{6}^{2-}( aq)\) Now we need to balance the equation. We see that there are 2 iodine atoms and 2 sulfur atoms on each side, but we have 3 oxygen atoms on the left side and 6 oxygen atoms on the right side. To balance the oxygens, we can use 2 thiosulfate ions on the left side. Now the balanced net ionic equation is: \(\displaystyle 2 \textrm{I}^{-}( aq) +\textrm{ S}_{4}\textrm{O}_{6}^{2-}( aq) \rightarrow \textrm{I}_{2}( s) + 2\textrm{ S}_{2}\textrm{O}_{3}^{2-}( aq)\)

Calculate moles of iodine

We are given 25.0 grams of iodine and we want to calculate the moles of iodine. To do this, we will use iodine's molar mass, which is 253.8 g/mol. Moles of iodine = \(\frac{\textrm{Mass of iodine}}{\textrm{Molar mass of iodine}} = \frac{25.0 \textrm{g}}{253.8 \textrm{g/mol}} = 0.0985 \textrm{mol}\)

Use stoichiometry to find moles of sodium thiosulfate

According to the balanced net ionic equation, 1 mole of iodine will react with 2 moles of thiosulfate ions. So, we must find out how many moles of thiosulfate ions are required to react with 0.0985 moles of iodine. Moles of sodium thiosulfate = moles of iodine \(\times\) \(\frac{\textrm{Moles of sodium thiosulfate}}{\textrm{Moles of iodine}}\) = \(0.0985 \textrm{mol} \times \frac{2 \textrm{mol}}{1 \textrm{mol}} = 0.197 \textrm{mol}\)

Calculate the volume of sodium thiosulfate solution needed

We are given the molarity (M) of the sodium thiosulfate solution, which is 0.244 M. We can use this information along with the moles of sodium thiosulfate calculated in the previous step to find the required volume of the solution. Volume of sodium thiosulfate solution = \(\frac{\textrm{Moles of sodium thiosulfate}}{\textrm{Molarity}} = \frac{0.197 \textrm{mol}}{0.244 \textrm{M}} = 0.807 \textrm{L}\) The required volume of 0.244 M sodium thiosulfate solution to react completely with 25.0 g of iodine is 0.807 L.

Key concepts

Net Ionic Equation

In chemical reactions, especially those that occur in aqueous solutions, we often come across the concept of net ionic equations. These are simplified versions of chemical equations that only include the species that change during the reaction. They help us focus on the essential parts of a chemical reaction, ignoring the spectator ions that do not participate directly.
To write a balanced net ionic equation, we start by writing the chemical formulas of the reactants and products, ensuring that all ions and elements are correctly represented. Next, we need to balance the number of atoms of each element and the charge on both sides of the equation.
In the iodine and thiosulfate reaction - \[ \mathrm{I_2}(s) + 2 \mathrm{S_2O_3^{2-}}(aq) \rightarrow 2 \mathrm{I^-}(aq) + \mathrm{S_4O_6^{2-}}(aq) \] - notice that the equation involves solid iodine and aqueous ions. Here, iodine is reduced to iodide ions and thiosulfate ions are oxidized to tetrathionate ions. Each side of the equation has the same number of each type of atom and the charge is balanced.

Stoichiometry

Stoichiometry is like a bridge that connects the quantities of reactants and products in a chemical reaction. It's essential for predicting how much of each substance is involved or produced. In this exercise, stoichiometry helps us calculate the amount of sodium thiosulfate needed to fully react with iodine.
We start by examining the balanced net ionic equation. From the equation \[ \mathrm{I_2}(s) + 2 \mathrm{S_2O_3^{2-}}(aq) \rightarrow 2 \mathrm{I^-}(aq) + \mathrm{S_4O_6^{2-}}(aq) \] we understand that 1 mole of iodine reacts with 2 moles of thiosulfate ions. This stoichiometric ratio is critical in calculating the amounts needed for the reaction.
Once we know how many moles of iodine we have (as calculated previously), using the stoichiometric ratio, we can find the moles of thiosulfate ions required by simply multiplying the moles of iodine by the ratio of thiosulfate to iodine.

Moles Calculation

Calculating moles is a foundational skill in chemistry that allows us to convert between mass and quantity in a reaction. Moles express amounts of a chemical substance that contains as many entities (atoms, molecules, etc.) as there are atoms in 12 grams of pure carbon-12. It's essential for understanding the proportion of substances involved in chemical reactions.
In our problem, we are given 25.0 grams of iodine. The first step is converting this mass into moles, using the molar mass of iodine, which is 253.8 g/mol. This molar mass acts as a bridge from grams to moles, calculated with the formula:\[\text{Moles of iodine} = \frac{\text{Mass of iodine}}{\text{Molar mass of iodine}} = \frac{25.0 \text{ g}}{253.8 \text{ g/mol}}\]This conversion allows us to know exactly how many moles of iodine are present, which can then be used to find out how many moles of other substances react with it.

Molarity and Solution Preparation

Molarity is a way of expressing the concentration of a solute in a solution. It tells us how many moles of a solute are present in one liter of solution. This concept is crucial for solution preparation and determining the volume of solution needed for reactions.
In this exercise, we know the molarity of the sodium thiosulfate solution is 0.244 M. Molarity is calculated using the formula:\[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]To find the volume of sodium thiosulfate required, we rearrange the formula to solve for volume:\[\text{Volume of sodium thiosulfate solution} = \frac{\text{Moles of sodium thiosulfate}}{\text{Molarity}}\]Substituting the moles of thiosulfate obtained from stoichiometry into this formula gives us the volume of solution necessary to ensure all the iodine reacts completely. This calculation shows how understanding molarity helps in preparing and using solutions effectively in chemistry.