Question

What is the molar concentration of \(\mathrm{Fe}^{2+}\) ion in an aqueous solution if \(31.50 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KBrO}_{3}\) is required for complete reaction with \(10.00 \mathrm{~mL}\) of the \(\mathrm{Fe}^{2+}\) solution? The net ionic equation is: $$ 6 \mathrm{Fe}^{2+}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 6 \mathrm{Fe}^{3+}(a q)+\mathrm{Br}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$

Short answer

The molar concentration of \( \text{Fe}^{2+} \) in the solution is 1.9845 M.

Step-by-step solution

Identify the Net Ionic Equation

The provided net ionic equation is: \[ 6 \text{Fe}^{2+}(aq) + \text{BrO}_3^{-}(aq) + 6 \text{H}^{+}(aq) \rightarrow 6 \text{Fe}^{3+}(aq) + \text{Br}^{-}(aq) + 3 \text{H}_2\text{O}(l) \] This equation shows that 6 moles of \( \text{Fe}^{2+} \) react with 1 mole of \( \text{BrO}_3^{-} \).

Calculate Moles of \( \text{BrO}_3^{-} \) Used

Use the molarity formula to calculate the moles of \( \text{BrO}_3^{-} \): \[ \text{molarity} = \frac{\text{moles}}{\text{liters}} \] Given that the volume of \( \text{KBrO}_3 \) solution is 31.50 mL (or 0.03150 L) and its molarity is 0.105 M, moles of \( \text{BrO}_3^{-} \) are:\[ \text{moles of } \text{BrO}_3^{-} = 0.105 \times 0.03150 \] Calculating gives: \[ 0.0033075 \text{ moles} \]

Use Stoichiometry to Find Moles of \( \text{Fe}^{2+} \)

From the balanced equation, we know the ratio of \( \text{Fe}^{2+} \) to \( \text{BrO}_3^{-} \) is 6:1. Therefore, the moles of \( \text{Fe}^{2+} \) is: \[ 6 \times 0.0033075 = 0.019845 \text{ moles} \]

Calculate Molar Concentration of \( \text{Fe}^{2+} \)

To find the molarity (M) of the \( \text{Fe}^{2+} \) solution, use the formula:\[ \text{molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]The moles of \( \text{Fe}^{2+} \) is 0.019845 and the volume of the \( \text{Fe}^{2+} \) solution is 10.00 mL (or 0.01000 L):\[ \text{molarity of } \text{Fe}^{2+} = \frac{0.019845}{0.01000} = 1.9845 \text{ M} \]

Review and Verify

Check everyone of the previous step to ensure that calculations are accurate, especially the conversion of units, the stoichiometry method used, and the division when calculating molarity. Confirm that the final answer seems reasonable given the context of moles and molarities involved.

Key concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that deals with the quantitative relationships between the reactants and products in a balanced chemical equation. In a chemical reaction, stoichiometry involves using these ratios to calculate the amounts of reactants and products.

It is like a recipe that tells us exactly how much of each ingredient we need to get the intended final product. For example, in the exercise mentioned, the net ionic equation illustrates a stoichiometric relationship where 6 moles of \( \text{Fe}^{2+} \) ions react with 1 mole of \( \text{BrO}_3^- \). This specific ratio is essential for calculating the correct amount of \( \text{Fe}^{2+} \) needed to react completely with the given amount of \( \text{BrO}_3^- \).

In practice, stoichiometry requires careful balancing of the chemical equation first. Then, you determine the mole ratio - a core step for calculating further quantities of substances involved. Using this ratio, you can predict how much product will form or how much of a reactant is required.

Net Ionic Equation

A net ionic equation simplifies a chemical reaction by only showing the species that directly participate in the reaction. It excludes the spectator ions, which are present in the solution but do not play a direct role in the reaction process.

In the problem exercise, the net ionic equation is: \[ 6 \text{Fe}^{2+}(aq) + \text{BrO}_3^-(aq) + 6 \text{H}^+(aq) \rightarrow 6 \text{Fe}^{3+}(aq) + \text{Br}^-(aq) + 3 \text{H}_2\text{O}(l)\]

This equation highlights the transformation of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \), the consumption of \( \text{BrO}_3^- \), and the generation of \( \text{Br}^- \) ions.

Net ionic equations are powerful for focusing on actual chemical changes. They help us simplify a balanced chemical reaction to its core components, making stoichiometric calculations easier. This is particularly useful in titrations or any reaction involving solutions, as it helps in tracking the actual changes occurring in the solution.

Moles Calculation

The concept of moles calculation is fundamental in chemistry as it connects the microscopic world of atoms and molecules to the macroscopic world that we can measure. The mole is a unit that represents \( 6.022 \times 10^{23} \) particles, ions, or molecules and helps us simplify calculations involving atoms and molecules.

In the provided exercise, the mole concept is used twice. First, to calculate the moles of \( \text{BrO}_3^- \) by using its molarity formula: \( \text{moles} = \text{molarity} \times \text{volume in liters} \). Given the molarity of \( \text{KBrO}_3 \) is 0.105 M, and the volume is 0.03150 L, the calculation becomes: \[ \text{moles of } \text{BrO}_3^- = 0.105 \times 0.03150 = 0.0033075 \text{ moles} \]

The stoichiometry of the reaction tells us that 6 moles of \( \text{Fe}^{2+} \) react with 1 mole of \( \text{BrO}_3^- \). Therefore, the moles of \( \text{Fe}^{2+} \) is calculated as: \[ 6 \times 0.0033075 = 0.019845 \text{ moles} \]

Finally, to find the molar concentration of \( \text{Fe}^{2+} \), we use: \( \text{molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). With 0.019845 moles of \( \text{Fe}^{2+} \) in a 0.01000 L solution, the molarity is \[ \frac{0.019845}{0.01000} = 1.9845 \text{ M} \].

Understanding moles calculations allows us to translate volumes and concentrations back into the number of molecules involved in a reaction, making this concept vital for practical chemistry applications.