Question

What is the molar concentration of As(III) in a solution if \(22.35 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KBr} \mathrm{O}_{3}\) is needed for complete reaction with \(50.00 \mathrm{~mL}\) of the As(III) solution? The balanced equation is: \(3 \mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\mathrm{BrO}_{3}^{-}(a q) \longrightarrow \mathrm{Br}^{-}(a q)+3 \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)\)

Short answer

0.1341 M As(III).

Step-by-step solution

Understand the Chemical Reaction

The balanced chemical equation for the reaction is: \[3 \text{H}_{3}\text{AsO}_{3}(aq) + \text{BrO}_{3}^{-}(aq) \longrightarrow \text{Br}^{-}(aq) + 3 \text{H}_{3}\text{AsO}_{4}(aq)\]This equation indicates that 1 mole of \(\text{BrO}_{3}^{-}\) reacts with 3 moles of H\(_{3}\)AsO\(_{3}\).

Calculate Moles of \(\text{BrO}_{3}^{-}\)

First, calculate the moles of \(\text{BrO}_{3}^{-}\) in the KBrO\(_3\) solution. Use the formula for moles:\[ \text{Moles} = \text{Concentration (M)} \times \text{Volume (L)} \]Given:- Concentration of KBrO\(_3\) \((\text{BrO}_{3}^{-})\): 0.100 M- Volume: 22.35 mL = 0.02235 LSo, Moles of \(\text{BrO}_{3}^{-}\) = 0.100 M \(\times\) 0.02235 L = 0.002235 \text{ moles}.

Determine Moles of As(III)

From the balanced chemical equation, 1 mole of \(\text{BrO}_{3}^{-}\) reacts with 3 moles of H\(_{3}\)AsO\(_{3}\). Thus, the moles of As(III) is:\[ \text{Moles of As(III)} = 3 \times \text{Moles of } \text{BrO}_{3}^{-} \]\[= 3 \times 0.002235 \text{ moles} = 0.006705 \text{ moles} \]

Calculate Molarity of As(III) Solution

The molarity of a solution is given by:\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} \]Here, we have 0.006705 moles of As(III) in 50.00 mL of solution, which is 0.05000 L.So, the molarity is:\[ \text{Molarity of As(III)} = \frac{0.006705}{0.05000} = 0.1341 \text{ M} \]

Key concepts

Chemical Reaction Stoichiometry

When dealing with chemical reactions, understanding the stoichiometry is crucial. Stoichiometry is the quantitative relationship between reactants and products in a chemical equation. It tells us how much of each substance is involved or produced. In this context, the balanced equation for the chemical reaction between H\(_3\)AsO\(_3\) and BrO\(_3^-\) states that it requires 1 mole of BrO\(_3^-\) to fully react with 3 moles of H\(_3\)AsO\(_3\). This translates to the stoichiometric coefficients in the equation – telling us the exact proportions of reactants needed for the reaction to go to completion.Understanding these ratios ensures that we can calculate the exact amount of As(III) necessary in the reaction. This means for every mole of BrO\(_3^-\) used, 3 times that amount of As(III) is consumed. By analyzing this relationship, we apply stoichiometry to solve many chemical problems effectively, especially when determining unknown concentrations in a solution.

Moles Calculation

The concept of moles is central in chemistry for quantifying matter at the molecular scale. In the given exercise, we calculate the moles of BrO\(_3^-\) as follows:

• First, we use the formula: \[ \text{Moles} = \text{Concentration (M)} \times \text{Volume (L)} \]This will give us the number of moles based on the molarity and volume of the solution.
• In the exercise, we are given a 0.100 M solution of BrO\(_3^-\) and the volume used was 22.35 mL. Convert the volume into liters by dividing by 1000, giving 0.02235 L.
• Now, compute the moles: \[ 0.100 \, \text{M} \times 0.02235 \, \text{L} = 0.002235 \, \text{moles of BrO}_{3}^{-} \]

These calculated moles represent how much BrO\(_3^-\) participated in the reaction. Our next step would be to apply these findings using stoichiometry to determine the moles of other reactants, specifically As(III), involved in the reaction. This is crucial to solving the problem tied to molar concentrations.

Solution Molarity

Solution molarity measures the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. In this exercise, after calculating the moles of As(III), we determine its molarity as follows:

• From the stoichiometry in the chemical equation-based calculation, we know the moles of As(III):\[ 3 \times 0.002235 = 0.006705 \, \text{moles of As(III)} \]
• Next, we'll calculate the molarity using:\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} \]
• For As(III) in 50.00 mL or 0.05000 L of solution, the molarity is:\[ \frac{0.006705}{0.05000} = 0.1341 \, \text{M} \]

This result tells us the molar concentration of As(III) in the solution. Molarity is a convenient way to express concentration, particularly in chemical reactions, since it directly correlates with the amounts needed for reactions to occur under specific conditions.