Problem 82 - General Chemistry

Question

Calculate the concentrations of each ion present in a solution that results from mixing \(50.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{NaClO}_{3}(a q)\) solution with \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\). Assume that the volumes are additive.

Step-by-Step Solution

Answer

[Na⁺] = 0.267 M, [ClO₃⁻] = 0.133 M, [SO₄²⁻] = 0.067 M.

Step 1: Determine Total Volume

The total volume of the mixed solution is the sum of the individual volumes: 50.0 mL + 25.0 mL = 75.0 mL. Since 1 L = 1000 mL, the total volume is 0.075 L.

Step 2: Calculate Initial Moles of Each Compound

For NaClO\(_3\):\[ Moles = 0.20 \, \text{mol/L} \times 0.050 \, \text{L} = 0.010 \, \text{mol} \]For Na\(_2\)SO\(_4\):\[ Moles = 0.20 \, \text{mol/L} \times 0.025 \, \text{L} = 0.005 \, \text{mol} \]

Step 3: Find Concentrations of Na⁺ Ions

NaClO\(_3\) gives 0.010 mol of Na\(^+\) ions per mole and Na\(_2\)SO\(_4\) gives 0.010 mol because each mole of Na\(_2\)SO\(_4\) contributes 2 moles of Na\(^+\) ions: \[ \text{Total moles of Na}^+ = 0.010 + 0.010 = 0.020 \, \text{mol} \]Calculate concentration:\[ [\text{Na}^+] = \frac{0.020}{0.075} = 0.267 \, \text{M} \]

Step 4: Find Concentrations of ClO₃⁻ Ions

NaClO\(_3\) gives 0.010 mol of ClO\(_3\)\(^-\) ions:\[ [\text{ClO}_3^-] = \frac{0.010}{0.075} = 0.133 \, \text{M} \]

Step 5: Find Concentrations of SO₄²⁻ Ions

Na\(_2\)SO\(_4\) gives 0.005 mol of SO\(_4\)\(^{2-}\) ions:\[ [\text{SO}_4^{2-}] = \frac{0.005}{0.075} = 0.067 \, \text{M} \]

Key Concepts

Moles Calculation

To start understanding moles calculation, envision a mole as a basic counting unit used in chemistry akin to a dozen, but for atoms or molecules. Calculating moles helps determine how much of a substance is present in a given solution.

In our exercise, we were tasked with calculating the moles of two different substances, NaClO\(_3\) and Na\(_2\)SO\(_4\). The formula used was:

• \( \text{Moles} = \text{Concentration} \times \text{Volume} \)

Given the concentration and volume of each solution, we performed multiplication to find the number of moles for each solute.

For NaClO\(_3\), \(0.20\) mol/L was multiplied by \(0.050\) L, resulting in \(0.010\) moles. Likewise, for Na\(_2\)SO\(_4\), \(0.20\) mol/L was multiplied by \(0.025\) L, resulting in \(0.005\) moles.

Calculating moles accurately is crucial as it's the foundation for determining various concentrations in a solution.

Ion Concentration

Ion concentration refers to the amount of specific ions present in a solution. This is pivotal in understanding the chemical properties and behaviors of solutions.

In our example exercise, once the moles were calculated, the next step was finding the concentration of each ion. Since each compound dissociates in water releasing ions, we need to consider the dissociation for calculating ion concentration:

• NaClO\(_3\): Dissociates to give \(1\) Na\(^+\) for each NaClO\(_3\)
• Na\(_2\)SO\(_4\): Dissociates to give \(2\) Na\(^+\) ions per molecule of Na\(_2\)SO\(_4\)

Using these, the total moles of Na\(^+\) ions were found to be \(0.020\) mol by adding contributions from both compounds.

To find the concentration of each ion in the new mixture, we used:

• \([\text{Na}^+] = \frac{0.020}{0.075}\) = \(0.267\) M
• \([\text{ClO}_3^-] = \frac{0.010}{0.075}\) = \(0.133\) M
• \([\text{SO}_4^{2-}] = \frac{0.005}{0.075}\) = \(0.067\) M

The accurate calculation of ion concentrations is essential for predicting reactions and properties of the solutions.

Chemistry Solution Mixing

Mixing solutions in chemistry involves combining two or more solutions to form a single solution. An important aspect in solution mixing is the understanding of how volumes and concentrations change.
In this task, when two solutions were mixed, the total solution volume was simply the sum of the two initial volumes, assuming volumes are additive (50.0 mL + 25.0 mL = 75.0 mL or 0.075 L). This is a key assumption needed to simplify calculations.

After mixing, the new concentrations of ions in the solution were calculated based on the total final volume. Just like cooking where ingredients need to be adjusted based on the serving size, in chemistry, concentrations are recalculated taking this new volume into account.

Being mindful of these principles ensures accuracy in experiments and solution preparations, maintaining consistency in lab results and real-world applications.