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 \(0.20 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) solution. Assume that the volumes are additive.

Short answer

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

Step-by-step solution

Calculate the total volume of the solution

When two solutions are mixed, their volumes are considered to add up. The total volume of the mixed solution is the sum of the volumes of the two solutions. Here, you mix \(50.0\, \text{mL}\) of \(\text{NaClO}_3\) with \(25.0\, \text{mL}\) of \(\text{Na}_2\text{SO}_4\). Thus, the total volume \(V_{\text{total}}\) is:\[V_{\text{total}} = 50.0\, \text{mL} + 25.0\, \text{mL} = 75.0\, \text{mL}\]

Calculate moles of each solute in the mixture

Calculate the moles of each solute before mixing. For \(\text{NaClO}_3\):\[\text{moles of } \text{NaClO}_3 = 0.20\, \text{M} \times 0.050\, \text{L} = 0.010\, \text{moles}\]For \(\text{Na}_2\text{SO}_4\):\[\text{moles of } \text{Na}_2\text{SO}_4 = 0.20\, \text{M} \times 0.025\, \text{L} = 0.005\, \text{moles}\]

Calculate concentration of ClO₃⁻ ions

Since the total volume of solution is \(75.0\, \text{mL} = 0.075\, \text{L}\), calculate the concentration of \(\text{ClO}_3^-\):\[[\text{ClO}_3^-] = \frac{\text{moles of } \text{ClO}_3^-}{V_{\text{total}}} = \frac{0.010\, \text{moles}}{0.075\, \text{L}} = 0.133\, \text{M}\]

Calculate concentration of Na⁺ ions

From \(\text{NaClO}_3\), you have \(0.010\, \text{moles of } \text{Na}^+\). From \(\text{Na}_2\text{SO}_4\), since there are 2 \(\text{Na}^+\) ions per formula unit, you have \(2 \times 0.005 = 0.010\, \text{moles of } \text{Na}^+\).The total moles of \(\text{Na}^+\) in the mixed solution is \(0.010 + 0.010 = 0.020\, \text{moles}\).Thus, the concentration of \(\text{Na}^+\) is:\[[\text{Na}^+] = \frac{0.020\, \text{moles}}{0.075\, \text{L}} = 0.267\, \text{M}\]

Calculate concentration of SO₄²⁻ ions

There are \(0.005\, \text{moles of } \text{SO}_4^{2-}\) from \(\text{Na}_2\text{SO}_4\).Thus, the concentration of \(\text{SO}_4^{2-}\) is:\[[\text{SO}_4^{2-}] = \frac{0.005\, \text{moles}}{0.075\, \text{L}} = 0.067\, \text{M}\]

Key concepts

Ion Concentration

When we talk about ion concentration, we're referring to how much of a specific ion is present in a solution. This concentration tells us the strength of that ion in a given volume of liquid. Understanding ion concentration is important in chemistry because ions are often directly involved in the reactions that occur in solutions.

• To find ion concentration, you need to know the moles of the ion and the total volume of the solution.
• The formula for calculating ion concentration is \([\text{Ion Concentration}] = \frac{\text{moles of ion}}{\text{Volume of solution in L}}\).
• This value is expressed as molarity (M), which means moles per liter.

Knowing the concentration of ions helps predict how they will behave in chemical reactions, like whether they will precipitate, remain in solution, or react further with other ions.

Sodium Chlorate

Sodium chlorate (\(\text{NaClO}_3\)) is a chemical compound that dissolves in water to release sodium ions (\(\text{Na}^+\)) and chlorate ions (\(\text{ClO}_3^-\)). In most chemistry scenarios, understanding how sodium chlorate dissociates in a solution is crucial to calculating the ion concentration.

• Sodium chlorate is used in various applications like bleaching and weed control.
• When it dissolves, it separates into its constituent ions: \( \text{NaClO}_3 (s) \rightarrow \text{Na}^+ (aq) + \text{ClO}_3^- (aq)\) .
• This process is known as dissociation, critical for calculating concentrations of individual ions.

In the given exercise, sodium chlorate provided chlorate ions whose concentration was determined as part of the solution calculation steps.

Sodium Sulfate

Sodium sulfate (\(\text{Na}_2\text{SO}_4\)) is another key substance mentioned in the exercise. It is a soluble ionic compound that releases sodium ions (\(\text{Na}^+\)) and sulfate ions (\(\text{SO}_4^{2-}\)) when dissolved in water. Identifying the resulting ion concentrations is essential for understanding the chemical behavior of the solution.

• Sodium sulfate is commonly used in manufacturing detergents and as a drying agent.
• When dissolved, it dissociates completely: \( \text{Na}_2\text{SO}_4 (s) \rightarrow 2 \text{Na}^+ (aq) + \text{SO}_4^{2-} (aq)\).
• For every one unit of sodium sulfate, two sodium ions are produced.

This two-to-one ratio is critical for calculating the concentration of sodium ions in the original exercise.

Molarity Calculation

Molarity refers to the number of moles of a solute present in one liter of solution. Calculating molarity accurately enables chemists to prepare solutions with desired concentrations and understand their potential reactive properties.

• Molarity is denoted by the symbol \(M\) and the formula is \([\text{Molarity}] = \frac{\text{moles of solute}}{\text{Volume of solution in L}}\).
• In the original exercise, this concept helped calculate concentrations of ions from mixed solutions of sodium chlorate and sodium sulfate.
• Keen attention needs to be paid to the volume, as it's considered in liters for these calculations.

Accurate molarity calculations are essential for understanding the strength and reacting behavior of solutions in various chemical processes.