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Chapter 4: Problem 76

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.

Short Answer

Expert verified
\([\text{Na}^+] = 0.267 \, \text{M}, \, [\text{ClO}_3^-] = 0.133 \, \text{M}, \, [\text{SO}_4^{2-}] = 0.067 \, \text{M}\).

Step by step solution

01

Find the total volume of the mixed solution

Add the volumes of the two solutions to determine the total volume.The total volume is:\[ V_{ ext{total}} = 50.0 ext{ mL} + 25.0 ext{ mL} = 75.0 ext{ mL} \]
02

Calculate the moles of each solute before mixing

Calculate the moles of each solute using their given molarities and volumes before they are combined.For \(\text{NaClO}_3\):\[\text{moles of } \text{NaClO}_3 = 0.20 ext{ M} \times 0.050 ext{ L} = 0.010 ext{ moles}\]For \(\text{Na}_2\text{SO}_4\):\[\text{moles of } \text{Na}_2\text{SO}_4 = 0.20 ext{ M} \times 0.025 ext{ L} = 0.005 ext{ moles}\]
03

Calculate concentrations of individual ions after mixing

Now that we have the total volume and moles of each solute, calculate the concentrations of each resulting ion.1. For \(\text{NaClO}_3\), it dissociates into \(\text{Na}^+\) and \(\text{ClO}_3^-\): - Concentration of \(\text{Na}^+\) contributed by \(\text{NaClO}_3\) is: \[ [\text{Na}^+] = \frac{0.010 ext{ moles}}{0.075 ext{ L}} \approx 0.133 ext{ M} \] - Concentration of \(\text{ClO}_3^-\): \[ [\text{ClO}_3^-] = \frac{0.010 ext{ moles}}{0.075 ext{ L}} \approx 0.133 ext{ M} \]2. For \(\text{Na}_2\text{SO}_4\), which dissociates into \(2\text{Na}^+\) and \(\text{SO}_4^{2-}\): - Concentration of \(\text{Na}^+\) contributed by \(\text{Na}_2\text{SO}_4\) is twice the moles of \(\text{Na}_2\text{SO}_4\): \[ [\text{Na}^+] = \frac{0.005 ext{ moles} \times 2}{0.075 ext{ L}} \approx 0.133 ext{ M} \] - Concentration of \(\text{SO}_4^{2-}\): \[ [\text{SO}_4^{2-}] = \frac{0.005 ext{ moles}}{0.075 ext{ L}} \approx 0.067 ext{ M} \]
04

Add contributions to find total ion concentrations

Combine contributions from each solution to find the total concentrations of ions in the solution.- Total \([\text{Na}^+]\) (from both \(\text{NaClO}_3\) and \(\text{Na}_2\text{SO}_4\)): \[ [\text{Na}^+] = 0.133 ext{ M} + 0.133 ext{ M} = 0.267 ext{ M} \]- \([\text{ClO}_3^-]\) stays as: \[ [\text{ClO}_3^-] = 0.133 ext{ M} \]- \([\text{SO}_4^{2-}]\) stays as: \[ [\text{SO}_4^{2-}] = 0.067 ext{ M} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ion Concentration
Ion concentration is crucial in solution chemistry. It refers to the amount of a particular ion in a given volume. We often express this concentration in molarity. In the exercise, we need to calculate the concentration of various ions when two solutions are mixed. Both solutions, containing different salts, dissociate into ions once they are in water. After finding the number of moles of each ion from the given volumes and molarities, we can divide these moles by the total solution volume to get the final concentration of each ion. This helps to determine the chemical environment of the solution, critical in many chemical processes.
  • 狈补颁濒翱鈧� and 狈补鈧係翱鈧� generate ions Na鈦�, ClO鈧冣伝, and SO鈧劼测伝 when in solution.
  • The final concentration of an ion considers all sources contributing that ion.
Dissolution
Dissolution is the process where a solute (like 狈补颁濒翱鈧� or 狈补鈧係翱鈧�) dissolves in a solvent, forming a homogeneous mixture. In the exercise, as each salt dissolves in water, it splits into its constituent ions. Dissolution is essential to fully understand how ions behave in a solution. Each salt dissociates into specific ions
. For 狈补颁濒翱鈧�, the dissolution produces Na鈦� and ClO鈧冣伝 ions. Similarly, 狈补鈧係翱鈧� splits into Na鈦� and SO鈧劼测伝 ions.
  • The ability of a salt to dissolve and dissociate into ions depends on its solubility.
  • Both the quantity of the substance and the total solution volume affect the resultant ion concentration.
Molarity
Molarity is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution. In this exercise, the molarity of the solutions before mixing is set at 0.20 M. That ensures we have precise initial conditions before mixing.
Molarity helps determine the concentration of ions post-dissolution and post-mixing.
  • Moles of a solute calculated from its molarity and volume help find the ion concentration.
  • It is important to maintain consistency in units (liters for volume) to avoid errors in calculations.
Volume Calculation
When calculating the total volume of a solution resulting from mixing, assume the volumes are additive, as instructed in this exercise. Volume calculation is integral in determining the final concentrations of ions, as concentration depends on the number of ions divided by total solution volume. Adding the volumes of the separate solutions, we find the combined volume necessary for further calculations.
In this problem, calculate the total volume by simply adding the initial volumes: 50.0 mL of 狈补颁濒翱鈧� solution and 25.0 mL of 狈补鈧係翱鈧� solution, which makes 75.0 mL.
  • The assumption of additive volumes is crucial for straightforward calculations.
  • Without accurate volume calculation, finding the precise molar concentration post-mixing would be difficult.

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Most popular questions from this chapter

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