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Chapter 4: Problem 82

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

Expert verified
[ClO鈧冣伝] = 0.133 M, [Na鈦篯 = 0.267 M, [SO鈧劼测伝] = 0.067 M.

Step by step solution

01

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:\[V_{\text{total}} = 50.0\, \text{mL} + 25.0\, \text{mL} = 75.0\, \text{mL}\]
02

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 \(\text{Na}_2\text{SO}_4\):\[\text{moles of } \text{Na}_2\text{SO}_4 = 0.20\, \text{M} \times 0.025\, \text{L} = 0.005\, \text{moles}\]
03

Calculate concentration of ClO鈧冣伝 ions

Since the total volume of solution is \(75.0\, \text{mL} = 0.075\, \text{L}\), calculate the concentration of \(\text{ClO}_3^-\):\[[\text{ClO}_3^-] = \frac{\text{moles of } \text{ClO}_3^-}{V_{\text{total}}} = \frac{0.010\, \text{moles}}{0.075\, \text{L}} = 0.133\, \text{M}\]
04

Calculate concentration of Na鈦� 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:\[[\text{Na}^+] = \frac{0.020\, \text{moles}}{0.075\, \text{L}} = 0.267\, \text{M}\]
05

Calculate concentration of SO鈧劼测伝 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:\[[\text{SO}_4^{2-}] = \frac{0.005\, \text{moles}}{0.075\, \text{L}} = 0.067\, \text{M}\]

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Key Concepts

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

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.

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Most popular questions from this chapter

Describe in words how you would do each of the following preparations. Then give the molecular equation for each preparation. (a) \(\mathrm{CuCl}_{2}(s)\) from \(\mathrm{CuSO}_{4}(s)\) (b) \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(s)\) from \(\mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{NaNO}_{3}(s)\) from \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) (d) \(\mathrm{MgCl}_{2}(s)\) from \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

Elemental bromine is the source of bromine compounds. The element is produced from certain brine solutions that occur naturally. These brines are essentially solutions of calcium bromide that, when treated with chlorine gas, yield bromine in a displacement reaction. What are the molecular equation and net ionic equation for the reaction? A solution containing \(40.0 \mathrm{~g}\) of calcium bromide requires \(14.2 \mathrm{~g}\) of chlorine to react completely with it, and \(22.2 \mathrm{~g}\) of calcium chloride is produced in addition to whatever bromine is obtained. How many grams of calcium bromide are required to produce 10.0 pounds of bromine?

A metal, M, was converted to the sulfate, \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). Then a solution of the sulfate was treated with barium chloride to give barium sulfate crystals, which were filtered off. \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{BaCl}_{2}(a q) \longrightarrow\) $$ 2 \mathrm{MCl}_{3}(a q)+3 \mathrm{BaSO}_{4}(s) $$ If \(1.200 \mathrm{~g}\) of the metal gave \(6.026 \mathrm{~g}\) of barium sulfate, what is the atomic weight of the metal? What is the metal?

Nickel(II) sulfate solution reacts with sodium hydroxide solution to produce a precipitate of nickel(II) hydroxide and a solution of sodium sulfate. Write the molecular equation for this reaction. Then write the corresponding net ionic equation.

Write the molecular equation and the net ionic equation for the reaction of solid iron(II) sulfide and hydrochloric acid. Add phase labels.
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