Problem 32 - Chemistry An Atoms First Approach

Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.0200 mole of sodium phosphate in \(10.0 \mathrm{mL}\) of solution b. 0.300 mole of barium nitrate in \(600.0 \mathrm{mL}\) of solution c. \(1.00 \mathrm{g}\) of potassium chloride in \(0.500 \mathrm{L}\) of solution d. \(132 \mathrm{g}\) of ammonium sulfate in \(1.50 \mathrm{L}\) of solution

Step-by-Step Solution

Answer

a. \(\mathrm{Na^+}\) ions: \(6.00 \mathrm{M}\), \(\mathrm{PO_4^{3-}}\) ions: \(2.00 \mathrm{M}\) b. \(\mathrm{Ba^{2+}}\) ions: \(0.500 \mathrm{M}\), \(\mathrm{NO_3^{-}}\) ions: \(1.00 \mathrm{M}\) c. \(\mathrm{K^+}\) ions: \(0.0268 \mathrm{M}\), \(\mathrm{Cl^{-}}\) ions: \(0.0268 \mathrm{M}\) d. \(\mathrm{NH_4^+}\) ions: \(1.33 \mathrm{M}\), \(\mathrm{SO_4^{2-}}\) ions: \(0.6659 \mathrm{M}\)

Step 1: Identify ions and stoichiometry

Sodium phosphate dissociates into ions as follows: \(\mathrm{Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-}}\). So, we have 3 moles of \(\mathrm{Na^+}\) and 1 mole of \(\mathrm{PO_4^{3-}}\) for every mole of sodium phosphate in the solution.

Step 2: Calculate the molarity of the electrolyte

To find the molarity, we can use the formula: \(M = \frac{mol}{L}\). First, convert the volume of the solution to liters: \(10.0 \mathrm{mL}=0.0100 \mathrm{L}\). Now, we can calculate the molarity of sodium phosphate: \(M = \frac{0.0200}{0.0100} = 2.00 \mathrm{M}\).

Step 3: Find the concentration of each ion using stoichiometry

According to the stoichiometry of the dissociation, we have 3 moles of \(\mathrm{Na^+}\) for every mole of sodium phosphate. Therefore, the concentration of \(\mathrm{Na^+}\) ions is: \(3 \times 2.00 = 6.00 \mathrm{M}\). Similarly, the concentration of \(\mathrm{PO_4^{3-}}\) ions is: \(1 \times 2.00 = 2.00 \mathrm{M}\). b. 0.300 mole of barium nitrate in 600.0 mL of solution

Step 1: Identify ions and stoichiometry

Barium nitrate dissociates into ions as follows: \(\mathrm{Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^{-}}\). We have 1 mole of \(\mathrm{Ba^{2+}}\) and 2 moles of \(\mathrm{NO_3^{-}}\) for every mole of barium nitrate in the solution.

Step 2: Calculate the molarity of the electrolyte

First, convert the volume of the solution to liters: \(600.0 \mathrm{mL}=0.600 \mathrm{L}\). Then, calculate the molarity of barium nitrate: \(M = \frac{0.300}{0.600} = 0.500 \mathrm{M}\).

Step 3: Find the concentration of each ion using stoichiometry

The concentration of \(\mathrm{Ba^{2+}}\) ions is: \(1 \times 0.500 = 0.500 \mathrm{M}\). The concentration of \(\mathrm{NO_3^{-}}\) ions is: \(2 \times 0.500 = 1.00 \mathrm{M}\). c. 1.00 g of potassium chloride in 0.500 L of solution

Step 1: Identify ions and stoichiometry

Potassium chloride dissociates into ions as follows: \(\mathrm{KCl \rightarrow K^+ + Cl^{-}}\). We have 1 mole of \(\mathrm{K^+}\) and 1 mole of \(\mathrm{Cl^{-}}\) for every mole of potassium chloride in the solution.

Step 2: Calculate the molarity of the electrolyte

First, convert the mass of potassium chloride to moles: \(\frac{1.00 \mathrm{g}}{74.55 \mathrm{g/mol}} = 0.0134 \mathrm{mol}\). Then, calculate the molarity of potassium chloride: \(M = \frac{0.0134}{0.500} = 0.0268 \mathrm{M}\).

Step 3: Find the concentration of each ion using stoichiometry

The concentration of \(\mathrm{K^+}\) ions is: \(1 \times 0.0268 = 0.0268 \mathrm{M}\). The concentration of \(\mathrm{Cl^{-}}\) ions is: \(1 \times 0.0268 = 0.0268 \mathrm{M}\). d. 132 g of ammonium sulfate in 1.50 L of solution

Step 1: Identify ions and stoichiometry

Ammonium sulfate dissociates into ions as follows: \(\mathrm{(NH_4)_2SO_4 \rightarrow 2NH_4^+ + SO_4^{2-}}\). We have 2 moles of \(\mathrm{NH_4^+}\) and 1 mole of \(\mathrm{SO_4^{2-}}\) for every mole of ammonium sulfate in the solution.

Step 2: Calculate the molarity of the electrolyte

First, convert the mass of ammonium sulfate to moles: \(\frac{132 \mathrm{g}}{132.14 \mathrm{g/mol}} = 0.9989 \mathrm{mol}\). Then, calculate the molarity of ammonium sulfate: \(M = \frac{0.9989}{1.50} = 0.6659 \mathrm{M}\).

Step 3: Find the concentration of each ion using stoichiometry

The concentration of \(\mathrm{NH_4^+}\) ions is: \(2 \times 0.6659 = 1.33 \mathrm{M}\). The concentration of \(\mathrm{SO_4^{2-}}\) ions is: \(1 \times 0.6659 = 0.6659 \mathrm{M}\).

Key Concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (\textbf{moles/Liter} or \textbf{M}). Molarity can be calculated using the formula: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]For instance, when calculating the concentration of ions in a sodium phosphate solution, we first need to determine the amount of moles of sodium phosphate, and then divide by the volume of the solution in liters to find its molarity. This value is crucial for subsequent calculations, like determining the concentration of individual ions.
In a classroom or homework setting, teaching molarity often begins with simple examples using pure compounds, then advances to mixtures and determining individual ion concentrations in electrolyte solutions, as in the textbook exercise provided.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's like a recipe that outlines how much of each ingredient you need. In terms of ion concentration calculations, stoichiometry tells us the ratio of ions that result when an electrolyte dissociates.
For example, sodium phosphate (\textbf{Na}\(_3\)\textbf{PO}\(_4\)) produces three sodium ions (\textbf{Na}\(^+\)) for every one phosphate ion (\textbf{PO}\(_4$$^{3-}\)). Understanding the stoichiometric coefficients is essential for determining the concentration of individual ions in solution. By using these ratios, we apply the initial molarity calculated for the electrolyte to find the molarity for each type of ion released upon dissociation.

Electrolyte Dissociation

Electrolyte dissociation refers to the separation of an electrolyte into its constituent ions when it dissolves in water. This process is essential for the conduction of electrical current in the solution. For instance, when potassium chloride (\textbf{KCl}) dissolves, it dissociates into potassium ions (\textbf{K}\(^+\)) and chloride ions (\textbf{Cl}\(^-\)). Each molecule of \textbf{KCl} yields one \textbf{K}\(^+\) and one \textbf{Cl}\(^-\).
Understanding the process of electrolyte dissociation is important for accurately calculating ion concentrations. The extent of dissociation will determine whether all the solute particles contribute to the total concentration of ions. For strong electrolytes, like the ones described in our exercise, complete dissociation is assumed, which simplifies the calculation of ion concentrations.

Strong Electrolytes

Strong electrolytes are substances that completely dissociate into ions when dissolved in water, resulting in a solution that conducts electricity very well. Examples of strong electrolytes include most salts, strong acids, and strong bases. In the context of the provided exercise examples like sodium phosphate (\textbf{Na}\(_3\)\textbf{PO}\(_4\)), barium nitrate (\textbf{Ba(NO}\(_3\))\(_2\)), potassium chloride (\textbf{KCl}), and ammonium sulfate ((\textbf{NH}\(_4\))\(_2\)\textbf{SO}\(_4\)), are all strong electrolytes.
Since they dissociate completely, the concentration of ions can be directly calculated from the stoichiometry of the electrolyte without considering any equilibrium constants. This property greatly simplifies many aspects of quantitative chemistry, including the calculation of ion concentrations in solutions.