Problem 31 - Chemistry An Atoms First Approach

Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{mL}\) of solution b. 2.5 moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 1.25 L of solution c. \(5.00 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{mL}\) of solution d. \(1.00 \mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}\) in \(250.0 \mathrm{mL}\) of solution

Step-by-Step Solution

Answer

The concentrations of ions in each solution are: a. \(\mathrm{Ca^{2+}}: 0.00100\, \mathrm{M}\), \(\mathrm{NO_3^{-}}: 0.00200\, \mathrm{M}\) b. \(\mathrm{Na^{+}}: 4.0\, \mathrm{M}\), \(\mathrm{SO_4^{2-}}: 2.0\, \mathrm{M}\) c. \(\mathrm{NH_4^{+}}: 0.187\, \mathrm{M}\), \(\mathrm{Cl^{-}}: 0.187\, \mathrm{M}\) d. \(\mathrm{K^{+}}: 0.0564\, \mathrm{M}\), \(\mathrm{PO_4^{3-}}: 0.0188\, \mathrm{M}\)

Step 1: Calculate the molarity of the solutions in each case

First, we need to find the molarity of the solution for each substance: a. Molarity \(a= \frac{0.100 \, \text{mol of} \, \mathrm{Ca(NO}_{3})_{2}}{100.0 \, \text{mL}} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} =0.00100\, \mathrm{M}\) b. Molarity \(b= \frac{2.5 \, \text{mol of} \, \mathrm{Na}_{2}\mathrm{SO}_{4}}{1.25 \, \text{L}} =2.0 \, \mathrm{M}\) c. First, we need to find the moles of \(\mathrm{NH}_{4}\mathrm{Cl}\): Moles \(c = \frac{5.00\, \mathrm{g}}{53.49 \, \mathrm{g/mol}} \approx 0.0935\, \mathrm{mol}\) Molarity \(c= \frac{0.0935 \, \text{mol of} \, \mathrm{NH}_{4}\mathrm{Cl}}{0.500 \, \text{L}} =0.187 \, \mathrm{M}\) d. First, we need to find the moles of \(\mathrm{K}_{3}\mathrm{PO}_{4}\): Moles \(d = \frac{1.00\, \mathrm{g}}{212.28 \, \mathrm{g/mol}} \approx 0.00471\, \mathrm{mol}\) Molarity \(d= \frac{0.00471 \, \text{mol of} \, \mathrm{K}_{3}\mathrm{PO}_{4}}{0.25 \, \text{L}} =0.0188 \, \mathrm{M}\)

Step 2: Identify the ions formed by the dissociation of each electrolyte and their stoichiometry and Calculate molarity of each ion

Now, we will calculate the molarity of the resulting ions in each solution: a. \(\mathrm{Ca(NO}_{3})_{2}\) dissociates into \(\mathrm{Ca^{2+}}\) and \(\mathrm{NO_3^-}\), with a stoichiometry of 1:2. \(\mathrm{M(Ca^{2+})} = 0.00100\, \mathrm{M}\) \(\mathrm{M(NO_3^{-})} = 2 \times 0.00100\, \mathrm{M} = 0.00200\, \mathrm{M}\) b. \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) dissociates into \(\mathrm{Na^{+}}\) and \(\mathrm{SO_4^{2-}}\), with a stoichiometry of 2:1. \(\mathrm{M(Na^{+})} = 2 \times 2.0\, \mathrm{M} = 4.0\, \mathrm{M}\) \(\mathrm{M(SO_4^{2-})} = 2.0\, \mathrm{M}\) c. \(\mathrm{NH}_{4}\mathrm{Cl}\) dissociates into \(\mathrm{NH_4^{+}}\) and \(\mathrm{Cl^-}\), with a stoichiometry of 1:1. \(\mathrm{M(NH_4^{+})} = 0.187\, \mathrm{M}\) \(\mathrm{M(Cl^{-})} = 0.187\, \mathrm{M}\) d. \(\mathrm{K}_{3}\mathrm{PO}_{4}\) dissociates into \(\mathrm{K^{+}}\) and \(\mathrm{PO_4^{3-}}\), with a stoichiometry of 3:1. \(\mathrm{M(K^{+})} = 3\times 0.0188\, \mathrm{M} = 0.0564\, \mathrm{M}\) \(\mathrm{M(PO_4^{3-})} = 0.0188\, \mathrm{M}\) Final concentrations of ions: a. \(\mathrm{Ca^{2+}}: 0.00100\, \mathrm{M}\), \(\mathrm{NO_3^{-}}: 0.00200\, \mathrm{M}\) b. \(\mathrm{Na^{+}}: 4.0\, \mathrm{M}\), \(\mathrm{SO_4^{2-}}: 2.0\, \mathrm{M}\) c. \(\mathrm{NH_4^{+}}: 0.187\, \mathrm{M}\), \(\mathrm{Cl^{-}}: 0.187\, \mathrm{M}\) d. \(\mathrm{K^{+}}: 0.0564\, \mathrm{M}\), \(\mathrm{PO_4^{3-}}: 0.0188\, \mathrm{M}\)

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry that represents the concentration of a solution. It is defined as the number of moles of solute dissolved in one liter of solution. The formula to calculate molarity (M) is:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the context of the given exercise, molarity is used to express the concentration of strong electrolytes in solution. Understanding molarity is crucial for predicting the behavior of the solution, such as its boiling point, vapor pressure, and osmotic pressure. When the solute is a strong electrolyte, it dissociates completely in water, and the molarity will help in determining the concentration of each ion present in the solution.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the realm of ionic concentration calculations, stoichiometry helps us understand how compounds dissociate in water to produce ions. For instance, calcium nitrate, \(\mathrm{Ca(NO_3)_2}\), dissociates into one \(\mathrm{Ca^{2+}}\) ion and two \(\mathrm{NO_3^-}\) ions. Hence, the stoichiometry of dissociation is 1:2, and this ratio is pivotal in calculating the concentration of each ion produced.
To illustrate, knowing the molarity of the original compound and the stoichiometry of its dissociation allows us to accurately determine the molarity of the resulting ions. This knowledge is a cornerstone in fields such as analytical chemistry, environmental science, and pharmaceuticals, where precise measurements of chemical composition are necessary.

Electrolyte Dissociation

Electrolyte dissociation is a process in which an ionic compound, known as an electrolyte, disintegrates into its constituent ions when dissolved in a solvent, such as water. This phenomenon is essential when calculating ionic concentrations, as it defines how many ions will be present in the solution. Strong electrolytes, like \(\mathrm{Na_2SO_4}\), dissociate completely, resulting in a specific number of ions that can be deduced from the chemical formula of the compound.
For example, when sodium sulfate (\(\mathrm{Na_2SO_4}\)) dissolves in water, it produces two sodium ions (\(\mathrm{Na^+}\)) and one sulfate ion (\(\mathrm{SO_4^{2-}}\)) per formula unit. The knowledge of electrolyte dissociation is crucial in fields such as medicine and environmental science, where ionic balance is vital. In the exercise given, understanding this concept allows students to calculate the molarity of individual ions after considering their stoichiometry in the dissociation process.