Question

Calculate the number of moles of each of the ions in the following solutions. (a) \(32.3 \mathrm{~mL}\) of \(0.455 \mathrm{M} \mathrm{CaCl}_{2}\) (b) \(50.0 \mathrm{~mL}\) of \(0.408 \mathrm{MAlCl}_{3}\)

Short answer

For (a) 0.0147 moles of Ca2+ and 0.0294 moles of Cl-. For (b) 0.0204 moles of Al3+ and 0.0612 moles of Cl-.

Step-by-step solution

Calculate Moles of CaCl2

Calculate the moles of CaCl2 using the molarity equation \( M = \frac{n}{V} \), where M is molarity, n is the number of moles, and V is the volume in liters. For the solution of CaCl2, first convert the volume from milliliters to liters by dividing by 1000: \( 32.3 \text{ mL} = 0.0323 \text{ L} \). Then use the molarity of CaCl2 to find the moles: \( n = M \times V = 0.455 \text{ M} \times 0.0323 \text{ L} \).

Calculate Moles of Ions in CaCl2 Solution

Since CaCl2 dissociates into one Ca2+ ion and two Cl- ions, calculate the moles of each ion by multiplying the moles of CaCl2 by the stoichiometric coefficients. For Ca2+, the moles are the same as the moles of CaCl2. For Cl-, the number of moles is twice that of CaCl2 because there are two Cl- ions for every one CaCl2 formula unit.

Calculate Moles of AlCl3

To calculate the moles of AlCl3, convert the volume from milliliters to liters: \( 50.0 \text{ mL} = 0.0500 \text{ L} \). Then use the molarity of AlCl3 to find the moles: \( n = M \times V = 0.408 \text{ M} \times 0.0500 \text{ L} \).

Calculate Moles of Ions in AlCl3 Solution

Since AlCl3 dissociates into one Al3+ ion and three Cl- ions, calculate the moles of each ion by multiplying the moles of AlCl3 by the stoichiometric coefficients. For Al3+, the moles are the same as the moles of AlCl3. For Cl-, the number of moles is three times that of AlCl3 because there are three Cl- ions for every one AlCl3 formula unit.

Key concepts

Stoichiometry

Stoichiometry is a sect of chemistry focusing on the quantitative relationship between reactants and products in a chemical reaction. In simple terms, it's about figuring out how much of each substance is needed to react properly with another, and what quantity of product will result. Let's apply this directly to the exercise involving the calculation of moles of ions in a solution.

Imagine baking a cake requiring 3 eggs for every 2 cups of flour. If you have 6 cups of flour, you'd need 9 eggs to maintain the proper ratio. Stoichiometry works similarly, but instead of eggs and flour, we're dealing with moles of compounds and their dissociated ions. When we multiply the moles of an ionic compound like CaCl2 by the stoichiometric coefficients (like the '2' for Cl- in CaCl2), we're essentially following a recipe for dissociation—ensuring that we know the amount of each ion produced.

Dissociation of Ionic Compounds

Understanding the dissociation of ionic compounds is essential when dealing with solutions in chemistry. Ionic compounds, like table salt (NaCl), are made of positively and negatively charged ions that are bound together. When these compounds dissolve in water, they 'dissociate,' or separate into individual ions that move independently throughout the solution.

Consider our example, where CaCl2 breaks down into Ca2+ and Cl- ions. This dissociation is crucial for calculating the number of moles of each ion in solution. If the recipe says one unit of CaCl2 gives us one Ca2+ ion and two Cl- ions, then knowing the moles of CaCl2 directly tells us how many moles of each ion we have—just like knowing the number of cake mixes could tell you how many cakes you can bake. This relationship allows us to provide a detailed, precise calculation of all ions present in a solution.

Mole Concept

The mole concept is a cornerstone of chemistry and defines the mole as a unit of measurement for the amount of substance. It's a bit like counting eggs by the dozen—we use moles because atoms and molecules are too small to count individually. One mole is Avogadro's number (\(6.022 \times 10^{23}\)) of particles, whether they're atoms, ions, or molecules.

In the tackled exercise, we use the mole concept to translate the volume and molarity of a solution into a countable amount. Just as you might use cups to measure the volume of milk in a recipe, in the lab, we use moles to gauge how many formula units of a compound we have in a given volume of solution. With molarity, the definition of concentration, expressed in moles per liter (\( \frac{moles}{L} \)), we can figure the moles of solute in a solution, leading us right back to stoichiometry and enabling us to calculate the discrete number of moles of each ion post-dissociation.