Question

Calculate the number of moles of each ion present in each of the following solutions. a. \(10.2 \mathrm{~mL}\) of \(0.451 \mathrm{M} \mathrm{AlCl}_{3}\) solution b. \(5.51 \mathrm{~L}\) of \(0.103 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution c. \(1.75 \mathrm{~mL}\) of \(1.25 \mathrm{M} \mathrm{CuCl}_{2}\) solution d. \(25.2 \mathrm{~mL}\) of \(0.00157 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) solution

Short answer

The number of moles of each ion present in each solution are: a. \(Al^{3+}\): \(4.601 \times 10^{-3}\) moles, \(Cl^{-}\): \(13.803 \times 10^{-3}\) moles b. \(Na^{+}\): 1.7019 moles, \(PO_{4}^{3-}\): 0.5673 moles c. \(Cu^{2+}\): \(2.188 \times 10^{-3}\) moles, \(Cl^{-}\): \(4.376 \times 10^{-3}\) moles d. \(Ca^{2+}\): \(3.964 \times 10^{-5}\) moles, \(OH^{-}\): \(7.928 \times 10^{-5}\) moles

Step-by-step solution

Write the dissociation equation for AlCl3

AlCl3 dissociates in water as follows: \[AlCl_{3} \rightarrow Al^{3+} + 3Cl^{-}\]

Calculate moles of AlCl3

Given the volume and molarity of AlCl3 solution, we can calculate the moles of AlCl3: moles of AlCl3 = volume × molarity moles of AlCl3 = \(10.2 \times 10^{-3} L \times 0.451 M\) moles of AlCl3 = \(4.601 \times 10^{-3}\) moles

Calculate the moles of each ion

Using the mole ratio from the dissociation equation, calculate the moles of each ion: moles of Al3+ = \(4.601 \times 10^{-3}\) moles moles of Cl- = 3 × \(4.601 \times 10^{-3}\) moles = \(13.803 \times 10^{-3}\) moles #b. 5.51 L of 0.103 M Na3PO4 solution#

Write the dissociation equation for Na3PO4

Na3PO4 dissociates in water as follows: \[Na_{3}PO_{4} \rightarrow 3Na^{+} + PO_{4}^{3-}\]

Calculate moles of Na3PO4

Given the volume and molarity of Na3PO4 solution, we can calculate the moles of Na3PO4: moles of Na3PO4 = volume × molarity moles of Na3PO4 = \(5.51 L \times 0.103 M\) moles of Na3PO4 = 0.5673 moles

Calculate the moles of each ion

Using the mole ratio from the dissociation equation, calculate the moles of each ion: moles of Na+ = 3 × 0.5673 moles = 1.7019 moles moles of PO43- = 0.5673 moles #c. 1.75 mL of 1.25 M CuCl2 solution#

Write the dissociation equation for CuCl2

CuCl2 dissociates in water as follows: \[CuCl_{2} \rightarrow Cu^{2+} + 2Cl^{-}\]

Calculate moles of CuCl2

Given the volume and molarity of CuCl2 solution, we can calculate the moles of CuCl2: moles of CuCl2 = volume × molarity moles of CuCl2 = \(1.75 \times 10^{-3} L \times 1.25 M\) moles of CuCl2 = \(2.188 \times 10^{-3}\) moles

Calculate the moles of each ion

Using the mole ratio from the dissociation equation, calculate the moles of each ion: moles of Cu2+ = \(2.188 \times 10^{-3}\) moles moles of Cl- = 2 × \(2.188 \times 10^{-3}\) moles = \(4.376 \times 10^{-3}\) moles #d. 25.2 mL of 0.00157 M Ca(OH)2 solution#

Write the dissociation equation for Ca(OH)2

Ca(OH)2 dissociates in water as follows: \[Ca(OH)_{2} \rightarrow Ca^{2+} + 2OH^{-}\]

Calculate moles of Ca(OH)2

Given the volume and molarity of Ca(OH)2 solution, we can calculate the moles of Ca(OH)2: moles of Ca(OH)2 = volume × molarity moles of Ca(OH)2 = \(25.2 \times 10^{-3} L \times 0.00157 M\) moles of Ca(OH)2 = \(3.964 \times 10^{-5}\) moles

Calculate the moles of each ion

Using the mole ratio from the dissociation equation, calculate the moles of each ion: moles of Ca2+ = \(3.964 \times 10^{-5}\) moles moles of OH- = 2 × \(3.964 \times 10^{-5}\) moles = \(7.928 \times 10^{-5}\) moles

Key concepts

Molar Concentration

When studying chemistry, one essential concept is molar concentration, often called molarity. It specifies the number of moles of a solute dissolved in one liter of solution and is expressed in moles per liter (mol/L or M). To calculate the number of moles from molarity, the formula used is:

\[\begin{equation}\text{Moles of solute} = \text{Volume of solution in liters} \times \text{Molarity (M)}\end{equation}\]
By knowing the volume and the molarity, you can quickly determine how many moles of the solute are presented in the solution. This is crucial for reactions where the amount of each chemical determines the course and outcome of the reaction.

Dissociation Equations

A dissociation equation represents how an ionic compound separates into its ions when dissolved in water. Ionic compounds, like salts, consist of positive and negative ions, which are held together by ionic bonds. In a dissolution process, these bonds are broken, and the individual ions are released into the solution. For example, when aluminum chloride (\(AlCl_3\)) dissolves, it dissociates into one aluminium ion (\(Al^{3+}\)) and three chloride ions (\(3Cl^{-}\)).

\[\begin{equation}AlCl_{3} \rightarrow Al^{3+} + 3Cl^{-}\end{equation}\]
Understanding the dissociation of compounds is fundamental for predicting the products of a reaction and calculating the number of moles of ions produced in solution.

Stoichiometry

In chemical equations, stoichiometry is the principle that describes the relative quantities of reactants and products in a reaction. It involves calculations based on the conservation of mass and the concept of moles, allowing chemists to predict the outcomes of chemical reactions. Through stoichiometry, one can determine the amounts of products produced from given amounts of reactants, and vice versa. This is achieved by using the coefficients of a balanced chemical equation, which provide the ratio of moles of each substance involved, known as the mole-to-mole ratio. These ratios are key to calculating quantities in chemical reactions.

Mole-to-Mole Ratio

The mole-to-mole ratio is the heart of stoichiometric calculations in chemistry. It comes from the coefficients in a balanced chemical equation, reflecting the ratio of moles of reactants used and moles of products formed. For example, in the dissociation of calcium hydroxide (\(Ca(OH)_2\)) into calcium ions (\(Ca^{2+}\)) and hydroxide ions (\(OH^{-}\)), the equation shows a 1:1:2 ratio:

\[\begin{equation}Ca(OH)_{2} \rightarrow Ca^{2+} + 2OH^{-}\end{equation}\]
For every one mole of \(Ca(OH)_2\) that dissociates, one mole of \(Ca^{2+}\) and two moles of \(OH^{-}\) ions are produced. Understanding these ratios is critical for calculating the amounts of reacting ions in solution, an important step when determining the stoichiometry of a reaction in an aqueous environment.