Question

Indicate the concentration of each ion present in the solution formed by mixing (a) \(42.0 \mathrm{~mL}\) of \(0.170 \mathrm{M} \mathrm{NaOH}\) and \(37.6 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH}\), (b) \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\), (c) \(3.60 \mathrm{~g} \mathrm{KCl}\) in \(75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.

Short answer

In the resulting solutions: (a) The concentration of \(Na^+\) and \(OH^-\) is 0.2783 M. (b) The concentrations are \(Na^+\) = 0.1275 M, \(SO_4^{2-}\) = 0.0638 M, \(K^+\) = 0.0543 M, and \(Cl^-\) = 0.0543 M. (c) The concentrations are \(K^+\) = 0.684 M, \(Cl^-\) = 1.184 M, and \(Ca^{2+}\) = 0.250 M.

Step-by-step solution

Calculate the moles of \(Na^+\) and \(OH^-\) ions in each solution

First, we need to find the amount of moles of each ion in both solutions. For the \( 0.170 \mathrm{M} \mathrm{NaOH} \): moles of \(Na^+\) = moles of \(OH^-\) = volume × molarity moles of \(Na^+\) = moles of \(OH^-\) = 0.042 L × 0.170 mol/L = 0.00714 mol For the \( 0.400 \mathrm{M} \mathrm{NaOH} \): moles of \(Na^+\) = moles of \(OH^-\) = volume × molarity moles of \(Na^+\) = moles of \(OH^-\) = 0.0376 L × 0.400 mol/L = 0.01504 mol

Add the moles of each ion

Now, we need to add the moles of each ion: Total moles of \(Na^+\) = 0.00714 mol + 0.01504 mol = 0.02218 mol Total moles of \(OH^-\) = 0.00714 mol + 0.01504 mol = 0.02218 mol

Calculate the total volume of the solution

Since the volumes are additive, we can simply add the individual volumes to find the total volume of the mixed solution: Total volume = 0.042 L + 0.0376 L = 0.0796 L

Calculate the concentration of each ion

Now, we can use the formula Concentration = (moles of ion) / (total volume of solution) to find the concentration of each ion: Concentration of \(Na^+\) = 0.02218 mol / 0.0796 L = 0.2783 M Concentration of \(OH^-\) = 0.02218 mol / 0.0796 L = 0.2783 M #Case (b)#

Calculate the moles of \(Na^+\), \(SO_4^{2-}\), \(K^+\), and \(Cl^-\) ions in each solution

For the \( 0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4} \): moles of \(Na^+\) = 2 × (volume × molarity) moles of \(Na^+\) = 2 × (0.044 L × 0.100 mol/L) = 0.0088 mol moles of \(SO_4^{2-}\) = volume × molarity moles of \(SO_4^{2-}\) = 0.044 L × 0.100 mol/L = 0.0044 mol For the \( 0.150 \mathrm{M} \mathrm{KCl} \): moles of \(K^+\) = moles of \(Cl^-\) = volume × molarity moles of \(K^+\) = moles of \(Cl^-\) = 0.025 L × 0.150 mol/L = 0.00375 mol

Add the moles of each ion

Now, we need to add the moles of each ion: Total moles of \(Na^+\) = 0.0088 mol Total moles of \(SO_4^{2-}\) = 0.0044 mol Total moles of \(K^+\) = 0.00375 mol Total moles of \(Cl^-\) = 0.00375 mol

Calculate the total volume of the solution

Total volume = 0.044 L + 0.025 L = 0.069 L

Calculate the concentration of each ion

We can use the formula Concentration = (moles of ion) / (total volume of solution) to find the concentration of each ion: Concentration of \(Na^+\) = 0.0088 mol / 0.069 L = 0.1275 M Concentration of \(SO_4^{2-}\) = 0.0044 mol / 0.069 L = 0.0638 M Concentration of \(K^+\) = 0.00375 mol / 0.069 L = 0.0543 M Concentration of \(Cl^-\) = 0.00375 mol / 0.069 L = 0.0543 M #Case (c)#

Calculate the moles of \(K^+\), \(Cl^-\), \(Ca^{2+}\), and \(Cl^-\) ions in each solution

For the \(3.60 \mathrm{~g} \mathrm{KCl}\): First, find the moles of \(KCl\): moles of \(KCl\) = mass / molar mass moles of \(KCl\) = 3.60 g / (39.10 g/mol + 35.45 g/mol) = 0.0513 mol Since there are equal moles of \(K^+\) and \(Cl^-\) in \(KCl\), we have: moles of \(K^+\) = moles of \(Cl^-\) = 0.0513 mol For the \(0.250 \mathrm{M} \mathrm{CaCl}_{2}\): moles of \(Ca^{2+}\) = volume × molarity moles of \(Ca^{2+}\) = 0.075 L × 0.250 mol/L = 0.01875 mol moles of \(Cl^-\) = 2 × (volume × molarity) moles of \(Cl^-\) = 2 × (0.075 L × 0.250 mol/L) = 0.0375 mol

Add the moles of each ion

Now, we need to add the moles of each ion: Total moles of \(K^+\) = 0.0513 mol Total moles of \(Cl^-\) = 0.0513 mol + 0.0375 mol = 0.0888 mol Total moles of \(Ca^{2+}\) = 0.01875 mol

Calculate the total volume of the solution

Since the weight of \(KCl\) was given, we assume that the volume it occupied was negligible, so the total volume is just the volume of the \(CaCl_2\) solution: Total volume = 0.075 L

Calculate the concentration of each ion

We can use the formula Concentration = (moles of ion) / (total volume of solution) to find the concentration of each ion: Concentration of \(K^+\) = 0.0513 mol / 0.075 L = 0.684 M Concentration of \(Cl^-\) = 0.0888 mol / 0.075 L = 1.184 M Concentration of \(Ca^{2+}\) = 0.01875 mol / 0.075 L = 0.250 M

Key concepts

Understanding Solution Chemistry

Solution chemistry is a fascinating area of study in the field of chemistry. It focuses on how substances dissolve, how solutions are formed, and the interactions between different species within a solution. A solution is a homogeneous mixture composed of two major components: the solute and the solvent. The solvent is the substance that dissolves the solute, while the solute is the substance that is dissolved.

When ions dissolve in water, they dissociate into their respective positive and negative ions. For instance, when sodium hydroxide (NaOH) dissolves, it breaks down into sodium ions ( Na^+ ) and hydroxide ions ( OH^- ). Understanding how these ions behave in a solution is critical. This knowledge helps in predicting reactions, understanding concentration, and manipulating solution properties for various applications.

• The solubility of a compound determines how much of it can dissolve in a given solvent.
• The nature of the solute and solvent, temperature, and pressure are factors affecting solubility.
• Solutions can be gases, liquids, or solids.

Exploring Ionic Compounds

Ionic compounds are made up of positive and negative ions held together by strong ionic bonds. These characters have complete transfers of electrons between atoms, resulting in charged species. For example, sodium chloride (NaCl) is an ionic compound that dissociates into sodium ions ( Na^+ ) and chloride ions ( Cl^- ) in solution.

The structure of ionic compounds renders them soluble in water, which means they readily separate into their ions when mixed with water. It's crucial to understand that ionic compounds do not exist as molecules but as a network of ions.

• Ionic compounds generally have high melting and boiling points due to strong ionic bonds.
• They conduct electricity when melted or dissolved in water.
• They tend to be hard and brittle.

Understanding the behavior of ionic compounds is essential for tasks involving calculations of ion concentrations in solutions.

Mastering Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (mol/L). It is a crucial concept in chemistry for preparing solutions and conducting experiments where precise concentrations are needed.

The formula for calculating molarity (M) is:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]

To find the moles, you can use:\[\text{moles} = \text{volume (L)} \times \text{molarity (mol/L)}\]

In practical applications, calculating the concentration of ions in a solution involves careful measurement of volumes and concentrations. Once you know the moles of each ion, you can find the total concentration by summing the moles and dividing by the new total volume.

• Always ensure that the volumes are in liters when performing molarity calculations.
• Remember to account for additive volumes if solutions are mixed.
• The units of molarity help standardize the comparison of solution concentrations.

Mastering molarity calculations enables precise control over chemical reactions and processes.