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

(a) In the mixed solution, the concentration of Na+ is \(0.233 \mathrm{M}\) and OH- is \(0.218 \mathrm{M}\). (b) In the mixed solution, the concentration of Na+ is \(0.0508 \mathrm{M}\), SO42- is \(0.0254 \mathrm{M}\), K+ is \(0.0339 \mathrm{M}\), and Cl- is \(0.0339 \mathrm{M}\). (c) In the mixed solution, the concentration of K+ is \(0.270 \mathrm{M}\), Cl- is \(0.527 \mathrm{M}\), Ca2+ is \(0.250 \mathrm{M}\), and Cl2- is \(0.500 \mathrm{M}\).

Step-by-step solution

Problem (a)

First, let's find the concentrations of the ions in a solution formed by mixing \(42.0 \mathrm{~mL}\) of \(0.170 \mathrm{M} \mathrm{NaOH}\) and \(37.6 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH}\). Step 1: Calculate the number of moles of each compound number of moles of the first solution = volume x molarity number of moles of the second solution = volume x molarity Step 2: Determine the total volume of the mixture total volume = volume of the first solution + volume of the second solution Step 3: Calculate the new concentration of ions in the mixed solution new concentration = total number of moles / total volume

Problem (b)

Now let's find the concentration of each ion in a solution formed by mixing \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na_{2}SO_{4}}\) and \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl}\). Step 1: Calculate the number of moles of each compound number of moles of the first solution = volume x molarity number of moles of the second solution = volume x molarity Step 2: Determine the total volume of the mixture total volume = volume of the first solution + volume of the second solution Step 3: Calculate the new concentration of ions in the mixed solution new concentration = total number of moles / total volume

Problem (c)

Finally, let's find the concentration of each ion in a solution formed by dissolving \(3.60 \mathrm{~g} \mathrm{KCl}\) in \(75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CaCl_{2}}\) solution. Step 1: Calculate the number of moles of KCl number of moles of KCl = mass of KCl / molar mass of KCl Step 2: Calculate the molarity of KCl in the initial solution molarity of KCl = number of moles of KCl / volume of the solution Step 3: Calculate the number of moles of each compound number of moles of KCl = volume x molarity number of moles of CaCl2 = volume x molarity Step 4: Determine the total volume of the mixture total volume = initial volume + volume of the added KCl solution Step 5: Calculate the new concentration of ions in the mixed solution new concentration = total number of moles / total volume

Key concepts

Molarity Calculations

When we talk about molarity, we're referring to a way of expressing how concentrated a solution is with respect to its solute. Molarity, abbreviated as \( M \), is calculated using the formula: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). This is crucial because it helps us understand the number of particles present in a given volume, which directly relates to the solution's chemical reactivity.
To find the molarity, you first need to determine the number of moles of solute in your mixture. You can do this by multiplying the volume (in liters) of the solution by its molarity. For example, if you have a \(42.0 \, \text{mL} \) solution of \(0.170 \, M \), the number of moles is calculated as \(0.042 \, L \times 0.170 \, M = 0.00714 \, \text{moles} \).
In multi-component systems where two or more solutions are mixed, the molarity of the final solution depends on the total number of moles of solute and the total volume of the solution, not just the initial concentrations of the individual components.

Total Volume in Solution

Total volume in a solution is pretty much what it sounds like: the combined volume of all the liquid components when they are mixed together. This is important in calculating how concentrated a solution is after mixing multiple components with different concentrations.
For mixtures, especially when solutions are added together, the assumption is made that the volumes are additive. This means you simply add up the volumes of each liquid component to find the total volume. For instance, if you combine \(42.0 \, \text{mL} \) of one solution with \(37.6 \, \text{mL} \) of another, the total volume will be \(42.0 \, \text{mL} + 37.6 \, \text{mL} = 79.6 \, \text{mL} \). Remember to always express the final total volume in liters when using it in molarity calculations.
Remember, being precise with your measurements in milliliters and converting them accurately to liters ensures that the molarity calculations remain reliable and accurate.

Ion Concentration

Ion concentration is fundamental in understanding the chemical dynamics of a solution, especially in reactions involving ionic compounds. When we dissolve ionic substances in water, they disassociate into ions, which are charged particles responsible for conducting electricity.
To find the concentration of each ion, you first calculate the total number of moles of each ion produced from the initial compounds. The calculated moles of ions are then divided by the total volume of the solution to determine their concentrations. For example, if you dissolve \(\text{Na}_2\text{SO}_4\), it separates into \(2 \text{Na}^+ \) ions and \(\text{SO}_4^{2-} \) ions. If you started with 0.0044 moles of \(\text{Na}_2\text{SO}_4\), you'd have 0.0088 moles of \(\text{Na}^+\) after disassociation.
Knowing the concentration of each ion is crucial for predicting the outcomes of chemical reactions and understanding the properties of the solution, such as its pH or ionic strength.