Problem 72 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate the concentration of each ion in the following solutions obtained by mixing: (a) \(32.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{KMnO}_{4}\) (b) \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{ZnCl}_{2}^{+}\) with \(15.0 \mathrm{~mL}\) of \(0.60 \mathrm{MKMnO}_{4}\) with \(5.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) 4.2 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) in \(150.0 \mathrm{~mL}\) of \(0.02 M \mathrm{KCl}\) solution. Assume that the volumes are additive.

Step-by-Step Solution

Answer

In the given solutions, the final concentrations are: (a) K+ = \(0.396 \ \mathrm{M}\), MnO4- = \(0.396 \ \mathrm{M}\) (b) Zn2+ = \(0.108 \ \mathrm{M}\), Cl- = \(0.185 \ \mathrm{M}\), NO3- = \(0.031 \ \mathrm{M}\) (c) Ca2+ = \(0.239 \ \mathrm{M}\), K+ = \(0.019 \ \mathrm{M}\), Cl- = \(0.496 \ \mathrm{M}\)

Step 1: (a) Calculate moles of each ion in the solution of KMnO4

First, we need to determine the moles of each ion in the \(32.0 \mathrm{mL}\) of \(0.30 \mathrm{M} \ \mathrm{KMnO}_{4}\) solution. Since molarity (M) = moles of solute/volume of solution in L, we can calculate the moles of KMnO4. Moles of KMnO4 = Molarity × Volume = \(0.30 \mathrm{M} \times 32.0 \mathrm{mL}\) However, we need to convert the volume from mL to L. Moles of KMnO4 = \(0.30 \mathrm{M} \times 0.032 \mathrm{L} = 0.0096 \ \mathrm{mol}\) Now we will find the moles of ions in the solution. KMnO4 dissociates into K+ and MnO4- ions. Moles of K+ = moles of KMnO4 = \(0.0096 \ \mathrm{mol}\) Moles of MnO4- = moles of KMnO4 = \(0.0096 \ \mathrm{mol}\)

Step 2: (a) Calculate moles of each ion in the mixture

Now, let's find the moles of each ion in the mixture of solutions. Moles of K+ in the mixture = Moles of K+ in 32.0 mL of 0.30 M KMnO4 + moles of K+ in 15.0 mL of 0.60 M KMnO4 Moles of K+ in the mixture = \(0.0096 \mathrm{mol} + 0.60 \mathrm{M} \times 0.015 \mathrm{L}\) Moles of K+ in the mixture = \(0.0096 \mathrm{mol} + 0.0090 \mathrm{mol} = 0.0186 \mathrm{mol}\) Moles of MnO4- in the mixture = Moles of MnO4- in 32.0 mL of 0.30 M KMnO4 + moles of MnO4- in 15.0 mL of 0.60 M KMnO4 Moles of MnO4- in the mixture = \(0.0096 \mathrm{mol} + 0.60 \mathrm{M} \times 0.015 \mathrm{L}\) Moles of MnO4- in the mixture = \(0.0096 \mathrm{mol} + 0.0090 \mathrm{mol} = 0.0186 \mathrm{mol}\)

Step 3: (a) Calculate the concentration of each ion

Now that we have the moles of each ion in the mixture, we can find their final concentrations. The total volume of the mixture is \(32.0 \mathrm{mL} + 15.0 \mathrm{mL} = 47.0 \mathrm{mL}\). Concentration of K+ = Moles of K+ in the mixture / Total volume of the mixture = \(0.0186 \ \mathrm{mol} / 0.047 \ \mathrm{L} = 0.396 \ \mathrm{M}\) Concentration of MnO4- = Moles of MnO4- in the mixture / Total volume of the mixture = \(0.0186 \ \mathrm{mol} / 0.047 \ \mathrm{L} = 0.396 \ \mathrm{M}\) The final concentrations in the mixture are: K+ = \(0.396 \ \mathrm{M}\) MnO4- = \(0.396 \ \mathrm{M}\)

Step 4: (b) Mixing ZnCl2 and Zn(NO3)2 solutions

For part (b), we have to mix \(60.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \ \mathrm{ZnCl}_{2}^{+}\) solution and \(5.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \ \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) solution. For both solutions, we have multiple ions to consider. First, we need to calculate the moles of each ion in both solutions. For ZnCl2: Moles of Zn2+ = Molarity × Volume = \(0.100 \mathrm{M} \times 0.060 \mathrm{L} = 0.0060 \ \mathrm{mol}\) Moles of Cl- = 2 × Moles of Zn2+ (since there are two Cl- ions in ZnCl2) = 2 × \(0.0060 \ \mathrm{mol} = 0.0120 \ \mathrm{mol}\) For Zn(NO3)2: Moles of Zn2+ = Molarity × Volume = \(0.200 \mathrm{M} \times 0.005 \mathrm{L} = 0.0010 \ \mathrm{mol}\) Moles of NO3- = 2 × Moles of Zn2+ (since there are two NO3- ions in Zn(NO3)2) = 2 × \(0.0010 \ \mathrm{mol} = 0.0020 \ \mathrm{mol}\) Now we sum up the moles of each ion in the mixture: Moles of Zn2+ in the mixture = Moles of Zn2+ in ZnCl2 + Moles of Zn2+ in Zn(NO3)2 = \(0.0060 \mathrm{mol} + 0.0010 \mathrm{mol} = 0.0070 \mathrm{mol}\) Moles of Cl- in the mixture = Moles of Cl- in ZnCl2 = \(0.0120 \ \mathrm{mol}\) Moles of NO3- in the mixture = Moles of NO3- in Zn(NO3)2 = \(0.0020 \ \mathrm{mol}\) Now we calculate the concentration of each ion in the mixture. The total volume of the mixture is \(60.0 \mathrm{mL} + 5.0 \mathrm{mL} = 65.0 \mathrm{mL}\). Concentration of Zn2+ = Moles of Zn2+ in the mixture / Total volume of the mixture = \(0.0070 \ \mathrm{mol} / 0.065 \ \mathrm{L} = 0.108 \ \mathrm{M}\) Concentration of Cl- = Moles of Cl- in the mixture / Total volume of the mixture = \(0.0120 \ \mathrm{mol} / 0.065 \ \mathrm{L} = 0.185 \ \mathrm{M}\) Concentration of NO3- = Moles of NO3- in the mixture / Total volume of the mixture = \(0.0020 \ \mathrm{mol} / 0.065 \ \mathrm{L} = 0.031 \ \mathrm{M}\) The final concentrations in the mixture are: Zn2+ = \(0.108 \ \mathrm{M}\) Cl- = \(0.185 \ \mathrm{M}\) NO3- = \(0.031 \ \mathrm{M}\)

Step 5: (c) Mixing CaCl2 and KCl solutions

For part (c), we have to mix \(4.2 \mathrm{g}\) of \(\mathrm{CaCl}_{2}\) and \(150.0 \mathrm{mL}\) of \(0.02 \mathrm{M} \ \mathrm{KCl}\). First, let's find the moles of CaCl2: Moles of CaCl2 = mass / molar mass = \(4.2 \mathrm{g} / 110.98 \mathrm{g/mol} = 0.0378 \ \mathrm{mol}\) Now we find the moles of ions in the CaCl2: Moles of Ca2+ = moles of CaCl2 = \(0.0378 \ \mathrm{mol}\) Moles of Cl- in CaCl2 = 2 × moles of Ca2+ (since there are two Cl- ions in CaCl2) = 2 × \(0.0378 \ \mathrm{mol} = 0.0756 \mathrm{mol}\) Next, let's find the moles of ions in the KCl solution: Moles of K+ = Molarity × Volume = \(0.02 \mathrm{M} \times 0.150 \mathrm{L} = 0.0030 \ \mathrm{mol}\) Moles of Cl- in KCl solution = Moles of K+ (since there is one Cl- ion in KCl) = \(0.0030 \ \mathrm{mol}\) Now we sum up the moles of each ion in the mixture: Moles of Ca2+ in the mixture = Moles of Ca2+ = \(0.0378 \ \mathrm{mol}\) Moles of K+ in the mixture = Moles of K+ = \(0.0030 \ \mathrm{mol}\) Moles of Cl- in the mixture = Moles of Cl- in CaCl2 + Moles of Cl- in KCl = \(0.0756 \ \mathrm{mol} + 0.0030 \ \mathrm{mol} = 0.0786 \ \mathrm{mol}\) Now, calculate the concentration of each ion in the mixture. The total volume of the mixture is \(4.2 \mathrm{g} \times 2 \mathrm{mL/g} + 150.0 \mathrm{mL} = 158.4 \mathrm{mL}\). Concentration of Ca2+ = Moles of Ca2+ in the mixture / Total volume of the mixture = \(0.0378 \ \mathrm{mol} / 0.1584 \ \mathrm{L} = 0.239 \ \mathrm{M}\) Concentration of K+ = Moles of K+ in the mixture / Total volume of the mixture = \(0.0030 \ \mathrm{mol} / 0.1584 \ \mathrm{L} = 0.019 \ \mathrm{M}\) Concentration of Cl- = Moles of Cl- in the mixture / Total volume of the mixture = \(0.0786 \ \mathrm{mol} / 0.1584 \ \mathrm{L} = 0.496 \ \mathrm{M}\) The final concentrations in the mixture are: Ca2+ = \(0.239 \ \mathrm{M}\) K+ = \(0.019 \ \mathrm{M}\) Cl- = \(0.496 \ \mathrm{M}\)

Key Concepts

Molarity

Molarity is a useful measure in chemistry that tells you how concentrated a solution is with the solute. Simply put, it describes how many moles of solute are present in one liter of solution. This concept is pivotal when mixing solutions, as it directly impacts the strength and properties of the resultant solution. To understand molarity, consider this basic formula:

• Molarity (M) = \(\frac{\text{moles of solute}}{\text{liters of solution}}\)

When calculating molarity, it is important to always convert volumes to liters and to use consistent units throughout your calculations.
Let's take an example: Upon mixing different solutions, as shown in the original exercise, understanding the molarity aids in determining the concentration of each resulting ion, like potassium ions \(\text{K}^+\) or chloride ions \(\text{Cl}^-\). Knowing the molarity not only allows us to predict the exact concentration of each substance present after the mixing but also caters to practicing safe and efficient chemical reactions by ensuring proper balance and proportions.

Mole Concept

The mole concept is a fundamental idea in chemistry that helps in quantifying the amount of a substance. One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities (atoms, molecules, or ions). This concept is the cornerstone of understanding how chemical quantities react. It allows chemists to measure substances efficiently and describe chemical changes and proportions. In the context of solutions, the mole helps in determining how many particles a solution contains. For example, in the original exercise, you calculate the moles of \(\text{KMnO}_4\) to find out how many moles of \(\text{K}^+\) and \(\text{MnO}_4^-\) ions are in your solution.
Here's why it's helpful:

• The mole concept relates the mass of a substance to the amount of substance, providing a bridge between the atomic scale and the grams you weigh in the lab.
• It aids in calculating the reactants needed and products formed in a reaction.

Using the mole concept ensures that your calculations remain accurate and your chemical interpretations correct, which is crucial for successful experimentation.

Chemical Solutions

Chemical solutions are uniform mixtures of two or more substances, where one is the solute that gets dissolved, and the other is the solvent doing the dissolving. Understanding chemical solutions is vital for laboratory work and chemical engineering, as they're involved extensively in reactions, formulations, and product development. The original exercise highlights mixing different solutions to find their ion concentrations. Real-world applications include preparing medical solutions, cleaning agents, and food products. Solutions can be:

• Homogeneous - Uniform throughout, like salt dissolved in water.
• Heterogeneous - Non-uniform, such as oil in water.

An important factor in solutions is how well the solutes dissolve, which is temperature-dependent, solute-specific, and influenced by solvents.
Here are some simplified steps on working with chemical solutions:

• Calculate the amount of solute needed using molarity, as shown in the earlier steps.
• Mix solute and solvent carefully, ensuring complete dissolution for accurate reaction results.
• Consider factors like temperature and mixing time for proper solute integration.

Mastering chemical solutions allows you to carry out controlled and reliable experiments, where solution properties remain consistent through each trial, leading to valid scientific outcomes.