Question

Suppose \(50.0 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

Short answer

After mixing the given solutions, the new concentrations of the ions present are: Co²⁺: \(0.1667 \mathrm{M}\) Ni²⁺: \(0.1167 \mathrm{M}\) Cl⁻: \(0.5667 \mathrm{M}\)

Step-by-step solution

Calculate the initial moles of each ion

To find the initial moles, we can use the following formula: Mi = Ci * Vi For Co²⁺: Mi(Co²⁺) = 0.250 M * 50.0 mL = 12.5 mmol For Ni²⁺: Mi(Ni²⁺) = 0.350 M * 25.0 mL = 8.75 mmol Since each mole of CoCl₂ compound produces 2 moles of Cl⁻ ions and each mole of NiCl₂ compound produces 2 moles of Cl⁻ ions, the initial moles of Cl⁻ ions are: Mi(Cl⁻) = 2 * (12.5 + 8.75) = 42.5 mmol

Calculate the total volume of the mixed solution

In this exercise, we assume that the volumes are additive. Thus, the total volume of the mixed solution equals the sum of the initial volumes of CoCl₂ and NiCl₂ solutions: Vt = 50.0 mL + 25.0 mL = 75.0 mL

Calculate the new concentrations

We can now use the formula mentioned in the analysis part to calculate the new concentration of each ion: New concentration = (Initial concentration × Volume) / (Total volume) For Co²⁺: New concentration(Co²⁺) = (12.5 mmol) / (75.0 mL) = 0.1667 M For Ni²⁺: New concentration(Ni²⁺) = (8.75 mmol) / (75.0 mL) = 0.1167 M For Cl⁻: New concentration(Cl⁻) = (42.5 mmol) / (75.0 mL) = 0.5667 M

Final Answer

After mixing the given solutions, the new concentrations of the ions present are: Co²⁺: 0.1667 M Ni²⁺: 0.1167 M Cl⁻: 0.5667 M

Key concepts

Understanding Molarity

Molarity is a commonly used unit in chemistry to measure the concentration of a solution. It tells us how many moles of solute are present in one liter of solution. This unit is crucial because it helps chemists to prepare solutions with precise concentrations.
To calculate molarity, you use the formula:

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

In simpler terms, molarity describes the number of molecules or moles of a substance in a specific volume, helping us understand how concentrated that substance is in a mixture.
For example, if we have a solution with a molarity of 0.250 M cobalt(II) chloride (CoCl₂), it means that there are 0.250 moles of CoCl₂ in every liter of that solution.
This measurement is particularly useful when performing experiments or chemical reactions, as it allows for precise predictions and outcomes.

Breaking Down Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the amounts of substances involved in chemical reactions. It's like being a detective, where you're piecing together clues about how much of each ingredient you need to mix, or what you’ll get after a reaction.
A vital part of stoichiometry is balancing chemical equations. This means ensuring that the number of atoms for each element is the same on both the reactant and product sides of an equation. This balance reflects the conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

• For instance, in the reaction of CoCl₂ dissolving in water, stoichiometry helps us calculate the number of chloride ions ( ext{Cl}⁻) produced.
• Each molecule of ext{CoCl}₂ produces two ext{Cl}⁻ ions, so the stoichiometry reveals how these ions appear in solutions.

Stoichiometry is fundamental when working with chemical mixtures, guiding the determination of the right amounts of reactants needed to obtain the desired product.

The Art of Solution Preparation

Solution preparation is more than just mixing two liquids; it requires careful calculation and attention to detail to ensure that the solution has the correct concentration. This process generally involves dissolving a known quantity of solute into a solvent to make a solution of desired molarity.
To prepare a solution, you first calculate how much solute you need. Then, measure and dissolve it in a specified volume of solvent. For example, to prepare 75.0 ext{mL} of a mixed solution with specific concentrations, you need to calculate the initial moles, as done in the given problem.

• Measure initial solutions accurately.
• Add volumes together if the problem assumes they are additive.
• Finally, calculate the molarity of each ion post-mixing.

With accurate preparation, solutions achieve the exact concentration required for experiments, making solution preparation a foundational skill in chemistry.

Exploring Chemical Mixtures

Chemical mixtures involve combining two or more substances where each keeps its own chemical properties. These mixtures can be homogeneous like solutions, where the composition is uniform, or heterogeneous, where the components are distinguishable.
In the context of the exercise, mixing ext{CoCl}₂ and ext{NiCl}₂ creates a homogeneous mixture because when dissolved, these ionic compounds distribute evenly throughout the solution.

• Homogeneous mixtures, or solutions, allow uniform distribution of ions or molecules.
• They ensure that the chemical properties are consistent throughout the entire sample.

Understanding chemical mixtures is crucial, especially for applications in stoichiometry, where the quantity and properties of mixed solutions affect outcomes in chemical reactions.