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

The concentrations of the ions present after mixing the solutions are: Co²⁺: \(0.167\, \mathrm{M}\), Ni²⁺: \(0.117\, \mathrm{M}\), and Cl⁻: \(0.333\, \mathrm{M}\).

Step-by-step solution

Identify the given information

We are given the following information: - Initial volume of CoCl₂ solution: \(50.0\, \mathrm{mL}\) - Concentration of CoCl₂ solution: \(0.250\, \mathrm{M}\) - Initial volume of NiCl₂ solution: \(25.0\, \mathrm{mL}\) - Concentration of NiCl₂ solution: \(0.350\, \mathrm{M}\)

Calculate the initial moles of ions

We will use the formula \(M_1V_1 = n\) to find the initial moles of each ion. Here, \(M_1\) is the molar concentration of the solution, \(V_1\) is the volume of the solution, and \(n\) is the number of moles of the ion. For Co²⁺ ions: \(n_{Co} = M_1V_1 = 0.250\, \mathrm{M} \times 50.0\, \mathrm{mL} = 12.5 \,\mathrm{mmol}\) For Ni²⁺ ions: \(n_{Ni} = M_1V_1 = 0.350\, \mathrm{M} \times 25.0\, \mathrm{mL} = 8.75 \,\mathrm{mmol}\) For Cl⁻ ions (remember there are a total two Cl⁻ ions in both CoCl₂ and NiCl₂): \(n_{Cl} = M_1V_1_{CoCl_{2}} + M_1V_1_{NiCl_{2}} = (2 \times 0.250\, \mathrm{M})(50.0\, \mathrm{mL}) + (2 \times 0.350\, \mathrm{M})(25.0\, \mathrm{mL}) = 25.0\, \mathrm{mmol}\)

Calculate the final volume of the mixed solution

We are given that the volumes of the solutions are additive, so the total volume of the mixed solution will be the sum of the individual volumes: \(V_{total} = V_{CoCl_{2}} + V_{NiCl_{2}} = 50.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 75.0\, \mathrm{mL}\)

Calculate the concentration of each ion in the final solution

To find the concentration of each ion in moles per liter, divide the number of moles of each ion by the volume of the final solution in liters. Then, convert the volume from milliliters to liters by dividing by 1000. For Co²⁺ ions: \[M_{Co} = \frac{n_{Co}}{V_{total}} = \frac{12.5 \,\mathrm{mmol} }{75.0\, \mathrm{mL}} = 0.167\, \mathrm{M}\] For Ni²⁺ ions: \[M_{Ni} = \frac{n_{Ni}}{V_{total}} = \frac{8.75 \,\mathrm{mmol} }{75.0\, \mathrm{mL}} = 0.117\, \mathrm{M}\] For Cl⁻ ions: \[M_{Cl} = \frac{n_{Cl}}{V_{total}} = \frac{25.0 \,\mathrm{mmol}}{75.0\, \mathrm{mL}} = 0.333\, \mathrm{M}\] So, after mixing the solutions, the concentrations of the ions present are: Co²⁺: \(0.167\, \mathrm{M}\), Ni²⁺: \(0.117\, \mathrm{M}\), and Cl⁻: \(0.333\, \mathrm{M}\).

Key concepts

Molarity

Molarity, represented by the symbol 'M', is the measure of concentration denoted in moles of solute per liter of solution. It allows chemists to relate the volume of a solution to the amount of substance present in that solution.

The general formula for calculating molarity is given by: \[ M = \frac{n}{V} \]where 'M' is the molarity, 'n' is the number of moles of the solute, and 'V' is the volume of the solution in liters. Remember that 1 liter equals 1000 milliliters, so converting volumes from milliliters to liters is a crucial step in molarity calculations.

For instance, when the solution concentration calculation is performed, such as combining two solutions of different molarities, total molarity of individual ions can be determined by first finding the moles of each solute, then summing their volumes and finally dividing the total moles of solute by the total volume in liters.

Stoichiometry

Stoichiometry is the section of chemistry that involves quantifying relationships between reactants and products in a chemical reaction. In solution stoichiometry, this often involves using molarity to convert between moles of solute and the volume of solution.

Understanding stoichiometry is critical when mixing chemical solutions to predict the concentrations of all species present after mixing. The stoichiometric coefficients, which refer to the numbers in front of compounds in a balanced chemical equation, also play a significant role in these calculations.

For example, when calculating the moles of chloride ions (\(Cl^-\)) in a solution of \(CoCl_2\) and \(NiCl_2\), the stoichiometry of the compounds shows that each formula unit of \(CoCl_2\) and \(NiCl_2\) releases two chloride ions. This means you must double the moles of the metal chloride solutions to get the total moles of \(Cl^-\) ions.

Chemical Solution Mixing

Chemical solution mixing involves combining two or more solutions to create a new solution. When solutions are mixed, their volumes are often additive; however, their molarities cannot be directly added together. Instead, calculations must account for the individual substances present and their proportions.

In the context of calculating ion concentrations after mixing, the resulting molarity of each ion depends on the initial molarities and volumes of the mixed solutions and the total volume of the resulting solution.

For example, adding a \(CoCl_2\) solution to a \(NiCl_2\) solution requires calculating the combined volume of the mixed solution, which in most cases will simply be the sum of the two individual volumes. Then, as seen in the provided solution steps, the concentration of each ion must be determined separately by dividing the moles of each ion by the final combined volume.