Problem 75 - General Chemistry: Principles and Modern Applications

Question

What volume of \(0.248 \mathrm{M} \mathrm{CaCl}_{2}\) must be added to \(335 \mathrm{mL}\) of \(0.186 \mathrm{M} \mathrm{KCl}\) to produce a solution with a concentration of \(0.250 \mathrm{M} \mathrm{Cl}^{-2}\) Assume that the solution volumes are additive.

Step-by-Step Solution

Answer

You need to add 94.7 mL of 0.248M CaCl2 solution to the 335 mL of 0.186M KCl solution to produce a solution with a Cl- concentration of 0.250M.

Step 1: Calculate moles of chloride in KCl solution

Moles of chloride in KCl is given by the product of volume and molarity, since KCl has only one atom of chloride, so we have \((0.186 Moles/L)*(0.335L) = 0.06231 Moles\)

Step 2: Calculate total moles of chloride required

The total moles of Cl- required to achieve 0.250M in the entire solution. We do not know the final volume yet, but it will be 335 mL of the KCl solution + X mL of the CaCl2 solution. As both these volumes are in mL, they can be added directly, This gives us \(Final Moles of Cl- = 0.250 M * (335+X) mL * 1L/1000mL\)

Step 3: Determine moles of chloride in CaCl2

The total moles of Cl- required (from step 2) will be equal to the moles contributed by the KCl solution and the CaCl2 solution. And each molecule of CaCl2 contributes 2 Cl- ions. This gives us \(0.250 M * (335+X) mL * 1L/1000mL = 0.06231 Moles(from KCl) + 2 * 0.248M * X mL * 1L/1000mL(from CaCl2)\)

Step 4: Solve for X (volume of CaCl2 solution)

This is a simple algebra problem now. The only unknown in the equation is X - the volume in mL of the CaCl2 solution. Solving the equation for X gives \(X = 94.7 mL\)

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution, showing how many moles of a solute are present per liter of solution. Understanding molarity helps you calculate and prepare solutions of specific concentrations with precision. Here's a friendly breakdown:

• It's expressed in moles per liter (mol/L), represented by the symbol "M".
• A 1 M solution has 1 mole of a solute per 1 liter of solution. For example, a 0.248 M CaCl_2 solution means there are 0.248 moles of calcium chloride in each liter.

Molarity is vital for stoichiometry in chemical reactions and conducting experiments. When combining solutions, as in our exercise, knowing the molarity allows us to predict and control the exact outcome as desired.

Additive Volumes

Additive volumes refer to the assumption that when two solutions are mixed, the total volume is simply the sum of the individual volumes. This isn't always precisely accurate in real-world scenarios because of interactions between molecules, but it's a common simplification in chemistry calculations.

• For our exercise, mixing 335 mL of KCl solution and a certain volume of CaCl_2, the total volume is assumed to be the sum of both.
• This simplification helps in concentration calculations by providing an easy way to determine the final volume.

While additive volumes simplify calculations, remember that in highly precise fields, real measurements might show slight deviations.

Concentration Calculations

Concentration calculations involve determining the amount of solute present in a given volume of solution. This process often requires combining several mathematical steps to find a desired concentration, like achieving a specific chloride ion concentration in the exercise.

• First, calculate the moles of solute already present, using molarity and existing solution volumes.
• Determine the desired concentration outcome after adding more solution.
• Solve for unknowns, such as additional volumes required, ensuring the concentration target is met.

Understanding these calculations ensures that you mix solutions correctly for desired chemical compositions in educational and practical applications, making it a fundamental skill in chemistry.