What is the molar concentration of chloride ions in a

solutionprepared by mixing 100mL of 2.0 M KCL with 50 L of a 1.5

MCaCl2 solution?

The concentration of chloride ions in the final solution is 3 M.

Explanation:

The number of moles present in a solution can be calculated as follows:

number of moles = concentration in molarity * volume

In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻

For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:

number of moles of Cl⁻ = 2 * number of moles of CaCl₂

number of moles of Cl⁻ = 2 (50 l * 1.5 mol / l) = 150 mol Cl⁻

The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol) 150.2 mol.

Assuming ideal behavior, the volume of the final solution will be (50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:

Concentration = number of moles of Cl⁻ / volume

Concentration = 150.2 mol / 50.1 l = 3.0 M