Molar Concentration of Ions Example Problem

This example problem demonstrates how to calculate the molarity of ions in an aqueous solution. Molarity is a concentration in terms of moles per liter of solution. Because an ionic compound dissociates into its components cations and anions in solution, the key to the problem is identifying how many moles of ions are produced during dissolution.

A solution is prepared by dissolving 9.82 grams of copper chloride (CuCl₂) in enough water to make 600 milliliters of solution. What is the molarity of the Cl ions in the solution?

To find the molarity of the ions, first determine the molarity of the solute and the ion-to-solute ratio.

Step 1: Find the molarity of the solute.

From the periodic table:

Atomic mass of Cu = 63.55
Atomic mass of Cl = 35.45
Atomic mass of CuCl₂ = 1(63.55) + 2(35.45)
Atomic mass of CuCl₂ = 63.55 + 70.9

Atomic mass of CuCl₂ = 134.45 g/mol

Number of moles of CuCl₂ = 9.82 g x 1 mol/134.45 g
Number of moles of CuCl₂ = 0.07 mol
M_solute = Number of moles of CuCl₂/Volume
M_solute = 0.07 mol/(600 mL x 1 L/1000 mL)
M_solute = 0.07 mol/0.600 L
M_solute = 0.12 mol/L

Step 2: Find the ion-to-solute ratio.

CuCl₂ dissociates by the reaction

CuCl₂ → Cu²⁺ + 2Cl^-

Ion/solute = Number of moles of Cl^-/number of moles of CuCl₂
Ion/solute = 2 moles of Cl^-/1 mole CuCl₂

Step 3: Find the ion molarity.

M of Cl^- = M of CuCl₂ x ion/solute
M of Cl^- = 0.12 moles CuCl₂/L x 2 moles of Cl^-/1 mole CuCl₂
M of Cl^- = 0.24 moles of Cl^-/L
M of Cl^- = 0.24 M

The molarity of the Cl ions in the solution is 0.24 M.

While this calculation is straightforward when an ionic compound completely dissolves in solution, it's a bit trickier when a substance is only partially soluble. You set up the problem the same way but then multiply the answer by the fraction that dissolves.