Molar Concentration of Ions Example Problem

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.

Molar Concentration of Ions Problem

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
Msolute = Number of moles of CuCl2/Volume
Msolute = 0.07 mol/(600 mL x 1 L/1000 mL)
Msolute = 0.07 mol/0.600 L
Msolute = 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.

A Note About Solubility

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.