Science, Tech, Math Science Molar Concentration of Ions Example Problem Share Flipboard Email Print GIPhotoStock / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry in Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate by Todd Helmenstine Updated December 03, 2018 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?Solution To find the molarity of the ions, we must first find 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 = 63.55Atomic mass of Cl = 35.45Atomic mass of CuCl2 = 1(63.55) + 2(35.45)Atomic mass of CuCl2 = 63.55 + 70.9Atomic mass of CuCl2 = 134.45 g/molNumber of moles of CuCl2 = 9.82 g x 1 mol/134.45 gNumber of moles of CuCl2 = 0.07 molMsolute = Number of moles of CuCl2/VolumeMsolute = 0.07 mol/(600 mL x 1 L/1000 mL)Msolute = 0.07 mol/0.600 LMsolute = 0.12 mol/LStep 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 CuCl2Ion/solute = 2 moles of Cl-/1 mole CuCl2Step 3: Find the ion molarity. M of Cl- = M of CuCl2 x ion/soluteM of Cl- = 0.12 moles CuCl2/L x 2 moles of Cl-/1 mole CuCl2M of Cl- = 0.24 moles of Cl-/LM of Cl- = 0.24 MAnswer 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 tiny bit trickier when a substance is only partially soluble. Basically, you set up the problem the same way, but then multiply the answer time the fraction that dissolves. Continue Reading