1. Concentration of Ions in Solution Remember: concentration IS molarity.

2. Ionic compounds Exist as crystals

3. Dissociation: when a salt crystal breaks up into ions when placed in water.

4. For example: NaCl  Na + + Cl - 1.0 mole NaCl 1.0 mole Na + 1.0 mole Cl - 1.0 L Molarity = moles M = 1 mol = 1.0M litres 1.0 L So every component is 1.0M.

5. CaBr 2  Ca +2 + 2Br -1 1.0 mole 1.0 mole 2.0 moles 1.0 L Calculate the molarity of each ion M= 1.0 moles Ca + = 1.0 M Ca + 1.0 litres solution 1.0 M CaBr 2 1.0M Ca +2 2.0M Br -1 Notice the mole ratios are equal to the molarity ratios

6. To find the concentration of ions in solution: 1.Write an equation for the dissociation of the ions. 2.Balance it. 3.Calculate the molarity of each ion.

7. Al(OH) 3  Al +3 + 3OH -1 1.0 mole 1.0 mole 3.0 moles 1.0 L M= 1.0 mol= 1.0 M Al +3 1.0 L M= 3.0 mol= 3.0 M OH - 1.0 L Notice the mole ratios are equal to the molarity ratios

8. Find the concentration of ions in a solution made by dissolving 2.0 moles of calcium phosphate 2.5 L of water. Ca 3 (PO 4 ) 2  3 Ca +2 + 2 PO 4 -3 2.0 moles 6.0 moles 4.0 moles M = 6.0 moles = 2.4 M Ca +2 2.5 L M = 4.0 moles = 1.6 M PO 4 -3 2.5 L Notice the mole ratios are equal to the molarity ratios

9. Find the concentration of Sr +2 & Br -1 in a solution made by dissolving 365g of strontium bromide in 836ml of water. SrBr 2(s)  Sr +2 (aq) + 2Br -1 (aq) 1 mole SrBr 2 = 247.4g x= 1.48 mol SrBr 2 x mol 365g M = moles = 1.48 mol = 1.76 M SrBr 2 litre 0.836 L Therefore 1.76 M Sr +2 and 3.52 M Br -1

10. What is the concentration of the sodium ion in a solution made by dissolving 23.0g of sodium phosphate in 712 cm 3 of water? Na 3 PO 4(s)  3Na + (aq) + PO 4 -3 (aq) 1 mole = 164g x 23.0g x= 0.140 mol Na 3 PO 4 M = 0.140 mol = 0.197 M Na 3 PO 4 (and PO 4 -3 ) 0.712 L 3 x 0.197 M = 0.591 M Na +