Concentration of Ions in Solution Remember: concentration IS molarity.

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Presentation transcript:

Concentration of Ions in Solution Remember: concentration IS molarity.

Ionic compounds Exist as crystals

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

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

CaBr 2  Ca Br 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 litres solution 1.0 M CaBr 2 1.0M Ca M Br -1 Notice the mole ratios are equal to the molarity ratios

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.

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

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 PO moles 6.0 moles 4.0 moles M = 6.0 moles = 2.4 M Ca L M = 4.0 moles = 1.6 M PO L Notice the mole ratios are equal to the molarity ratios

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 L Therefore 1.76 M Sr +2 and 3.52 M Br -1

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= mol Na 3 PO 4 M = mol = M Na 3 PO 4 (and PO 4 -3 ) L 3 x M = M Na +