Copyright © Houghton Mifflin Company. All rights reserved. 15 | 1 Solutions Solutions are uniform mixtures. Solvent: the substance present in the highest.

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Copyright © Houghton Mifflin Company. All rights reserved. 15 | 1 Solutions Solutions are uniform mixtures. Solvent: the substance present in the highest percentage Solute: the dissolved substance, which is present in lesser amount Aqueous solutions: solutions with water as the solvent

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 2 Types of Solutions Solid in liquid (seawater). Gas in liquid (carbonated water). Liquid in liquid (antifreeze in water). Solid in solid (zinc in copper = brass).

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 3 The Solution Process: Ionic Compounds When ionic compounds dissolve in water they dissociate into ions.

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 4 Solution: Solid in Liquid Salt in water - separates into ions

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 5 Solubility When one substance dissolves in another, it is said to be soluble. –Salt is soluble in water When one substance does not dissolve in another, it is said to be insoluble. –Oil is insoluble in water

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 6 Figure 15.6: An oil layer floating on water.

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 7 How concentrated? A concentrated solution has a high proportion of solute to solution. A dilute solution has a low proportion of solute to solution.

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 8 How concentrated? (cont.) A saturated solution has the maximum amount of solute that will dissolve in the solvent. An unsaturated solution has less than the saturation limit.

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 9 How much? Concentration: the amount of solute in a given amount of solution

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 10 Solution Concentration- Molarity Moles of solute per 1 liter of solution If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc. molarity = moles of solute liters of solution

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 11 Molarity of ions CaCl 2 (aq) = Ca +2 (aq) + 2 Cl -1 (aq) A 1.0 M CaCl 2 (aq) solution contains 1.0 moles of CaCl 2 in each liter of solution Because each CaCl 2 dissociates to one Ca +2, 1.0 M CaCl 2 = 1.0 M Ca +2 Because each CaCl 2 dissociates to 2 Cl -1, 1.0 M CaCl 2 = 2.0 M Cl -1

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 12 Dilution Dilution: adding solvent to decrease the concentration of a solution The amount of solute stays the same, but the concentration decreases.

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 13 Dilution (cont.) Dilution Formula M 1 x V 1 = M 2 x V 2 # Moles/L · # L = # moles –In dilution we take a certain number of moles of solute and dilute to a bigger volume.

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 14

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 15 Molarity #15.3 Molarity = moles of solute liters of solution Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution M Molarity = mol = 1.25 L 15.6 g KBr (1 mol /119.0g) = mol KBr

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 16 Molarity #15.4 Molarity = moles of solute liter of solution Calculate the molarity of a solution prepared by dissolving 2.80 g of solid NaCl in enough water to make 135 mL of solution M Molarity = mol = L 2.80 g NaCl (1 mol /58.44g) = mol KBr 135 mL (1 L /1000 mL) = L

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 17 Calculating Ion Concentration from Molarity #15.5 Give the concentrations of all ions in each of the following solutions: M Na 2 SO M K 2 CrO 4 Na 2 SO 4  Na + (aq) + SO 4 2- (aq) 1.20 Na 2 SO 4  2.40 Na + (aq) SO 4 2- (aq) M Na +, 1.20M SO 4 2- K 2 CrO 4  K + (aq) + CrO 4 2- (aq) K 2 CrO 4  1.50 K + (aq) CrO 4 2- (aq) M K +, 0.750M CrO 4 2-

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 18 Dilution #15.6 A M 1 x V 1 = M 2 x V 2 What volume of 19M NaOH must be used to prepare 1.0 L of a 0.15M NaOH solution? 19 M x V 1 = 0.15M x 1.0 L V 1 = 0.15M x 1.0 L = L = 7.9 mL 19 M

Copyright © Houghton Mifflin Company. All rights reserved. 15 | 19 Dilution #15.6 B M 1 x V 1 = M 2 x V 2 What volume of 12M HCl must be used to prepare 250mL of a 1.0M HCl solution? 12 M x V 1 = 1 M x 0.25 L V 1 = 1M x 0.25 L = L = 21 mL 12 M

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