Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Nature of Solutes in Solutions Spread evenly throughout the solution. Cannot be separated by filtration. Can be separated by evaporation. Are not visible, but can give a color to the solution. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Examples of Solutions The solute and solvent can be a solid, liquid, and/or a gas. Table 12.3 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Water Water Is the most common solvent. Is a polar molecule. Forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Formation of a Solution Na+ and Cl- ions On the surface of a NaCl crystal are attracted to polar water molecules. In solution are hydrated as several H2O molecules surround each. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Equations for Solution Formation When NaCl(s) dissolves in water, the process can be written as: H2O NaCl(s) Na+(aq) + Cl- (aq) solid separation of ions The Na+ ions are attracted to the oxygen atom ( -) of water. The Cl- ions are attracted to the hydrogen atom (+) of water.

Like Dissolves Like Two substances form a solution When there is an attraction between the particles of the solute and solvent. When a polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl. When a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease.

Water and a Polar Solute Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2 CH2Cl2(nonpolar) (polar) I2 (nonpolar) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Electrolytes and Nonelectrolytes Chapter 12 Solutions Electrolytes and Nonelectrolytes Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Solutes and Ionic Charge In water, Strong electrolytes produce ions and conduct an electric current. Weak electrolytes produce a few ions. Nonelectrolytes do not produce ions. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Strong Electrolytes Strong electrolytes Dissociate in water producing positive and negative ions. Conduct an electric current in water. In equations show the formation of ions in aqueous (aq) solutions. H2O 100% ions NaCl(s) Na+(aq) + Cl− (aq) H2O CaBr2(s) Ca2+(aq) + 2Br− (aq)

Weak Electrolytes A weak electrolyte Dissociates only slightly in water. In water forms a solution of only a few ions and mostly undissociated molecules. HF(g) + H2O(l) H3O+(aq) + F- (aq) NH3(g) + H2O(l) NH4+(aq) + OH- (aq) Note: Unequal lengths of the arrows

Nonelectrolytes Nonelectrolytes Dissolve as molecules in water. Do not produce ions in water. Do not conduct an electric current. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Solutions Solubility Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Solubility Solubility Is the maximum amount of solute that dissolves in a specific amount of solvent. Can be expressed as grams of solute in 100 grams of solvent, usually water. g of solute 100 g water

Effect of Temperature on Solubility Depends on temperature. Of most solids increases as temperature increases. Of gases decreases as temperature increases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Unsaturated Solutions Contain less than the maximum amount of solute. Can dissolve more solute. Dissolved solute Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Saturated Solutions Saturated solutions Contain the maximum amount of solute that can dissolve. Have undissolved solute at the bottom of the container. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Supersaturated Solutions http://www.stevespanglerscience.com/content/science-video/super-saturated-solution

Soluble and Insoluble Salts Ionic compounds that Dissolve in water are soluble salts. Do not dissolve in water are insoluble salts. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Solubility Rules Soluble salts Typically contain at least one ion from Groups 1A(1) or NO3−, or C2H3O2− (acetate). Table 12.3 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Using the Solubility Rules The solubility rules predict if a salt Is soluble or Insoluble in water. Table 12.4 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Formation of a Solid When solutions of salts are mixed, A solid forms if ions of an insoluble salt are present. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Equations for Forming Solids A molecular equation shows the formulas of the compounds. Pb(NO3)(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq) An ionic equation shows the ions of the compounds. Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq) PbCl2(s) + 2Na+(aq) + 2NO3−(aq) A net ionic equation shows only the ions that form a solid. Ions remaining in solution are spectator ions. Pb2+(aq) + 2Cl−(aq) PbCl2(s)

Equations for the Insoluble Salt STEP 1 Observe the ions in the reactants. Pb2+(aq) + 2NO3−(aq) 2Na+(aq) + 2Cl−(aq) STEP 2 Determine if any new ion combinations are insoluble salts. Yes. PbCl2(s) STEP 3 Ionic equation with insoluble salt product. Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq) PbCl2(s) + 2Na+(aq) + 2NO3−(aq) STEP 4 Net ionic equation. Pb2+(aq) + 2Cl−(aq) PbCl2(s)

Chapter 12 Solutions Molarity and Dilution Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molarity (M) Molarity (M) is A concentration term for solutions. The moles of solute in 1 L solution. moles of solute liter of solution

Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared By weighing out 58.5 g NaCl (1.00 mol) and Adding water to make 1.00 liter of solution. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Calculation of Molarity What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mol/L) STEP 2 Plan g NaOH mol NaOH molarity STEP 3 Conversion factors 1 mol NaOH = 40.00 g 1 mol NaOH and 40.00 g NaOH 40.00 g NaOH 1 mol NaOH

Calculation of Molarity (cont.) STEP 4 Calculate molarity. 6.00 g NaOH x 1 mol NaOH = 0.150 mol 40.00 g NaOH 0.150 mol = 0.300 mol = 0.300 M NaOH 0.500 L 1 L

Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. Table 2.6 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molarity in Calculations How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl mol KCl g KCl

Molarity in Calculations (cont.) STEP 3 Conversion factors 1 mol KCl = 74.55 g 1 mol KCl and 74.55 g KCl 74.55 g KCl 1 mol KCl 1 L KCl = 0.720 mol KCl 1 L and 0.720 mol KCl 0.720 mol KCl 1 L STEP 4 Calculate grams. 0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g KCl 1 L 1 mol KCl

Dilution In a dilution Water is added. Volume increases. Concentration decreases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Comparing Initial and Diluted Solutions In the initial and diluted solution The moles of solute are the same. The concentrations and volumes are related by the equation M1V1 = M2V2 initial diluted

Dilution Calculations What is the molarity if 0.180 L of 0.600 M KOH is diluted to a final volume of 0.540 L? STEP 1 Prepare a table: M1= 0.600 M V1 = 0.180 L M2= ? V2 = 0.540 L STEP 2 Solve dilution equation for unknown. M1V1 = M2V2 M1V1/ V2 = M2 STEP 3 Set up and enter values: M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M V2 0.540 L