Section 15.1 Forming Solutions Steven S. Zumdahl Susan A. Zumdahl Donald J. DeCoste Gretchen M. Adams University of Illinois at Urbana-Champaign Chapter 15 Solutions

Section 15.1 Forming Solutions 1.To understand the process of dissolving 2.To learn why certain substances dissolve in water 3.To learn qualitative terms describing the concentration of a solution 4.To understand the factors that affect the rate at which a solid dissolves Objectives

Section 15.1 Forming Solutions What is a solution? Solution – homogeneous mixture  Solvent – substance present in largest amount  Solutes – other substances in the solution  Aqueous solution – solution with water as the solvent

Section 15.1 Forming Solutions

Section 15.1 Forming Solutions A. Solubility Solubility of Ionic Substances Ionic substances breakup into individual cations and anions.

Section 15.1 Forming Solutions A. Solubility Solubility of Ionic Substances Polar water molecules interact with the positive and negative ions of a salt.

Section 15.1 Forming Solutions A. Solubility Solubility of Polar Substances Ethanol is soluble in water because of the polar OH bond.

Section 15.1 Forming Solutions A. Solubility Solubility of Polar Substances Why is solid sugar soluble in water?

Section 15.1 Forming Solutions A. Solubility Substances Insoluble in Water Nonpolar oil does not interact with polar water. Water-water hydrogen bonds keep the water from mixing with the nonpolar molecules.

Section 15.1 Forming Solutions A. Solubility How Substances Dissolve A “hole” must be made in the water structure for each solute particle. The lost water-water interactions must be replaced by water-solute interactions. “like dissolves like”

Section 15.1 Forming Solutions B. Solution Composition: An Introduction The solubility of a solute is limited.  Saturated solution – contains as much solute as will dissolve at that temperature  Unsaturated solution – has not reached the limit of solute that will dissolve

Section 15.1 Forming Solutions B. Solution Composition: An Introduction  Supersaturated solution – occurs when a solution is saturated at an elevated temperature and then allowed to cool but all of the solid remains dissolved Contains more dissolved solid than a saturated solution at that temperature Unstable – adding a crystal causes precipitation

Section 15.1 Forming Solutions B. Solution Composition: An Introduction Solutions are mixtures. Amounts of substances can vary in different solutions.  Specify the amounts of solvent and solutes  Qualitative measures of concentration concentrated – relatively large amount of solute dilute – relatively small amount of solute

Section 15.1 Forming Solutions B. Solution Composition: An Introduction Which solution is more concentrated?

Section 15.1 Forming Solutions B. Solution Composition: An Introduction Which solution is more concentrated?

Section 15.1 Forming Solutions C. Factors Affecting the Rate of Dissolving Surface area Stirring Temperature

Section 15.2 Describing Solution Composition 1. To understand mass percent and how to calculate it 2. To understand and use molarity 3. To learn to calculate the concentration of a solution made by diluting a stock solution Objectives

Section 15.2 Describing Solution Composition A. Solution Composition: Mass Percent

Section 15.2 Describing Solution Composition Exercise What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6%

Section 15.2 Describing Solution Composition B. Solution Composition: Molarity Concentration of a solution is the amount of solute in a given volume of solution.

Section 15.2 Describing Solution Composition B. Solution Composition: Molarity Consider both the amount of solute and the volume to find concentration.

Section 15.2 Describing Solution Composition Exercise You have 1.00 mol of sugar in mL of solution. Calculate the concentration in units of molarity M

Section 15.2 Describing Solution Composition Exercise You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? L

Section 15.2 Describing Solution Composition Exercise A g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 1.57 M

Section 15.2 Describing Solution Composition Exercise Consider separate solutions of NaOH and KCl made by dissolving g of each solute in mL of solution. Calculate the concentration of each solution in units of molarity M NaOH 5.37 M KCl

Section 15.2 Describing Solution Composition Concept Check A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution? a)Add water to the solution. b)Pour some of the solution down the sink drain. c)Add more sodium chloride to the solution. d)Let the solution sit out in the open air for a couple of days. e)At least two of the above would decrease the concentration of the salt solution.

Section 15.2 Describing Solution Composition B. Solution Composition: Molarity To find the moles of solute in a given volume of solution of known molarity, use the definition of molarity.

Section 15.2 Describing Solution Composition B. Solution Composition: Molarity Standard solution - a solution whose concentration is accurately known To make a standard solution  Weigh out a sample of solute.  Transfer to a volumetric flask.  Add enough solvent to the mark on flask.

Section 15.2 Describing Solution Composition C. Dilution Water can be added to an aqueous solution to dilute the solution to a lower concentration. Only water is added in the dilution – the amount of solute is the same in both the original and final solution.

Section 15.2 Describing Solution Composition D. Dilution Diluting a solution  Transfer a measured amount of original solution to a flask containing some water.  Add water to the flask to the mark (with swirling) and mix by inverting the flask.

Section 15.2 Describing Solution Composition Exercise What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution? 60.0 mL

Section 15.3 Properties of Solutions 1.To learn to solve stoichiometric problems involving solution reactions 2.To do calculations involving acid-base reactions 3.To learn about normality and equivalent weight 4.To use normality in stoichiometric calculations 5.To understand the effect of a solute on solution properties Objectives

Section 15.3 Properties of Solutions A. Stoichiometry of Solution Reactions

Section 15.3 Properties of Solutions Concept Check 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).  What precipitate will form? lead(II) phosphate, Pb 3 (PO 4 ) 2  What mass of precipitate will form? 1.1 g Pb 3 (PO 4 ) 2

Section 15.3 Properties of Solutions Where do we want to go?  Find the mass of solid Pb 3 (PO 4 ) 2 formed. How do we get there?  What are the ions present in the combined solution?  What is the balanced net ionic equation for the reaction?  What are the moles of reactants present in the solution?  Which reactant is limiting?  What moles of Pb 3 (PO 4 ) 2 will be formed?  What mass of Pb 3 (PO 4 ) 2 will be formed? Let’s Think About It

Section 15.3 Properties of Solutions B. Neutralization Reactions An acid-base reaction is called a neutralization reaction. Steps to solve these problems are the same as before.

Section 15.3 Properties of Solutions Concept Check For the titration of sulfuric acid (H 2 SO 4 ) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to neutralize 1.00 L of M sulfuric acid? 1.00 mol NaOH

Section 15.3 Properties of Solutions Where do we want to go?  Find the moles of NaOH required for the reaction. How do we get there?  What are the ions present in the combined solution? What is the reaction?  What is the balanced net ionic equation for the reaction?  What are the moles of H + present in the solution?  How much OH – is required to react with all of the H + present? Let’s Think About It

Section 15.3 Properties of Solutions C. Normality Unit of concentration  One equivalent of acid – amount of acid that furnishes 1 mol of H + ions  One equivalent of base – amount of base that furnishes 1 mol of OH  ions  Equivalent weight – mass in grams of 1 equivalent of acid or base

Section 15.3 Properties of Solutions C. Normality

Section 15.3 Properties of Solutions C. Normality

Section 15.3 Properties of Solutions C. Normality To find number of equivalents

Section 15.3 Properties of Solutions C. Normality Advantage of equivalents

Section 15.3 Properties of Solutions D. Boiling Point and Freezing Point The presence of solute “particles” causes the liquid range to become wider.  Boiling point increases  Freezing point decreases

Section 15.3 Properties of Solutions D. Boiling Point and Freezing Point Why does the boiling point of a solution increase?  Solute particles block some of the water molecules trying to enter the bubble.  Need higher pressure to maintain the bubble.  Forming a bubble in a solution

Section 15.3 Properties of Solutions D. Boiling Point and Freezing Point  Comparing bubbles

Section 15.3 Properties of Solutions D. Boiling Point and Freezing Point Colligative property – a solution property that depends on the number of solute particles present