Chapter 15 Solutions

Objectives To understand the process of dissolving To learn why certain substances dissolve in water To learn qualitative terms describing the concentration of a solution To understand the factors that affect the rate at which a solid dissolves   

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

A. Solubility Review Solubility of Ionic Substances Ionic substances breakup into individual cations and anions.

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

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

A. Solubility Solubility of Polar Substances Why is solid sugar soluble in water?

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.

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”

H2O, The Universal Solvent Much of chemistry that affects each of us occurs among substances dissolved in water. Virtually all chemistry that makes life possible occurs in an aqueous environment. Aqueous Solution: a solution in which water is the dissolving medium or solvent. Solvent: the dissolving medium in a solution. Solute: a substance dissolved in a liquid to form a solution.

H2O, The Bent Universal Solvent Waters polar characteristics: Electrons that are shared spend more of their time with the oxygen atom

H2O, The Universal Solvent Hydration: hydration of ions tends to cause salts to “fall apart” or dissolve. Water pulls salts apart using it’s polarity http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/nacl2.gif

Some Properties of Water 1. Water is “bent” or V-shaped. 2. The O-H bonds are covalent. 3. Water is a polar molecule. 4. Hydration occurs when salts dissolve in water.

A Solute: Dissolves in water (or other “solvent”) Changes phase (if different from the solvent) Is present in lesser amount (if the same phase as the solvent)

A Solvent: Retains its phase (if different from the solute) Is present in greater amount (if the same phase as the solute)

Electrical Conductivity: the ability to conduct electricity in solution Strong Electrolytes: (HCl) Solution that conducts electricity efficiently Weak Electrolytes: (Vinegar) Solution that conducts electricity inefficiently Non Electrolytes: (Sugar) Solution that prevents the flow of electricity

Electrolytes in the Body Carry messages to and from the brain as electrical signals Maintain cellular function with the correct concentrations electrolytes

pure water, sugar solution Electrolytes Strong - conduct current efficiently NaCl, HNO3 Weak - conduct only a small current vinegar, tap water Non - no current flows pure water, sugar solution

hydrochloric and sulfuric acid Acids Strong acids -dissociate completely to produce H+ in solution hydrochloric and sulfuric acid Weak acids - dissociate to a slight extent to give H+ in solution acetic and formic acid

Bases Strong bases - react completely with water to give OH ions. sodium hydroxide Weak bases - react only slightly with water to give OH ions. ammonia

H2O, The Universal Solvent It is very important to recognize that when an ionic substance dissolves in water they break up into the individual cations and anions. Example: Ammonium Nitrate in water: Can you draw the ions associated with ammonium nitrate? When cations and anions have been dissolved in solution they move around independently

H2O, The Universal Solvent Solubility: the amount of substance that dissolves in a given volume of solvent at a given temperature The solubility of substances in water varies greatly. For ionic compounds it depends on the relative attraction between the ions. “Like dissolves Like”, in general polar or ionic compounds are expected to be soluble in water than non polar substances

Solubility Review: Rules for Solubility 1. Most nitrate (NO3-) salts are soluble 2. Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion (NH4+) are soluble 3. Most Chloride, Bromide, and Iodide salts are soluble. Exceptions: those containing Ag+, Pb2+, and Hg22+ 4. Most sulfate ions are soluble. Exceptions: BaSO4, PbSO4, Hg2SO4 and CaSO4 5. Most hydroxide salts are insoluable. Exceptions: NaOH and KOH are soluble 6. Most Sulfide (S2-), carbonate (CO32-), chromate (CrO42-) and phosphates (PO43-) are insoluable

AgNO3(aq)  Na3PO4(aq)  BaCl2(aq)  FeCl3(aq)  NaOH(aq)  Solubility Practice Show the ions . AgNO3(aq)  Na3PO4(aq)  BaCl2(aq)  FeCl3(aq)  NaOH(aq)  CuSO4(aq) 

Solubility Practice Answers Show the ions. AgNO3(aq)  Ag+ + NO3- Na3PO4(aq) PO43- + 3 Na+ BaCl2(aq)  2 Cl- + Ba2+ FeCl3(aq)  Fe3+ + 3Cl- NaOH(aq)  Na+ + OH- CuSO4(aq)  SO42- + Cu2+

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

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

A solution is a homogeneous mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

Solution Definitions Classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: Warm the solvent so that it will dissolve more, then cool the solution Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

Supersaturated : Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”

How to interpret a graphical representation of solute in solvent. Solubility Curves How to interpret a graphical representation of solute in solvent.

Solubility curve Saturated Supersaturated Unsaturated

Any point on a line represents a saturated solution. In a saturated solution, the solvent contains the maximum amount of solute. Example At 90oC, 40 g of NaCl(s) in 100g H2O(l) represent a saturated solution.

Any point below a line represents an unsaturated solution. In an unsaturated solution, the solvent contains less than the maximum amount of solute. Example At 90oC, 30 g of NaCl(s) in 100g H2O(l) represent an unsaturated solution. 10 g of NaCl(s) have to be added to make the solution saturated.

Solubility Curve Any point above a line represents a supersaturated solution. In a supersaturated solution, the solvent contains more than the maximum amount of solute. A supersaturated solution is very unstable and the amount in excess can precipitate or crystallize. Example At 90oC, 50 g of NaCl(s) in 100g H2O(l) represent a supersaturated solution. Eventually, 10 g of NaCl(s) will precipitate.

Solubility curve Any solution can be made saturated, unsaturated, or supersaturated by changing the temperature.

Answer: 70 grams of KBr can dissolve in 100g of water at 20°C How many grams of potassium bromide (KBr) can dissolve in 100 grams of water at 20°C? Answer: 70 grams of KBr can dissolve in 100g of water at 20°C 70g

How many grams of potassium nitrate (KNO3) can dissolve in 100 g of water at 60°C?

How many grams of potassium bromide (KBr) can dissolve in 100 g of water at 20°C?

How many grams of sodium chloride (NaCl) can dissolve in 100 g of water at 100°C?

200x2=400 g NaClO3 can be dissolved in 200 g of water at 80°C How many grams of sodium chlorate (NaClO3) can dissolve in 200 g of water at 80°C? 200 g 200g per 100 g of water, so in 200 g of water we will have to double it: 200x2=400 g NaClO3 can be dissolved in 200 g of water at 80°C

At what temperature can 150 grams of potassium nitrate (KNO3) dissolve in 100 g of water? Answer: 150 grams of Potassium nitrate can be dissolved in 100 g of water at 65°C

At what temperature can 100 grams of potassium bromide (KBr) dissolve in 100 g of water? Answer: 100 g of potassium bromide can dissolve in 100 g of water at 82°C

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

B. Solution Composition: An Introduction Which solution is more concentrated?

B. Solution Composition: An Introduction Which solution is more concentrated?

C. Factors Affecting the Rate of Dissolving Surface area Stirring Temperature

Objectives To understand mass percent and how to calculate it To understand and use molarity To learn to calculate the concentration of a solution made by diluting a stock solution

A. Solution Composition: Mass Percent

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% [5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6%

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

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

Exercise You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M 1.00 mol / (125.0 / 1000) = 8.00 M

Exercise You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 0.200 L 2.00 mol / 10.0 M = 0.200 L

Exercise A 500.0-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 500.0 g is equivalent to 2.355 mol K3PO4 (500.0 g / 212.27 g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L).

Exercise Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl [100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 M NaOH [100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 M KCl

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? Add water to the solution. 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. For letter a), adding water to the solution will increase the total volume of solution and therefore decrease the concentration. For letter b), pouring some of the solution down the drain will not change the concentration of the salt solution remaining. For letter c), adding more sodium chloride to the solution will increase the number of moles of salt ions and therefore increase the concentration. For letter d), water will evaporate from the solution and decrease the total volume of solution and therefore increase the concentration. Therefore, since only letter a) would decrease the concentration, letter e) cannot be correct.

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

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.

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.

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.

Exercise What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? 60.0 mL The minimum volume needed is 60.0 mL. M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL)

Objectives To learn to solve stoichiometric problems involving solution reactions To do calculations involving acid-base reactions To learn about normality and equivalent weight To use normality in stoichiometric calculations To understand the effect of a solute on solution properties   

A. Stoichiometry of Solution Reactions

What precipitate will form? lead(II) phosphate, Pb3(PO4)2 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, Pb3(PO4)2 What mass of precipitate will form? 1.1 g Pb3(PO4)2 The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s). 0.0030 mol Na3PO4 present to start and 0.0040 mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore 0.0013 mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is 811.54 g/mol, 1.1 g of Pb3(PO4)2 will form.

Where do we want to go? How do we get there? Let’s Think About It Find the mass of solid Pb3(PO4)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 Pb3(PO4)2 will be formed? What mass of Pb3(PO4)2 will be formed?

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

Concept Check For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to neutralize 1.00 L of 0.500 M sulfuric acid? 1.00 mol NaOH The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO4. 0.500 moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid. 1.00 mol of sodium hydroxide would be required.

Where do we want to go? How do we get there? Let’s Think About It 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?

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

C. Normality

C. Normality

C. Normality To find number of equivalents

C. Normality Advantage of equivalents

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

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

D. Boiling Point and Freezing Point Comparing bubbles

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