Natural Approach to Chemistry Chapter 9 Water & Solutions 9.1 Solutes, Solvents, and Water 9.2 Concentration and Solubility 9.3 Properties of Solutions

9.1 Assignments 290/1-11 (complete sentences, please),36,40-42

Special properties of water Cohesive nature Ability to moderate temperature Expands upon freezing Versatile solvent

Cohesive nature There is a strong attraction among water molecules due to hydrogen bonding Substances are generally denser in the solid phase than in the liquid phase. Water is different In ice, hydrogen bonds force water molecules to align in a crystal structure where molecules are farther apart than they are in a liquid.

Moderates temperature If ice did not float, ponds would freeze from the bottom up killing everything inside. Water boils at 100oC because hydrogen bonding keeps the molecules together and they cannot separate easily Called the universal solvent because it dissolves both ionic and covalent compounds.

Water as a solvent Not chemically bonded hydration: the process of molecules with any charge separation to collect water molecules around them.

Tap water contains dissolved salts and minerals. (Tap water will conduct electricity) Distilled water and deionized water have been processed to remove dissolved salts and minerals. Deionization is a specific filtration process to remove all ions. (Won’t conduct electricity) Distillation boils water to steam which is then condensed back to liquid water (Won’t conduct electricity)

Phases and Chemical Reactions Solids – chem rxn occur, but very slowly Gases – occur and VERY rapidly Low density and high mobility of molecules Fire needs oxygen to support burning

Chemical reactions in Liquids – occur easily because of high density and mobility.

Reactions in liquids Life involves many complex chemical reactions that only occur in aqueous solutions! A step in the Krebs cycle – this is how energy is extracted from glucose

Not everything dissolves in water. Why not? In general, “like” dissolves “like” Polar solvents dissolve polar solutes Nonpolar solvents dissolve nonpolar solutes

9.2 Assignments 290/76-80

9.2 Concentration & Solubility concentration: the amount of each solute compared to the total solution.

molality, m = moles solute kg solvent There are several ways to express concentration molality, m = moles solute kg solvent

Suppose you dissolve 10. 0 g of sugar in 90. 0 g of water Suppose you dissolve 10.0 g of sugar in 90.0 g of water. What is the mass percent concentration of sugar in the solution? Asked: The mass percent concentration Given: 10 g of solute (sugar) and 90 g of solvent (water) Relationships: Solve:

Asked: Molarity of solution Given: Volume of solution = 100.0 mL, Calculate the molarity of a salt solution made by adding 6.0 g of NaCl to 100 mL of distilled water. Asked: Molarity of solution Given: Volume of solution = 100.0 mL, mass of solute (NaCl) = 6.0 g Relationships: M = moles L Formula mass NaCl = 22.99 + 35.45 = 58.44 g/mole 1,000 mL = 1.0 L, therefore 100 ml = 0.10 L Moles NaCl = 6.0g NaCl x 1 mole NaCl = 0.103 moles NaCl 58.44 g NaCl Answer: M = 0.103 moles = 1.03 M solution of NaCl 0.100 L

Calculate the molality, m of a solution containing 350 Calculate the molality, m of a solution containing 350.9 g of NaCl in 750.0 g of water. (Density of water is 1g/L.) Find the molar mass of NaCl: 22.99 + 35.45 = 58.44 g/mol Mass to mole conversion: 350.9 g NaCl 1 mole = 6.004 mole NaCl 58.44g m = moles solute = 6.004 mole NaCl = 8.005 m kg solvent 0.7500kg

What happens when you add 10 g of sugar to 100 mL of water? H2O Water molecules dissolve sugar molecules Conc. (%) = 10 g/110 g

What happens when you add 10 g of sugar to 100 mL of water? But when two sugar molecules find each other, they will become “undissolved” (solid) again… … then, they become redissolved in water again.

This is an aqueous equilibrium! What happens when you add 10 g of sugar to 100 mL of water? Equilibrium This is an aqueous equilibrium!

dissolving Equilibrium “undissolving” saturation: situation that occurs when the amount of dissolved solute in a solution gets high enough that the rate of “undissolving” matches the rate of dissolving.

Temperature has an effect on solubility Temperature and solubility 20oC 30oC 210 g sugar 210 g sugar All the sugar is dis- solved Undis- solved sugar 100 mL H2O 100 mL H2O Temperature has an effect on solubility

solubility: the amount of a solute that will dissolve in a particular solvent at a particular temperature and pressure.

Temperature and solubility Temperature does not have the same effect on the solubility of all solutes

Temperature affects: - the solubility of solutes how much - the rate of solubility how fast

Slow (cold) molecules are not as effective as fast (hot) molecules Dissolving is a collision process Slow (cold) molecules are not as effective as fast (hot) molecules Salt dissolves faster in hot water

- with an increase in temperature The rate of solubility increases: - with an increase in temperature - with an increase in surface area of the solute At higher temperatures: - solid solutes (like salt and sugar) are more soluble - gases are less soluble

Seltzer water is a supersaturated solution of CO2 in water This solution is unstable, and the gas “undissolves” rapidly (bubbles escaping) supersaturation: term used to describe when a solution contains more dissolved solute than it can hold.

Determine the formula mass of the solute. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. M, molarity = moles solute / liter solution Determine the formula mass of the solute. Molar mass of CaCl2

Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. Molar mass of CaCl2: 110.98 g/mole We need 0.5 moles CaCl2

Preparing a solution Molar mass of CaCl2: 110.98 g/mole How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Molar mass of CaCl2: 110.98 g/mole Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. We need 0.5 moles CaCl2 We need 55.49 g CaCl2

Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. Weigh the grams of solute on the balance.

Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. Weigh the grams of solute on the balance. Add the solute to a volumetric flask or graduated cylinder.

Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. Weigh the grams of solute on the balance. Add the solute to a volumetric flask or graduated cylinder. Fill the flask about two thirds of the way up with distilled water. 500.0 mL mark Do not fill all the way up

Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. Weigh the grams of solute on the balance. Add the solute to a volumetric flask or graduated cylinder. Fill the flask about two thirds of the way up with distilled water. Mix the solution until the solid dissolves completely.

Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Determine the formula mass of the solute. Use the formula mass of the solute to determine the grams of solute needed. Weigh the grams of solute on the balance. Add the solute to a volumetric flask or graduated cylinder. Fill the flask about two thirds of the way up with distilled water. Mix the solution until the solid dissolves completely. Fill the volumetric flask or graduated cylinder up to the correct volume marker.

Ways to express concentration: Molality, m = moles solute kg solvent A higher temperature causes higher: - solubility of solutes how much - rates of solubility how fast

9.3 Assignments 290/84-87.

9.3 Properties of Solutions Reaction rate is generally dependent upon concentration – greater concentration means reaction occurs faster Heat of solution – energy absorbed or released when a solute dissolves in a particular solvent exothermic, loss of energy (gives off energy) or negative heat of solution (feels hot) endothermic, energy absorbed (feels cold) or positive heat of solution

Exothermic – energy lost Endothermic – energy gained

Heat loss must equal heat gained: net change is zero The energy inside the system is constant

What changes is the enthalpy NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = +25.7 kJ/mole HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole enthalpy: the energy potential of a chemical reaction measured in joule per mole (J/mole) or kilojoules per mole (kJ/mole).

Enthalpy Endothermic reaction Exothermic reaction “∆” means “change” NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = +25.7 kJ/mole Positive value Exothermic reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Negative value

= HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Heat released by the reaction Heat gained by the solution = The energy inside the system is constant

= ∆Hreaction = –56 kJ/mole ∆Hsolution = +56 kJ/mole HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Heat released by the reaction Heat gained by the solution = ∆Hreaction = –56 kJ/mole ∆Hsolution = +56 kJ/mole Opposite signs!

When a student mixes 40. 0 mL of 1. 0 M NaOH and 40. 0 mL of 1 When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water. Break down the problem! Experimental setup Given: 40.0 mL of NaOH (1.0 M) + 40.0 mL of HCl (1.0 M) NaOH + HCl  NaCl + H2O Tinitial = 22.0oC and Tfinal = 27oC

When a student mixes 40. 0 mL of 1. 0 M NaOH and 40. 0 mL of 1 When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water. Break down the problem! Experimental setup What is asked Asked: Amount of heat change (DH) for NaOH and HCl reaction

Given: Isolated system: ∆Hreaction = ∆Hsolution When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water. Given: Isolated system: ∆Hreaction = ∆Hsolution Density (H2O) = 1.0 g/mL Break down the problem! Experimental setup What is asked Assumptions

Relationships: Solve: First note that the temperature increased, so the reaction released energy to the solution. This means the reaction is exothermic and will have a negative DH. Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mL Total mass of solution is 80.0 g using the densitywater (1.0 g/mL).

Relationships: Solve: First note that the temperature increased, so the reaction released energy to the solution. This means the reaction is exothermic and will have a negative DH. Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mL Total mass of solution is 80.0 g using the densitywater (1.0 g/mL). The positive sign indicates heat is absorbed. We reverse the sign as heat gained by the solution is lost by the reaction. Therefore qrxn = –1.67 kJ.

Solve: qrxn = –1.67 kJ To find heat on a per mole basis, we use the molarity times the volume in liters to calculate moles; 1.0 M x 0.040 L = 0.040 moles of both reactants (where NaOH and HCl are in equimolar amounts).

Solve: qrxn = –1.67 kJ To find heat on a per mole basis, we use the molarity times the volume in liters to calculate moles; 1.0 M x 0.040 L = 0.040 moles of both reactants (where NaOH and HCl are in equimolar amounts). Lastly, Since the temperature increased, heat was released form the reaction, making DH negative. Answer: DH = –41.8 kJ/mole

What we have seen so far… Reaction rates increase with: increasing concentrations increasing temperatures

Solution vs. pure solvent Salt dissociates into ions, which fit in between water molecules Volumes of solute and solvent do not add up to the volume of solution 20 g salt 80 mL water 87 mL solution!

Why does ice melt when salt is sprinkled on it?

Freezing point depression Why does ice melt when salt is sprinkled on it? Pure water freezes at 0oC, but a water and salt solution freezes at a lower temperature.

The freezing point is lowered in the presence of salt Freezing point depression more less Pure solvent Solid formation is not hindered Order Entropy Solution Solute particles “get in the way” of solid formation less more The freezing point is lowered in the presence of salt

Freezing point depression is a colligative property more less Pure solvent Solid formation is not hindered Freezing point depression is a colligative property Order Entropy Solution Solute particles “get in the way” of solid formation less more colligative property: physical property of a solution that depends only on the number of dissolved solute particles not on the type (or nature) of the particle itself.

D Tf = Kf x m To calculate the freezing point of a solution: depression constant D Tf = Kf x m Change in freezing molality Point, oC Do not get confused with molarity, M (moles solute / L of solution)

Calculate the freezing point of a 1 Calculate the freezing point of a 1.8 m aqueous solution of antifreeze that contains ethylene glycol (C2H6O2) as the solute. Asked: The freezing point of a 1.8 m solution of ethylene glycol Given: molality, m = 1.8 m; Kf = 1.86oC/m (Kf , freezing point depression for water, the solvent) Relationships: Solve: Answer: The freezing point is lowered by 3.35oC.

Electrolyte solutions Aqueous solutions containing dissolved ions are able to conduct electricity 1 mole of solute → 2 moles of ions 1 mole of solute → 3 moles of ions electrolyte: solute capable of conducting electricity when dissolved in an aqueous solution.

Electrolyte solutions Aqueous solutions containing dissolved ions are able to conduct electricity 1 mole of solute → 2 moles of ions 1 mole of solute → 3 moles of ions The greater the number of particles in solution, the greater the effects. colligative property: physical property of a solution that depends only on the number of dissolved solute particles not on the type (or nature) of the particle itself.

Reaction rates increase with: increasing concentrations increasing temperatures Chemical reactions are accompanied by changes in enthalpy, ΔH Solution vs. pure solvent density (solution) > density (pure solvent) colligative properties: freezing point depression is an example