# CHAPTER 9 Water and Solutions 9.3 Properties of Solutions.

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CHAPTER 9 Water and Solutions 9.3 Properties of Solutions

Which acid will dissolve the limestone fastest?
Reaction rates Which acid will dissolve the limestone fastest?

Reaction rates Which acid will dissolve the limestone fastest?
Acid with the higher concentration Higher concentration generally means a faster reaction rate

Higher temperature generally means a faster reaction rate
Reaction rates Higher temperature generally means a faster reaction rate

Reaction rates Faster reaction rate Higher concentration
Higher temperature Faster reaction rate

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

In an exothermic process, energy is released
The heat comes from calcium chloride dissolving

In an endothermic process, energy is absorbed
The cooling effect comes from ammonium nitrate absorbing heat as it dissolves

Energy released Energy absorbed heat of solution: the energy absorbed or released when a solution dissolves in a particular solvent.

From Chapter 3.2 first law of thermodynamics: energy can neither be created nor destroyed.

From Chapter 3.2 first law of thermodynamics: energy can neither be created nor destroyed. The energy inside an isolated system is constant.

From Chapter 3.2 first law of thermodynamics: energy can neither be created nor destroyed. The energy inside an isolated system is constant. The energy lost by a system must be gained by the surroundings or another system.

Calorimetry A coffee cup calorimeter is an isolated system

Calorimetry Calori-metry
thermometer “heat” “measure” Remember: Heat and temperature are related A coffee cup calorimeter is an isolated system

The energy inside the system is constant

What changes is the enthalpy
NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = 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 = kJ/mole Positive value Exothermic reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Negative value

= The energy inside the system is constant
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!

If we can calculate ∆Hsolution, we can determine ∆Hreaction
qsolution = (grams of solution) x (specific heat of solution) x ΔT qsolution = –qreaction ΔHreaction = qreaction / moles If we can calculate ∆Hsolution, we can determine ∆Hreaction

qsolution = (grams of solution) x (specific heat of solution) x ΔT
qsolution = –qreaction ΔHreaction = qreaction / moles Seen in Chapter 3.2

qsolution = (grams of solution) x (specific heat of solution) x ΔT
qsolution = –qreaction ΔHreaction = qreaction / moles thermometer

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.

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!

Break down the problem! Experimental setup
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) mL of HCl (1.0 M) NaOH + HCl  NaCl + H2O Tinitial = 22.0oC and Tfinal = 27oC

Break down the problem! Experimental setup What is asked
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

Break down the problem! Experimental setup What is asked Assumptions
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 Assumptions Given: Isolated system: ∆Hreaction = ∆Hsolution Density (H2O) = 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 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 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 L = 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 L = 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

What we have seen so far…
Reaction rates increase with: increasing concentrations increasing temperatures Chemical reactions are accompanied by changes in enthalpy, ΔH

What’s next… Reaction rates increase with: increasing concentrations
increasing temperatures Chemical reactions are accompanied by changes in enthalpy, ΔH Solution vs. pure solvent

Solution vs. pure solvent
Volumes of solute and solvent do not add up to the volume of solution 20 g salt 80 mL water 87 mL solution!

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!

The density of a solution increases as more solute is added
Solution vs. pure solvent The density of a solution increases as more solute is added

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)

What’s next… 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

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.

To calculate the freezing point of a solution:
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.

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 = 1.8 m; Kf = 1.86oC/m (for water the solvent) Relationships:

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 = 1.8 m; Kf = 1.86oC/m (for water the solvent) Relationships: Solve:

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 = 1.8 m; Kf = 1.86oC/m (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 as an example