CHAPTER 9 Water and Solutions 9.3 Properties of Solutions.

CHAPTER 9

Water and Solutions

9.3 Properties of Solutions

2 9.3 Properties of Solutions

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

Reaction rates


Higher temperature generally means a faster reaction rate

Reaction rates


Reaction rates

Higher concentration

Higher temperature

Faster reaction rate


hydration: the process of molecules with any charge separation to collect water molecules around them.

Not chemically bonded


The heat comes from calcium chloride dissolving

In an exothermic process, energy is released


In an endothermic process, energy is absorbed

The cooling effect comes from ammonium nitrate absorbing heat as it dissolves


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

Energy released

Energy absorbed


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



The energy inside an isolated system is constant.

The energy lost by a system must be gained by the surroundings or

another system.

From Chapter 3.2



Calorimetry

A coffee cup calorimeter is an isolated system


Calorimetry

A coffee cup calorimeter is an isolated system

Calori-metry

“heat” “measure”

thermometer

Remember: Heat and temperature are related


The energy inside the system is constant


What changes is the enthalpy

enthalpy: the energy potential of a chemical reaction measured in joule per mole (J/mole) or kilojoules per mole (kJ/mole).

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


“∆” means “change”

Enthalpy

Endothermic reaction

Exothermic reaction

Positive value

Negative value

NH4NO3(s) + H2O(l) → NH4+(aq) + NO3

–(aq) ∆H = +25.7 kJ/mole



The energy inside the system is constant

Heat released by the reaction = Heat gained by

the solution



∆Hreaction = –56 kJ/mole

∆Hsolution = +56 kJ/mole

Opposite signs!

Heat released by the reaction = Heat gained by

the solution



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





4.18p o

JC water

g C

Seen in Chapter 3.2





thermometer


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!





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

- Experimental setup




- Experimental setup- What is asked

Asked: Amount of heat change (H)

for NaOH and HCl reaction




- Experimental setup- What is asked- Assumptions 4.18p p o

JC solution C water

g C

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 H. Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mLTotal mass of solution is 80.0 g using the densitywater (1.0 g/mL).

80.0 4.18 27.0 22.0

1,672 1.67

o o osolution

solution k

q g J g C C C

J Jq

solution solution p solutionq m C T


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 H. Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mLTotal 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.

solution solution p solutionq m C T

80.0 4.18 27.0 22.0

1,672 1.67

o o o

solution

solutionq g J g C

kJq J

C C


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 H negative.

Answer: H = –41.8 kJ/mole

1.67

0.04041.8

kJ

moleskJ mole H


What we have seen so far…

Reaction rates increase with:

increasing concentrations

increasing temperatures


What we have seen so far…




Chemical reactions are accompanied by changes in enthalpy, ΔH

,

#

reaction solution solution p solution

reactionreaction

q q m C T

qH

of moles


What’s next…





Solution vs. pure solvent

,

#


reactionreaction

q q m C T

qH

of moles


Volumes of solute and solvent do not add up to the volume of solution

20 g salt

80 mL water

87 mL solution!



Volumes of solute and solvent do not add up to the volume of solution

20 g salt

80 mL water

87 mL solution!

Salt dissociates into ions, which fit in between water molecules



The density of a solution increases as more solute is added








density (solution) > density (pure solvent)

,

#


reactionreaction

q q m C T

qH

of moles








colligative properties

What’s next…

,

#


reactionreaction

q q m C T

qH

of moles


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.


Freezing point depression

The freezing point is lowered in the presence of salt

Pure solventSolid formation is not hindered

SolutionSolute particles “get in the way” of solid formation

Ord

er

En

tro

py

more

less more

less


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.

Pure solventSolid formation is not hindered

SolutionSolute particles “get in the way” of solid formation

Ord

er

En

tro

py

more

less more

less

Freezing point depression is a

colligative property


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 1.8 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: f fT K m





Relationships:

Solve:

f fT K m

1.86 1.8 3.3

0 3.35 3.3

5

5o o

of f

o

o

Freezing point of antifreeze solution

T K m C m m

C C

C

C





Relationships:

Solve:

Answer: The freezing point is lowered by 3.35oC.

f fT K m

1.86 1.8 3.3

3.

5

30 3.35 5

o of

o

f

o oFreezing point of antifreeze solution

T K m C m m

C C

C

C


Electrolyte solutions

electrolyte: solute capable of conducting electricity when dissolved in an aqueous solution.

Aqueous solutions containing dissolved ions are able to conduct electricity

1 mole of solute → 2 moles of ions



Electrolyte solutions

Aqueous solutions containing dissolved ions are able to conduct electricity



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.








colligative properties: freezing point depression as an example

,

#


reactionreaction

q q m C T

qH

of moles

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CHAPTER 9 Water and Solutions 9.3 Properties of Solutions.

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