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

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Natural Approach to ChemistryChapter 9

Water & Solutions

9.1 Solutes, Solvents, and Water9.2 Concentration and Solubility

9.3 Properties of Solutions

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9.1 Assignments

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

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Special properties of water

• Cohesive nature• Ability to moderate temperature• Expands upon freezing• Versatile solvent

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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.

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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.

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Water as a solvent

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

Not chemically bonded

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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)

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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

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

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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

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In general,“like” dissolves

“like”

Polar solvents dissolve polar solutes

Nonpolar solvents dissolve nonpolar solutes

Not everything dissolves in water. Why not?

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9.2 Assignments

• 290/76-80

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concentration: the amount of each solute compared to the total solution.

9.2 Concentration & Solubility

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( )( )

( )

( )( / )

( )

( )(%) 100

( )

,

mass of solute gconcentration g L

volume of solution L

mass of solute gconcentration

m

mol

ass of solution

es of solute mmol

oleconcentration

volume of solarity M

n

g

utio L

There are several ways to express concentration

molality, m = moles solute kg solvent

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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 concentrationGiven: 10 g of solute (sugar) and 90 g of

solvent (water)Relationships:

Solve:100%

mass of soluteconcentration

total mass of solution

10

100%10 90

10%g sugar

concentrationg of solution

sugar

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Calculate the molarity of a salt solution made by adding 6.0 g of NaCl to 100 mL of distilled water.Asked: Molarity of solutionGiven: Volume of solution = 100.0 mL,

mass of solute (NaCl) = 6.0 gRelationships: M = moles

LFormula 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

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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

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100 mLH2O

What happens when you add 10 g of sugar to 100 mL of water?

Conc. (%) = 10 g/110 g

Water molecules dissolve sugar molecules

sugar10 g

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But when two sugar molecules find each other, they will become “undissolved” (solid) again…… then, they become redissolved in water again.

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This is an aqueous equilibrium!

Equilibrium

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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.

Equilibrium

dissolving“undissolving”

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Temperature has an effect on solubility

Temperature and solubility

210 gsugar

100 mLH2O

20oC

Undis-solved sugar

210 gsugar

100 mLH2O

30oC

All the sugar is dis-solved

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

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Temperature and solubility

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

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Temperature affects:

- the solubility of solutes how much

- the rate of solubility how fast

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Dissolving is a collision process

Slow (cold) molecules are not as effective as fast (hot) molecules

Salt dissolves faster in hot water

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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

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Seltzer water is a supersaturated solution of CO2 in water

supersaturation: term used to describe when a solution contains more dissolved solute than it can hold.

This solution is unstable, and the gas “undissolves” rapidly (bubbles escaping)

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Preparing a solutionHow to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.M, molarity = moles solute / liter solution

1.Determine the formula mass of the solute.

Molar mass of CaCl2

40.078 2 35.43

110.98 /g mole

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Preparing a solution

1. Determine the formula mass of the solute.

2. Use the formula mass of the solute to determine the grams of solute needed.

Molar mass of CaCl2:110.98 g/mole

How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.

1.0 0.5

1.01.0

0.5

mole mM

ole

L L

We need 0.5 moles CaCl2

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Molar mass of CaCl2:110.98 g/mole


1.0 0.51.0

1.0 0.5

mole moleM

L L

We need 0.5 moles CaCl2

2

110.980.5

1

55.49

gmoles

mole

g CaCl

We need 55.49 g CaCl2

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3. Weigh the grams of solute on the balance.


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4. Add the solute to a volumetric flask or graduated cylinder.


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5. Fill the flask about two thirds of the way up with distilled water.


Do not fill all the way up

500.0 mL mark

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6. Mix the solution until the solid dissolves completely.


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6. Mix the solution until the solid dissolves completely.

7. Fill the volumetric flask or graduated cylinder up to the correct volume marker.


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Ways to express concentration:

A higher temperature causes higher:- solubility of solutes how much- rates of solubility how fast

Molality, m = moles solute kg solvent

( )( / )

( )

( )(%) 100

( )

( )( , )

( )

mass of solute gconcentration g L


mass of solute gconcentration

mass of solution g

moles of solute moleconcentration molarity M


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9.3 Assignments

• 290/84-87.

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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 solventexothermic, loss of energy (gives off energy) or negative heat of solution (feels hot)endothermic, energy absorbed (feels cold) or positive heat of solution

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Exothermic – energy lost

Endothermic – energy gained

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The energy inside the system is constant

Heat loss must equal heat gained: net change is zero

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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

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).

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“∆” means “change”

Enthalpy

Endothermic reaction

Exothermic reaction

Positive value

Negative value

NH4NO3(s) + H2O(l) → NH4+(aq) + NO3

–(aq) ∆H = +25.7 kJ/mole


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The energy inside the system is constant

Heat released by the reaction = Heat gained by

the solution


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∆Hreaction = –56 kJ/mole

∆Hsolution = +56 kJ/mole

Opposite signs!

Heat released by the reaction = Heat gained by

the solution


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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

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Break down the problem!- Experimental setup- What is asked

Asked: Amount of heat change (DH) for

NaOH and HCl reaction


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Break down the problem!

- 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


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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 D 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 D

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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 D 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 D

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

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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).

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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 D H negative.

Answer: D H = –41.8 kJ/mole

1.67

0.04041.8

kJ

moleskJ mole H

D

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What we have seen so far…

Reaction rates increase with:

increasing concentrations

increasing temperatures

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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

Solution vs. pure solvent

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Why does ice melt when salt is sprinkled on it?

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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.

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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

Entr

opy

more

less more

less

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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

Entr

opy

more

less more

less

Freezing point depression is a

colligative property

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To calculate the freezing point of a solution:

Do not get confused with molarity, M (moles solute / L of solution)

Freezing point depression constant

D Tf = Kf x mChange in freezing molalityPoint, oC

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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 glycolGiven: 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.

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

D

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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


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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.

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Reaction rates increase with:

increasing concentrations

increasing temperatures

Chemical reactions are accompanied by changes in enthalpy, ΔH

Solution vs. pure solventdensity (solution) > density (pure solvent)colligative properties: freezing point depression is an

example

,

#

reaction solution solution p solution

reactionreaction

q q m C T

qH

of moles

D

D

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

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