Solutions

Chemistry 5th & 8th

Hall©Hall2010

Solutions

• Solution– Homogenous mixture containing 2 or more

substances called solute and solvent

• Solute– Substance being dissolved

• Solvent– Dissolving medium

• Example– Salt into Water

Solutions Vocabulary

• Soluble– Substance that

dissloves in a solvent– E.g. salt into water,

sugar into water

• Insoluble– Substance that does

not dissolve into water

– E.g. sand into water

• Immiscible– 2 liquids that are not

soluble with each other– E.g. oil/water, oil/vinegar

• Miscible– 2 liquids that do dissolve

into each other– E.g. antifreeze,

water/vinegar

Solvation

• To form a solution, solute particles must separate.

• Solvation is the process of surrounding solute particles with solvent particles to form a solution

Ionic Compounds

• Water molecules are polar and in constant motion

• When an ionic compound is placed in water, the water molecules collide with the ionic crystal until entire crystal dissolves.

• Table salt into water

Molecular Compounds

• Like ionic, but the molecules are much bigger

• Sugar C12H22O11 into water

Factors that Affect Rate of Solvation

Increase in collisions

1. Agitation

2. Increase in surface area of solute

3. Increase Temperature of solvent

Heat of Solution

• During solvation, solute must separate into particles

• Solvent particles also must move apart• Energy is required to break attractive forces

with in solvent and solute• When solute and solvent particles mix, they

are attracted to each other and energy is released (exothermic)

• Heat of Solution is the overall change in energy

Solubility

• Maximum amount of solute that will dissolve in a given amount of solvent at a specific Temperature and Pressure.– Usually in

• Example: 10g of NaCl in 100g of H2O

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gramsofsolute

100gofsolvent

Saturation

• Saturated– Max amount of dissolved solute for a given

amount of solvent at a specific T & P

• Unsaturated– Contains less dissolved solute for a given

T & P than a saturated solution

Factors that affect Saturation

• Temperature– Increase in temperature=more soluble– Not true for gases though

• Supersaturation– Contains more dissolved solute than a saturated

solution at the same Temperature• To make, saturate a solution at high temperature and

allow to cool slowly• If a tiny amount of solute is added or the solution is

agitated, a precipitate will form

Pressure and Solubility

• Affects gases

• Solubility Pressure

• Henry’s Law– At a given temperature, solubility (S) of a

gas in a liquid is directly proportional to the pressure (P) of gas above the liquid

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S1P1=S2P2

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S1P2 = S2P1

Henry’s Law Example

• If 0.85g of a gas at 4.0 atm of pressure dissolves in 1.0L of water at 25C, how much will dissolve in 1.0L of water at 1.0 atm of pressure and the same temperature

S1= 0.85 g/L

P1= 4.0 atm

S2= ? g/L

P2= 1.0 atm

Henry’s Law Example

S1= 0.85 g/L

P1= 4.0 atm

S2= ? g/L

P2= 1.0 atm

Work problem 2 on pg. 461

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S1P2 = S2P1

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S2 =S1P2P1

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S2 =(0.85g /L)(1.0atm)

4.0atm=0.85g /L

4.0= 0.21g /L

Concentration

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Massofsolute

massofsolution•100

Percent by Mass

Whether it is mass or volume, the solution is the solvent + the solute Example:

Mass of Solute= 3.6 g

Mass of Solvent = 100.0 g

Find mass of solution first

Then Percent by Mass

Now you work 8,9,10 and worksheet is homework

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3.6g+100.0g =103.6g

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3.6g

103.6g• 100 = 3.5%

Concentration

Percent by Volume

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Volumeofsolute

Volumeofsolution• 100

Example:

Volume of solute= 35 mL

Volume of solvent = 115 mL

Now you do problems 11 and 12 on page 464

Find volume of Solution

Next find the Percent by Volume

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35mL +115mL =150mL

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35mL

150mL• 100 = 23%

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Volumeofsolute

Volumeofsolution• 100

Molarity

Number of Moles of solute dissolved per liter of solution

Know your UNIT CONVERSIONS and HOW TO GO FROM GRAMS TO MOLES

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Molarity(M) =Molesofsolute

litersofsolution

Molarity Example

A 100.5 mL intravenous solution contains 5.10 g glucose (C6H12O6). What is the molarity of this solution? Molar mass of glucose is 180.16 g/mol.

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5.10gC6H12O6 ∗1molC6H12O6180.16gC6H12O6

= 0.0283molC6H12O6

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100.5mL∗1L

1000mL= 0.1005L

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M =0.0283molC6H12O6

0.1005L= 0.282M

Molarity Example 2

What is the molarity of a bleach solution containing 9.5 g of NaOCl per liter of bleach?

First change grams to moles

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9.5gNaOCl∗1molNaOCl

74.352gNaOCl= 0.13molNaOCl

Next, find molarity. The volume of the solution is 1 L.

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M =0.13molNaOCl

1Lsolution= 0.13M