Solutions are homogeneous (1 phase) mixtures where 1 of the components (solvent) is found in larger quantities than the rest. All other components are

Solutions are homogeneous (1 phase) mixtures where 1 of the components (solvent) is found in larger quantities than the rest. All other components are said to be dissolved in the solvent. These components are called solutes.

Both solutes and solvents can be liquids, solids, or gases.Create a chart with solutes along the side and solvents at the top which shows examples of: gas in gas, gas in solid, gas in liquid, liquid in gas, etc.

Gas Liquid Solid

Gas in

Liquid in

Solid in

SOLVENT

SOLUTE

oxygen in air

(nitrogen)

oxygen in water

air bubbles in ice

water in air

alcohol in water

mercury in silver

Sugar in water

(syrup)

Invisible dust in air

tin in copper

(bronze)

When solutes are dissolved in solvents the solutes formula is written followed by a bracketed subscript which follows.Examples:magnesium chloride is dissolved in waterMgCl2(aq)

iodine is dissolved in alcoholI2(al)

Aqueous solutions have water as the solvent.They are always indicated by (aq) after the formula.

The ability to conduct electricity can be used to classify solutions.Electrolytes are substances which conduct electricity when dissolved in water. Ionic compounds are electrolytes and most molecular compounds are non-electrolytes.Solutions can also be categorized as acidic, basic or neutral. Litmus paper can be used in this determination.Questions - pg 269 # 1-8

Properties of Solutions

Why does solid NaCl dissolve easily in water?

H and O atoms in water molecules do not share electron pairs equally.

OH

H


H and O atoms in water molecules do not share electron pairs equally.

OH

H

-ve

+veHO

H+ve

-ve

Water moleculeshave oppositelycharged ends.They are polar molecules


HO

H

H

O H

HO

H

HOH

Moving water molecules collide with the ions of Na and Cl in solid NaCl crystals.

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

Here is a small crystal of NaCl

Drop the crystal in a container of water

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

Here is a small crystal of NaCl

Drop the crystal in a container of water

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HOH

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

H

OH

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

H OH

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

H

O H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HOH

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

H

OH

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

H

OH

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO H

Na1+ Cl1-

Na1+ Cl1-Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

Cl1-

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

Cl 1-HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

Cl 1-

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

Cl 1-

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

Cl

1-

H

OH

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

Cl1- H

OH

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+H

OH

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+H

OH

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1-Na1+ Na1+H OH

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1- Na1+Na1+H OH

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1- Na1+Na 1+

HO

H

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1- Na1+

Na 1+

HO

H

If the positive end of a water molecule strikes a chloride ion with enough energy it pulls it away. The same thing happens if the negative oxygen end of a water molecule strikes a Na1+ ion.

Na1+ Cl1-

Na1+ Cl1-

Na1+

Cl1- Na1+

Na 1+

HO

H

These types of interactions are called intermolecular and the NaCl crystal is dissociating. In reality each ion of Na and Cl become surrounding by a number of water molecules.

Cl1-

Na1+ Cl1-

Na1+

Cl1-

Na1+

Na 1+

HO

H

H

OH

HO

H

HO

HHO

H

HO

H

H OH

HO H

These complexes are called hydrated ions.All ions in water become hydrated.

Some substances do not easily dissolve in water.

When air is exhaled in water it does not easily dissolve.Why?

Air is made up mostly of nitrogen and oxygen.N2 and O2. Since they don't dissolve easily in water they must be non-polar.

Both N2 and O2 molecules are non-polar so they are not strongly attracted by polar water molecules.

N N Since all three pairs of electrons are equally shared this molecule is non-polar.

In general like dissolves like.Polar materials dissolve easily in polar solvents and non-polar materials dissolve easily in non-polar solvents.Water is often called the universal solvent because it dissolves so many different substances.This is due to the strong forces of attraction water molecules have on each other and on positive and negative particles in other substances.

OH

H

-ve

+ve

OH

H

-ve

+ve

The H end of one water molecule is strongly attracted to the O end of another water molecule.The special force of attraction is called a hydrogen bond and it occurs between molecules of substances with H and O, or H and N, or H and F.

Any molecular substance containing O atoms bonded to H atoms has polar regions which exert these attractive H bonds.For instance alcohols have OH groups. This allows them to easily mix with water.

CH3OH

HO

HAttractive force+ve -ve

Any molecular substance containing O atoms bonded to H atoms has polar regions which exert these attractive H bonds.For instance alcohols have OH groups. This allows them to easily mix with water.

CH3OHH

OH

H bond

Alcohol will dissolve in water but this solution does not conduct electricity. Why?There are no mobile ions present.

CH3OHH

OH

H bond

Why does sugar dissolve in water and not conduct electricity.

CHCH

O

CHCH

CH

OH

OH

OH OH

CH2OH

The OH groups form H bonds with water molecules.

HO

H

HO

H

HO

HO

HO

H

C6H12O6

No free ions are present.

Non-polar substances will dissolve in other non-polar substances.Non-polar molecules have no strongly positive or negative regions but they still have positive nuclei and negative electron clouds around them.

+ +


+ +


+ +


+ +


+ +


+ +


+ + + +

If 2 non-polar molecules are side by side the electron clouds of one affect the electron clouds of the other


+ + + +



+ + + +


+

A region of slight positive charge is created

Attractive force

+ + + ++Attractive force

This slight attractive force brought about by the influence of one electron cloud of 1 molecule on the electron cloud of a molecule beside it is called London Dispersion Force (LDF).It explains why non-polar solvents dissolve non-polar solutes.

Questions - Pg. 277 # 3-5Pg. 278 # 6,7Pg. 279 # 8-10

Solutions are homogeneous mixtures. Solutions involve 2 components. The substance doing the dissolving (solvent) and the substance being dissolved (solute).Typically the amount of solute dissolved is measured and compared to the total volume of solution. This quantity is known as the concentration of solution. A 710 mL bottle of coke has 30 g of sugar.What is the concentration in g/L (M/V)?30 g / 0.710 L = 42 g / L

What mass of sugar is there in a 355 mL can of coke?42 g / L x 0.355 L = 15 g of sugarSome solutions, like alcohol mixtures, list the quantity of alcohol as a percentage by volume since this number is bigger than the percentage by mass for solutes with a density smaller than water.A can of regular beer is 5% (V/V) alcohol by volume. What volume of alcohol is their in a 355 mL can of beer?5/100 x 355 mL = 18 mL of alcohol

Which has more alcohol45 mL of 40% (V/V) rye whiskey (typical shot)310 mL of 7% (V/V) vodka cooler341 mL of 4% (V/V) Coors light bottle of beer180 mL of 12% (V/V) glass of red wine45 mL x 40/100 = 18 mL in shot of rye310 mL x 7/100 = 21.7 mL in the cooler341 mL x 4/100 = 13.6 mL in the beer180 mL x 12/100 = 21.6 mL in the wine

A 1.0 L sample of water is found to have 0.0012 g of lead.The molar concentration works out to be a very small number. To avoid using really small numbers for concentrations of dilute solutions another more practical scale is used. This scale is called parts per million.What is the ppm of lead for the example above?1000 L would have 1.2 g of lead so it is 1.2 g in 1000 L or 1200 mg / 1000 L or 1.2 mg / 1.0 L

ppm can be expressed in a variety of ways1 ppm = 1 g/1000 L or1 ppm = 1 g / 1000 000 mL or1 ppm = 1 g / 106 mL or1 ppm = 1000 mg / 1000 L or1 ppm = 1 mg / LCalculating ppmIn a chemical analysis 3.4 mg of lead was found in 100 mL of tap water. Find the ppm of lead.ppm = 1 mg/L = 3.4 mg / 0.1 L = 34 ppm

What fraction of a part per million (ppm) is a part per billion (ppb)?1/1000So 1 ppm = ? ppb

1 ppm = 1000 ppbAn even smaller concentration unit is a part per trillion (ppt)1000 ppb = 1 ppt1 mg in 1.0 L is 1 ppm1 mg in 1000 L is a ppb1 mg in 1 000 000 L is a pptQuestions pg. 290 # 1-10

Measuring Quantities of Solutes in Solutions

The quantity of solute can be measured in grams or moles. The total volume of the solution is measured in L. The amount of solute in a given volume of solution is measured using these units:

g

L

mol

Lor =

kmol

m3=

mol

dm3

or molL-1 kmolm-3 moldm-3

M=

This leads to the development of the following equation:

Concentrationof a solution =

# of moles of solute

Volume, in L, of solution

C = n

V

Preparing Solutions From Solid ReagentsSample ProblemDescribe how to prepare 500 mL of a 0.035 M solution of sodium thiosulfate.

Given: V = 500 mL = 0.500 L

C = 0.035 M

Asked to Find: Mass of Na2S2O3

Mass

Mole

Concentration

Use n= m/MM

Use C = n/V

Step 1 - Find n using C = n/V

Rearrange this equation to getn = CVn = 0.035 M x 0.500 L = 0.0175 mol

Step 2- Find m using n = m/MM

Rearrange this equation to getm= n x MM = 0.0175 mol x 158.1 g/mol = 2.8 g

1. get a 500 mL Volumetric Flask2. Place it on a mass balance and

tare the balance (zero it).3. Mass out 2.8 g of Na2S2O3

4. Fill the flask up, almost to the top and dissolve the solute.

5. Top up with distilled water to the calibration mark and stopper it.

Questions p 284 # 1-8

Preparing Solutions From SolutionsDetermining Concentrations of Concentrated ReagentsConcentrations of solutions, in molL-1, can be determined from density and percentage composition.

Sample ProblemA solution of concentrated (conc.) HCl (hydrochloric acid) has a density of 1.25 g/mL and it is 35% HCl by mass. Find the concentration of the HCl.

density = 1.25 g/mL, 35% HCl

Change density into units of mass and volumem = 1.25 g, V = 1 mL = 0.00100 L Mass

Mole

Concentration

Given:

Step 1 - Find mass of HCl35% of 1.25 g = 0.4375 gStep 2 - Find nHCl = m/mm = 0.4375 g/ 36.45 g/mol = 0.01199 molStep 3 - Find C = n/V = 0.01199 mol/ 0.00100 L= 12 M

Describe how to prepare 1.5 L of 0.75 M HCl from this concentrated reagent.Solution: Find the volume of the concentrated reagent needed to prepare the solution.Given:Cd = 0.75 M , Vd = 1.5 LCc = 12 M, Vc = ?

# of moles Concentrated Reagent

=# of moles Diluted Reagent

CcVc = CdVdCc Cc

Vc = 0.75 M x 1.5 L12 M

= 0.094 L = 94 mL

1. Get a 1.5 L Volumetric Flask2. Measure 94 mL of concentrated HCl

using gloves, apron, shield3. Half fill the 1.5 L Volumetric flask

with distilled water4. Add the 94 mL of conc. HCl5. Top up with distilled water to the

calibration mark.

AW not WA

Describe how to prepare 2.0 L of a 1.5 M solution of ammonium hydroxide from a concentrated reagent which is 14.5 M.

# of moles Concentrated Reagent

=# of moles Diluted Reagent

CcVc = CdVdCc Cc

Vc = 1.5 M x 2.0 L14.5 M

= 0.207 L = 210 mL

1. Get a 2.0 L Volumetric Flask2. Measure 210 mL of concentrated

NH4OH using gloves, apron, shield3. Half fill the 2.0 L Volumetric flask

with distilled water4. Add the 210 mL of conc. NH4OH.

Top up with distilled water to the calibration mark.

What is the concentration of a H3PO4 concentrated reagent if its density is 1.4 g/mL and it is 45% phosphoric acid by mass.Describe how to prepare 250 mL of a 0.45 M solution of phosphoric acid from this concentrated reagent.

If 45 mL of a 0.15 molL-1 solution of lead(II) nitrate is combined with an excess of sodium iodide what mass of lead(II) iodide is formed. Assume a double displacement reaction.

Pb(NO3)2(aq) + NaI(aq) PbI2(aq) + NaNO3(aq)

If 85 mL of a 1.6 molL-1 solution of hydrochloric acid is combined with an excess of magnesium what volume of hydrogen gas is formed at 24oC and 101 kPa? Assume a single displacement reaction.

If 35 mL of 1.2 mol/L sulfuric acid is combined with 65 mL of a 0.95 mol/L solution of potassium hydroxide what mass of potassium sulfate is formed. The other product is water.

Prepare the following solutions1.100 mL of a 1.0 M NaOH from solid2. 500 mL of a 1.0 M HCl from 6.0 M solution3. 500 mL of 0.223 M Na2S2O3

.5H2O from a solid4. 500 mL of 0.5 M NaOH from a solid5. 1 L of a 1.0 M HCl from a 6.0 M solution6. 1 L of 0.25 M Na2S2O3

.5H2O from a solid7. 1 L of 1.0 M NaOH from solid8. 1 L of a 0.5 M HCl from a 6.0 M solution9. 50 mL of a 0.5 M HCl from a 6.0 M solution10. 1L of 2.0 M HCl for a 6.0 M solution

Prepare the following solutions1.100 mL of a 1.0 M NaOH from solid2. 500 mL of a 1.0 M HCl from 6.0 M solution3. 500 mL of 0.223 M Na2S2O3

.5H2O from a solid4. 1L of 2.0 M HCl for a 6.0 M solution5. 500 mL of 0.5 M NaOH from a solid6. 1 L of a 1.0 M HCl from a 6.0 M solution7. 1 L of 0.25 M Na2S2O3

.5H2O from a solid8. 50 mL of a 0.5 M HCl from a 6.0 M solution9. 1 L of 1.0 M NaOH from solid10. 1 L of a 0.5 M HCl from a 6.0 M solution

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Solutions are homogeneous (1 phase) mixtures where 1 of the components (solvent) is found in larger quantities than the rest. All other components are