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SolutionsSolutionsChemistry I2 – Chapters 7 & 8

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Some DefinitionsSome DefinitionsA solution is a A solution is a

homogeneous homogeneous mixture of 2 or more mixture of 2 or more substances . substances .

One constituent is One constituent is usually regarded as usually regarded as the the SOLVENTSOLVENT and and the others as the others as SOLUTESSOLUTES..

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Parts of a Solution• SOLUTE – the part

of a solution that is being dissolved (usually the lesser amount)

• SOLVENT – the part of a solution that dissolves the solute (usually the greater amount)

• Solute + Solvent = Solution

Solute Solvent Example

solid solid Metal alloys

solid liquid Kool aid

liquid liquid Alcoholic drinks

gas liquid Pepsi

gas gas air

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DefinitionsDefinitions

Solutions can be classified as Solutions can be classified as saturatedsaturated or or ununsaturatedsaturated..

A A saturatedsaturated solution contains solution contains the maximum quantity of the maximum quantity of solute that dissolves at that solute that dissolves at that temperature.temperature.

An An unsaturatedunsaturated solution solution contains less than the contains less than the maximum amount of solute maximum amount of solute that can dissolve at a that can dissolve at a particular temperatureparticular temperature

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DefinitionsDefinitions

SUPERSATURATED SOLUTIONSSUPERSATURATED SOLUTIONS contain more solute than is contain more solute than is possible to be dissolvedpossible to be dissolved

Supersaturated solutions are Supersaturated solutions are unstable. The supersaturation is unstable. The supersaturation is only temporary, and usually only temporary, and usually accomplished in one of two ways:accomplished in one of two ways:

1.1. Warm the solvent so that it will Warm the solvent so that it will dissolve more, then cool the dissolve more, then cool the solution solution

2. Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

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Miscibility

• Miscible: compounds that dissolve readily in each other in any proportion.

Ex: water and alcohol.

• Immiscible: Liquids that do not readily dissolve in each other

Ex: oil and water

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Solubility

• Solubility is relative, most substances will dissolve by extremely small amounts.

• Soluble: the solute will dissolve in a particular solvent greater than 1 g / 100 mL

• Insoluble: the solute will dissolve only 0.1 g / 100 mL, or not dissolve in the solvent

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IONIC COMPOUNDSIONIC COMPOUNDSCompounds in Aqueous SolutionCompounds in Aqueous Solution

Many reactions involve ionic Many reactions involve ionic compounds, especially reactions in compounds, especially reactions in water — water — aqueous solutions.aqueous solutions.

KMnOKMnO44 in water in water KK++(aq) + MnO(aq) + MnO44--(aq)(aq)

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How do we know ions are How do we know ions are present in aqueous present in aqueous solutions?solutions?

The solutions The solutions __________________________________________________

They are called They are called ELECTROLYTESELECTROLYTES

HCl, MgClHCl, MgCl22, and NaCl are , and NaCl are strong electrolytesstrong electrolytes. . They dissociate completely They dissociate completely (or nearly so) into ions.(or nearly so) into ions.

Aqueous Aqueous SolutionsSolutions

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Aqueous Aqueous SolutionsSolutions

Some compounds Some compounds dissolve in water but do dissolve in water but do not conduct electricity. not conduct electricity. They are called They are called nonelectrolytes.nonelectrolytes.

Examples include:Examples include:sugarsugarethanolethanolethylene glycolethylene glycol

Examples include:Examples include:sugarsugarethanolethanolethylene glycolethylene glycol

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Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a solution The amount of solute in a solution is given by its is given by its concentrationconcentration.

Molarity (M) = moles soluteliters of solution

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1.0 L of 1.0 L of water was water was

used to used to make 1.0 L make 1.0 L of solution. of solution. Notice the Notice the water left water left

over.over.

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PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.

PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22•6H•6H22OO

5.00 g • 1 mol

237.7 g = 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Step 2: Calculate MolarityCalculate Molarity

[NiClNiCl22•6 H•6 H22OO ] = 0.0841 M

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Step 1: Step 1: Change mL to L.Change mL to L.

250 mL * 1L/1000mL = 0.250 L250 mL * 1L/1000mL = 0.250 L

Step 2: Step 2: Calculate.Calculate.

Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesMoles = (0.0500 mol/L) (0.250 L) = 0.0125 moles

Step 3: Step 3: Convert moles to grams.Convert moles to grams.

(0.0125 mol)(90.00 g/mol) = (0.0125 mol)(90.00 g/mol) = 1.13 g1.13 g

USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

moles = M•Vmoles = M•V

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, is

required to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?

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Practice For homework:

• Pg 242 # 1-6, 9

• Pg 268 # 19 & 20 (Read pg 266-268)

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Solutions (continued)

Day 2

Mrs. Kay

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Learning Check

How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

1) 12 g

2) 48 g

3) 300 g

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An An IDEAL SOLUTIONIDEAL SOLUTION is is one where the properties one where the properties depend only on the depend only on the concentration of solute.concentration of solute.

Need conc. units to tell us the Need conc. units to tell us the number of solute particles number of solute particles per solvent particle.per solvent particle.

The unit “molarity” does not The unit “molarity” does not do this!do this!

Concentration UnitsConcentration Units

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Two Other Concentration Two Other Concentration UnitsUnits

grams solutegrams solutegrams solutiongrams solution

MOLALITY, mMOLALITY, m

% by mass% by mass = =

% by mass% by mass

m of solution = mol solute

kilograms solvent

20Calculating Calculating ConcentrationsConcentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate molality and % by mass of O. Calculate molality and % by mass of

ethylene glycol.ethylene glycol.

21Calculating Calculating ConcentrationsConcentrations

Calculate molalityCalculate molality

Calculate molalityCalculate molality

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate m & % of ethylene glycol (by mass).O. Calculate m & % of ethylene glycol (by mass).

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate m & % of ethylene glycol (by mass).O. Calculate m & % of ethylene glycol (by mass).

conc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molalconc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molal

%glycol = 62.1 g

62.1 g + 250. g x 100% = 19.9%%glycol =

62.1 g62.1 g + 250. g

x 100% = 19.9%

Calculate weight %Calculate weight %

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Learning Check

A solution contains 15 g Na2CO3 and 235 g of

H2O? What is the mass % of the solution?

1) 15% Na2CO3

2) 6.4% Na2CO3

3) 6.0% Na2CO3

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Using mass %

How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

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Try this molality problem

• 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution.

m = mol solute / kg solvent

25 g NaCl 1 mol NaCl

58.5 g NaCl= 0.427 mol NaCl

Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg

0.427 mol NaCl

5 kg water= 0.0854 molal salt water

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Preparing SolutionsPreparing SolutionsPreparing SolutionsPreparing Solutions

• Weigh out a solid Weigh out a solid solute and dissolve in a solute and dissolve in a given quantity of given quantity of solvent.solvent.

• Dilute a concentrated Dilute a concentrated solution to give one solution to give one that is less that is less concentrated.concentrated.

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Practice for homework:

• Read pg 255-263

• Do Pg 261 #5-8 (%m/m)

• Do Pg 268 # 19-20 (molarity)

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Changing Concentration:

• Our calculations for this unit are based on solutions being carefully made with known concentrations.

• Two ways to have a known concentration:

1. Dissolving a measured mass of pure solute in a certain volume of solution. (We practiced this as molarity questions)

2. Diluting a solution of known concentration to the new concentration you’re trying to achieve.

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Diluting a Standard solution:• Use the formula:

C1v1 = C2v2

You will be given 3 of the 4 variables and need to solve for the unknown.

C1 = the original concentrated solution

v1= the amount of original concentrated solution used.

C2= the new diluted concentration of solution

v2= the new volume of diluted solution

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Practice:

• I have 2.0 L of 0.10 M of sulfuric acid. This usually sold as an 18 M concentrated solution. How much of the concentrated solution do I need to make my new solution?

YOU KNOW: C1=18 M; C2=0.10 M; v2= 2.0 L

So, v1 = C2v2

C1

v1 = (0.10 M)(2.0 L) (18 M)

v1 = 0.011 L or 11 mL

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More Practice:

• Page 273 # 25-27

• Finish the Handout called “Molarity Review ” # 5-9

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CHEM 11ASolution Stoichiometry

You can solve for amounts of product made or reactants needed even though they are stated as mol/L rather than g or moles.

You need to use the basics that you learned in stoichiometry from grade 11 chemistry and apply it to what you’ve learned about solutions

Let’s work through an example

32How many grams of Ca(NO3)2 can be prepared by reacting 75 mL of 2.6 mol/L HNO3 with an excess of

Ca(OH)2?2 HNO3(l) + Ca(OH)2(s) Ca(NO3)2(aq) + 2H2O(l)

• What are you given? V HNO3= 0.075L and M=2.6 mol/L

• Find the moles of HNO3 used:

2.6 mol/L x 0.075 L = 0.195 mol

3. Use stoichiometry to find the ratio of acid moles to calcium nitrate.

0.195 mol (acid) x 1 mol (calcium nitrate) = 0.0975 mol

2 mol (acid)

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2 HNO3 + Ca(OH)2 Ca(NO3)2 + 2H2O

4. 0.0975 mol calcium nitrate needs to be converted to mass.

0.0975 mol Ca(NO3)2 X 164.04 g = 15.99 g

mol Ca(NO3)2

So, we will make 15.99 g of calcium nitrate during this reaction.

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Practice:

• Handout # 56-60