CHM101 Chapter15 Re

7/31/2019 CHM101 Chapter15 Re

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Chapter 15

Solutions


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15.1 Solubility

15.2 Solution Composition: An Introduction

15.3 Solution Composition: Mass Percent

15.4 Solution Composition: Molarity

15.5 Dilution15.6 Stoichiometry of Solution Reactions

15.7 Neutralization Reactions

15.8 Solution Composition: Normality
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What is a Solution?

Solution homogeneous mixture

Solventsubstance present in largest

amount

Solutesother substances in the solution

Aqueous solutionsolution with water as the

solvent


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Various Types of Solutions


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Ionic substances breakup into individual

cations and anions.

Solubility of Ionic Substances


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Polar water molecules interact with the positive

and negative ions of a salt.

Solubility of Ionic Substances


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Ethanol is soluble in water because of the

polar OH bond.

Solubility of Polar Substances


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Why is solid sugar soluble in water?

Solubility of Polar Substances


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Nonpolar oil does not interact with polar water.

Water-water hydrogen bonds keep the water

from mixing with the nonpolar molecules.

Substances Insoluble in Water


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A hole must be made in the water structure

for each solute particle.

The lost water-water interactions must be

replaced by water-solute interactions.

like dissolves like

How Substances Dissolve


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Concept Check

Which of the following solutes will generally not

dissolve in the specified solvent? Choose the best

answer. (Assume all of the compounds are in theliquid state.)

a) CCl4 mixed with water (H2O)

b) NH3 mixed with water (H2O)c) CH3OH mixed with water (H2O)

d) N2 mixed with methane (CH4)


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The solubility of a solute is limited.

Saturated solution contains as much

solute as will dissolve at that

temperature.

Unsaturated solution has not

reached the limit of solute that will

dissolve.


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Supersaturated solution occurs when a

solution is saturated at an elevated

temperature and then allowed to cool but

all of the solid remains dissolved. Contains more dissolved solid than a

saturated solution at that temperature.

Unstable

adding a crystal causesprecipitation.


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Solutions are mixtures.

Amounts of substances can vary in different

solutions.

Specify the amounts of solvent and

solutes.

Qualitative measures of concentration

concentrated relatively large amountof solute

dilute relatively small amount of solute


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mass of soluteMass percent = 100%

mass of solution

grams of soluteMass percent = 100%

grams of solute + grams of solvent


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Exercise

What is the percent-by-mass concentration of

glucose in a solution made my dissolving 5.5 g

of glucose in 78.2 g of water?

6.6%

[5.5 g / (5.5 g + 78.2 g)] 100 = 6.6% glucose


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Molarity (M) = moles of solute per

volume of solution in liters:

moles of solute= Molarity =liters of solution

M

6 moles of HCl3 HCl =2 liters of solution

M


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Exercise

You have 1.00 mol of sugar in 125.0 mL of

solution. Calculate the concentration in units

ofmolarity.

8.00 M

1.00 mol / (125.0 / 1000) = 8.00 M


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Exercise

A 500.0-g sample of potassium phosphate

is dissolved in enough water to make 1.50 L

of solution. What is the molarity of thesolution?

1.57 M

500.0 g is equivalent to 2.355 mol K3PO4 (500.0 g /

212.27 g/mol). The molarity is therefore 1.57 M

(2.355 mol/1.50 L).


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Exercise

You have a 10.0 Msugar solution. What

volume of this solution do you need to have

2.00 mol of sugar?

0.200 L

2.00 mol / 10.0 M= 0.200 L


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Exercise

Consider separate solutions of NaOH and KCl

made by dissolving 100.0 g of each solute in

250.0 mL of solution. Calculate theconcentration of each solution in units of

molarity.

10.0 MNaOH[100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 MNaOH

5.37 MKCl[100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 MKCl


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Concept Check

You have two HCl solutions, labeled Solution A andSolution B. Solution A has a greater concentration thanSolution B. Which of the following statements are true?

a) If you have equal volumes of both solutions,Solution B must contain more moles of HCl.

b) If you have equal moles of HCl in both solutions,Solution B must have a greater volume.

c) To obtain equal concentrations of both solutions,you must add a certain amount of water toSolution B.

d) Adding more moles of HCl to both solutions will makethem less concentrated.


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For a 0.25 MCaCl2 solution:

CaCl2 Ca2+ + 2Cl

Ca2+: 1 0.25 M= 0.25 MCa2+

Cl: 2 0.25 M= 0.50 MCl.

Concentration of Ions


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Concept Check

Which of the following solutions contains

the greatest number of ions?

a) 400.0 mL of 0.10 M NaCl.

b) 300.0 mL of 0.10 M CaCl2.

c) 200.0 mL of 0.10 M FeCl3.d) 800.0 mL of 0.10 M sucrose.


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Where are we going?

To find the solution that contains the greatest

number of moles of ions.

How do we get there? Draw molecular level pictures showing each

solution. Think about relative numbers of ions.

How many moles of each ion are in each

solution?

Lets Think About It


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The solution with the greatest number of

ions is not necessarily the one in which:

the volume of the solution is the

largest.

the formula unit has the greatest

number of ions.

Notice


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A solution whose concentration is

accurately known.

Standard Solution


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Weigh out a sample of solute.

Transfer to a volumetric flask.

Add enough solvent to mark on flask.

To Make a Standard Solution


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The process of adding water to a

concentrated orstock solution to achieve

the molarity desired for a particular

solution. Dilution with water does not alter the

numbers of moles of solute present.

Moles of solute before dilution = moles ofsolute after dilution

M1V1 = M2V2


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Transfer a measured amount of original solution

to a flask containing some water.

Add water to the flask to the mark (with swirling)

and mix by inverting the flask.

Diluting a Solution


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Concept Check

A 0.50 Msolution of sodium chloride in an open

beakersits on a lab bench. Which of the

following would decrease the concentration of

the salt solution?

a) Add water to the solution.

b) Pour some of the solution down the sink drain.

c) Add more sodium chloride to the solution.d) Let the solution sit out in the open air for a

couple of days.

e) At least two of the above would decrease the

concentration of the salt solution.


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Exercise

What is the minimum volume of a 2.00 M

NaOH solution needed to make 150.0 mL of

a 0.800 MNaOH solution?

60.0 mL

M1V1 = M2V2

(2.00 M)(V1) = (0.800 M)(150.0 mL)


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1. Write the balanced equation for the reaction. For

reactions involving ions, it is best to write the net

ionic equation.

2. Calculate the moles of reactants.

3. Determine which reactant is limiting.

4. Calculate the moles of other reactants or

products, as required.

5. Convert to grams or other units, if required.

Steps for Solving Stoichiometric Problems Involving Solutions


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Concept Check (Part I)

10.0 mL of a 0.30 Msodium phosphate

solution reacts with 20.0 mL of a 0.20 M

lead(II) nitrate solution (assume no volumechange).

What precipitate will form?

lead(II) phosphate, Pb3(PO4)2 What mass of precipitate will form?

1.1 g Pb3(PO4)2


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Where are we going? To find the mass of solid Pb3(PO4)2 formed.

How do we get there?

What are the ions present in the combined solution? What is the balanced net ionic equation for the

reaction?

What are the moles of reactants present in the

solution?

Which reactant is limiting?

What moles of Pb3(PO4)2 will be formed?

What mass of Pb3(PO4)2 will be formed?

Lets Think About It


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Where are we going?

To find the concentration of nitrate ions left in

solution after the reaction is complete.

How do we get there? What are the moles of nitrate ions present in the

combined solution?

What is the total volume of the combined

solution?

Lets Think About It


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Concept Check (Part III)

10.0 mL of a 0.30 Msodium phosphate

solution reacts with 20.0 mL of a 0.20 M

lead(II) nitrate solution (assume no volumechange).

What is the concentration of phosphateions left in solution after the reaction is

complete?

0.011 M


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Where are we going? To find the concentration of phosphate ions left in

solution after the reaction is complete.

How do we get there? What are the moles of phosphate ions present in

the solution at the start of the reaction?

How many moles of phosphate ions were used

up in the reaction to make the solid Pb3(PO4)2? How many moles of phosphate ions are left over

after the reaction is complete?

What is the total volume of the combined

solution?

Lets Think About It


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An acid-base reaction is called a

neutralization reaction.

Steps to solve these problems are the same

as before. For a strong acid and base reaction:

H+(aq) + OH(aq) H2O(l)


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Concept Check

For the titration of sulfuric acid (H2SO4) with

sodium hydroxide (NaOH), how many moles

of sodium hydroxide would be required toreact with 1.00 L of 0.500 Msulfuric acid?

1.00 mol NaOH


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Where are we going?

To find the moles of NaOH required for the

reaction.

How do we get there? What are the ions present in the combined

solution? What is the reaction?

What is the balanced net ionic equation for the

reaction? What are the moles of H+ present in the solution?

How much OH is required to react with all of the

H+ present?

Lets Think About It


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One equivalent of acid amount of acid

that furnishes 1 mol of H+ ions.

One equivalent of base amount of base

that furnishes 1 mol of OH ions Equivalent weight mass in grams of 1

equivalent of acid or base.

Unit of Concentration


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To find number of equivalents:

number of equivalents equivalents equivNormality = = = =

1 liter of solution liter LN

equiv

V = L = equivLN


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Concept Check

If Ba(OH)2 is used as a base, how many

equivalents of Ba(OH)2 are there in 4 mol

Ba(OH)2?

a) 2

b) 4

c) 8d) 16

CHM101 Chapter15 Re

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