Chapter 4 Chemical Quantities and Aqueous Reactions · Chapter 4 Chemical Quantities and Aqueous ... to form 16 molecules of CO 2 and 18 molecules of H 2 O ... 4? 23Tro: Chemistry:

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Copyright © 2011 Pearson Education, Inc.

Chapter 4

Chemical

Quantities

and Aqueous

Reactions

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA

Chemistry: A Molecular Approach, 2nd Ed.

Nivaldo Tro


Global Warming

• Scientists have measured an average 0.6 °C rise

in atmospheric temperature since 1860

• During the same period atmospheric CO2 levels

have risen 25%

• Are the two trends causal?

2 Tro: Chemistry: A Molecular Approach, 2/e


The Sources of Increased CO2 • One source of CO2 is the combustion reactions of fossil

fuels we use to get energy

• Another source of CO2 is volcanic action

• How can we judge whether global warming is natural or due to our use of fossil fuels?

3 Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.

Quantities in Chemical Reactions

• The amount of every substance used and

made in a chemical reaction is related to

the amounts of all the other substances in

the reaction

Law of Conservation of Mass

Balancing equations by balancing atoms

• The study of the numerical relationship

between chemical quantities in a chemical

reaction is called stoichiometry



Reaction Stoichiometry

• The coefficients in a balanced chemical

equation specify the relative amounts in moles

of each of the substances involved in the

reaction

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

2 molecules of C8H18 react with 25 molecules of O2

to form 16 molecules of CO2 and 18 molecules of H2O

2 moles of C8H18 react with 25 moles of O2

to form 16 moles of CO2 and 18 moles of H2O

2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O


Making Pizza

• The number of pizzas you can make depends

on the amount of the ingredients you use

• This relationship can be expressed mathematically

1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza

1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza

• If you want to make more or less than one pizza,

you can use the amount of cheese you have to

determine the number of pizzas you can make

assuming you have enough crusts and tomato sauce


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Predicting Amounts from Stoichiometry

• The amounts of any other substance in a

chemical reaction can be determined from the

amount of just one substance

• How much CO2 can be made from 22.0 moles of

C8H18 in the combustion of C8H18?

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

2 moles C8H18 : 16 moles CO2


Practice

• According to the following equation, how

many moles of water are made in the

combustion of 0.10 moles of glucose?

C6H12O6 + 6 O2 6 CO2 + 6 H2O



Practice How many moles of water are made in the

combustion of 0.10 moles of glucose?

0.6 mol H2O = 0.60 mol H2O because 6x moles of H2O as C6H12O6, the number makes

sense

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 1 mol C6H12O6 : 6 mol H2O

0.10 moles C6H12O6

moles H2O

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

mol H2O mol C6H12O6


Example: Estimate the mass of CO2 produced in

2007 by the combustion of 3.5 x 1015 g gasolne

• Assuming that gasoline is octane, C8H18, the

equation for the reaction is

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

• The equation for the reaction gives the mole

relationship between amount of C8H18 and CO2,

but we need to know the mass relationship, so the

conceptual plan will be

g C8H18 mol CO2 g CO2 mol C8H18



Example: Estimate the mass of CO2 produced in

2007 by the combustion of 3.5 x 1015 g gasoline

because 8x moles of CO2 as C8H18, but the molar mass of

C8H18 is 3x CO2, the number makes sense

1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2

3.4 x 1015 g C8H18

g CO2

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

g C8H18 mol CO2 g CO2 mol C8H18


Which Produces More CO2;

Volcanoes or Fossil Fuel Combustion?

• Our calculation just showed that the world

produced 1.1 x 1016 g of CO2 just from

petroleum combustion in 2007

1.1 x 1013 kg CO2

• Estimates of volcanic CO2 production are

2 x 1011 kg/year

• This means that volcanoes produce less

than 2% of the CO2 added to the air annually

12

%.%.

.81100

1011

1002

yrkg13

yrkg11

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Example 4.1: How many grams of glucose can be

synthesized from 37.8 g of CO2 in photosynthesis?

because 6x moles of CO2 as C6H12O6, but the molar mass

of C6H12O6 is 4x CO2, the number makes sense

1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 : 6 mol CO2

37.8 g CO2, 6 CO2 + 6 H2O C6H12O6+ 6 O2

g C6H12O6

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

g 44.01

mol 1

6 12 6

6 12 6

6 12 6

2

6 12 6

2

2 2

O H C g 5.8 2

O H C mol 1

O H C g 180.2

CO mol 6

O H C mol 1

CO g 44.01

CO mol 1 CO g 7.8 3

2

6 12 6

CO mol 6

O H C mol 1

g C6H12O6 mol CO2 g CO2 mol C6H12O6

mol 1

g 180.2


Practice — How many grams of O2 can be made

from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g)

(PbO2 = 239.2, O2 = 32.00)



Practice — How many grams of O2 can be made from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g)

because ½ moles of O2 as PbO2, and the molar mass of

PbO2 is 7x O2, the number makes sense

1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2

100.0 g PbO2, 2 PbO2 → 2 PbO + O2

g O2

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

g O2 mol PbO2 g PbO2 mol O2


Stoichiometry Road Map a A b B

Moles A Moles B

mass A mass B

volume A (l) volume B(l)

Pu

re

Su

bsta

nce

S

olu

tio

n

Volume A(g) Volume B(g)

% A(aq)

ppm A(aq)

% B(aq)

ppm B(aq) M A(aq) M B(aq)

MM MM

density density

equation

equation

22.4 L 22.4 L

M = moles

L



More Making Pizzas

17

• We know that

1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza

• But what would happen if we had 4 crusts,

15 oz. tomato sauce, and 10 cu cheese?

Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.

More Making Pizzas, Continued • Each ingredient could potentially make a

different number of pizzas

• But all the ingredients have to work together!

• We only have enough tomato sauce to make

three pizzas, so once we make three pizzas,

the tomato sauce runs out no matter how much

of the other ingredients we have.


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More Making Pizzas, Continued

• The tomato sauce limits the amount of pizzas

we can make. In chemical reactions we call

this the limiting reactant.

also known as the limiting reagent

• The maximum number of pizzas we can

make depends on this ingredient. In

chemical reactions, we call this the

theoretical yield.

it also determines the amounts of the other

ingredients we will use!


The Limiting Reactant • For reactions with multiple reactants, it is likely

that one of the reactants will be completely used before the others

• When this reactant is used up, the reaction stops and no more product is made

• The reactant that limits the amount of product is called the limiting reactant

sometimes called the limiting reagent

the limiting reactant gets completely consumed

• Reactants not completely consumed are called excess reactants

• The amount of product that can be made from the limiting reactant is called the theoretical yield



Limiting and Excess Reactants in the

Combustion of Methane

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

• Our balanced equation for the combustion of

methane implies that every one molecule of CH4

reacts with two molecules of O2


Limiting and Excess Reactants in the

Combustion of Methane

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

since less CO2

can be made

from the O2

than the CH4,

so the O2 is

the limiting

reactant

22

• If we have five molecules of CH4 and eight molecules

of O2, which is the limiting reactant?



Practice — How many moles of Si3N4 can be

made from 1.20 moles of Si and 1.00 moles

of N2 in the reaction 3 Si + 2 N2 Si3N4?


Limiting

reactant

Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction

3 Si + 2 N2 Si3N4?

2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4

1.20 mol Si, 1.00 mol N2

mol Si3N4

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

mol N2 mol Si3N4

mol Si mol Si3N4 Pick least

amount Limiting

reactant and

theoretical

yield

Theoretical

yield


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More Making Pizzas • Let’s now assume that as we are making

pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield.

• We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.


Theoretical and Actual Yield

• As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make

• The theoretical yield will always be the least possible amount of product

the theoretical yield will always come from the limiting reactant

• Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield



Example 4.4:

Finding limiting reactant,

theoretical yield, and

percent yield


Example:

• When 28.6 kg of C are allowed to react with 88.2 kg

of TiO2 in the reaction below, 42.8 kg of Ti are

obtained. Find the limiting reactant, theoretical

yield, and percent yield.

TiO

2(s) 2 C(s) Ti(s) 2 CO(g)



Example:

When 28.6 kg of C reacts with

88.2 kg of TiO2, 42.8 kg of Ti are

obtained. Find the limiting

reactant, theoretical yield, and

percent yield

TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)

• Write down the given quantity and its units

Given: 28.6 kg C

88.2 kg TiO2

42.8 kg Ti produced


Example:

Find the limiting

reactant, theoretical

yield, and percent yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

• Write down the quantity to find and/or its units

Find: limiting reactant

theoretical yield

percent yield

Information

Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti


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Example:

Find the limiting


yield, and percent

yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

• Write a conceptual plan

Information


Find: lim. rct., theor. yld., % yld.

kg

TiO2

kg

C } smallest

amount is

from

limiting

reactant

smallest

mol Ti


Example:

Find the limiting


yield, and percent

yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

• Collect needed relationships

1000 g = 1 kg

Molar Mass TiO2 = 79.87 g/mol

Molar Mass Ti = 47.87 g/mol

Molar Mass C = 12.01 g/mol

1 mole TiO2 : 1 mol Ti (from the chem. equation)

2 mole C : 1 mol Ti (from the chem. equation)

Information


Find: lim. rct., theor. yld., % yld.

CP: kg rct g rct mol rct mol Ti

pick smallest mol Ti TY kg Ti %Y Ti



• Apply the conceptual plan

Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct g rct mol rct mol Ti pick smallest mol Ti TY kg Ti %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti

Example:

Find the limiting

reactant,

theoretical yield,

and percent yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)

smallest moles of Ti limiting reactant



theoretical yield


Example:

Find the limiting

reactant,

theoretical yield,

and percent yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)





Example:

Find the limiting

reactant,

theoretical yield,

and percent yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)


• Check the solutions

limiting reactant = TiO2

theoretical yield = 52.9 kg

percent yield = 80.9%

Because Ti has lower molar mass than TiO2, the T.Y. makes

sense and the percent yield makes sense as it is less than 100%


Example:

Find the limiting

reactant,

theoretical yield,

and percent yield

TiO2(s) + 2 C(s)

Ti(s) + 2 CO(g)


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Practice — How many grams of N2(g) can be made

from 9.05 g of NH3 reacting with 45.2 g of CuO?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

If 4.61 g of N2 are made, what is the percent yield?


Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting

with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)


1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g

2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2

9.05 g NH3, 45.2 g CuO

g N2

Conceptual

Plan:

Relationships:

Given:

Find:

g NH3 mol N2 mol NH3

g N2

g CuO mol N2 mol CuO

Choose

smallest



because the percent yield is less than 100,

the answer makes sense

Check:

Solution:

Theoretical

yield

39

Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting

with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)



Solutions

• When table salt is mixed with water, it seems to

disappear, or become a liquid – the mixture is

homogeneous

the salt is still there, as you can tell from the taste, or

simply boiling away the water

• Homogeneous mixtures are called solutions

• The component of the solution that changes state is

called the solute

• The component that keeps its state is called the

solvent

if both components start in the same state, the major

component is the solvent



Describing Solutions

• Because solutions are mixtures, the composition

can vary from one sample to another

pure substances have constant composition

saltwater samples from different seas or lakes have

different amounts of salt

• So to describe solutions accurately, we must

describe how much of each component is

present

we saw that with pure substances, we can describe

them with a single name because all samples are

identical


Solution Concentration

• Qualitatively, solutions are

often described as dilute or

concentrated

• Dilute solutions have a

small amount of solute

compared to solvent

• Concentrated solutions

have a large amount of

solute compared to solvent


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Concentrations—Quantitative

Descriptions of Solutions

• A more precise method for describing a

solution is to quantify the amount of solute in a

given amount of solution

• Concentration = amount of solute in a given

amount of solution

occasionally amount of solvent


Solution Concentration

Molarity

• Moles of solute per 1 liter of solution

• Used because it describes how many

molecules of solute in each liter of solution



Preparing 1 L of a 1.00 M NaCl

Solution


Example 4.5: Find the molarity of a solution that has

25.5 g KBr dissolved in 1.75 L of solution

because most solutions are between 0 and 18 M, the

answer makes sense

1 mol KBr = 119.00 g, M = moles/L

25.5 g KBr, 1.75 L solution

molarity, M

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

g KBr mol KBr

L sol’n

M



Practice — What Is the molarity of a solution containing

3.4 g of NH3 (MM 17.03) in 200.0 mL of solution?


Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution?

the unit is correct, the number is reasonable because

the fraction of moles is less than the fraction of liters Check:

Solve:

M = mol/L, 1 mol NH3 = 17.03 g, 1 mL = 0.001 L

Conceptual

Plan:

Relationships:

3.4 g NH3, 200.0 mL solution

M

Given:

Find:

g NH3 mol NH3

mL sol’n L sol’n M

48

0.20 mol NH3, 0.2000 L solution

M


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Using Molarity in Calculations

• Molarity shows the relationship between the

moles of solute and liters of solution

• If a sugar solution concentration is 2.0 M, then

1 liter of solution contains 2.0 moles of sugar

2 liters = 4.0 moles sugar

0.5 liters = 1.0 mole sugar

• 1 L solution : 2 moles sugar


Example 4.6: How many liters of 0.125 M

NaOH contain 0.255 mol NaOH?

because each L has only 0.125 mol NaOH, it makes

sense that 0.255 mol should require a little more than 2 L

0.125 mol NaOH = 1 L solution

0.125 M NaOH, 0.255 mol NaOH

liters, L

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

mol NaOH L sol’n



Practice — Determine the mass of CaCl2

(MM = 110.98) in 1.75 L of 1.50 M solution


Practice — Determine the mass of CaCl2

(MM = 110.98) in 1.75 L of 1.50 M solution

because each L has 1.50 mol CaCl2, it makes sense

that 1.75 L should have almost 3 moles

1.50 mol CaCl2 = 1 L solution; 110.98 g CaCl2 = 1 mol

1.50 M CaCl2, 1.75 L

mass CaCl2, g

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

mol CaCl2 L sol’n g CaCl2



Example: How would you prepare 250.0 mL of a

1.00 M solution CuSO45 H2O(MM 249.69)?

the unit is correct, the magnitude seems

reasonable as the volume is ¼ of a liter Check:

Solution:

1.00 L sol’n = 1.00 mol; 1 mL = 0.001 L; 1 mol = 249.69 g

Conceptual

Plan:

Relationships:

250.0 mL solution

mass CuSO4 5 H2O, g

Given:

Find:

mL sol’n L sol’n g CuSO4 mol CuSO4

Dissolve 62.4 g of CuSO4∙5H2O in enough water to total 250.0 mL


Practice – How would you prepare 250.0 mL

of 0.150 M CaCl2 (MM = 110.98)?


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Practice – How would you prepare 250.0 mL

of 0.150 M CaCl2?

the unit is correct, the magnitude seems

reasonable as the volume is ¼ of a liter Check:

Solution:

1.00 L sol’n = 0.150 mol; 1 mL = 0.001L; 1 mol = 110.98 g

Conceptual

Plan:

Relationships:

250.0 mL solution

mass CaCl2, g

Given:

Find:

mL sol’n L sol’n g CaCl2 mol CaCl2

Dissolve 4.16 g of CaCl2 in enough water to total 250.0 mL


Dilution

• Often, solutions are stored as concentrated stock

solutions

• To make solutions of lower concentrations from

these stock solutions, more solvent is added

the amount of solute doesn’t change, just the volume of

solution

moles solute in solution 1 = moles solute in solution 2

• The concentrations and volumes of the stock and

new solutions are inversely proportional

M1∙V1 = M2∙V2



Example 4.7: To what volume should you dilute

0.200 L of 15.0 M NaOH to make 3.00 M NaOH?

because the solution is diluted by a factor of 5, the volume

should increase by a factor of 5, and it does

M1V1 = M2V2

V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M

V2, L

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

V1, M1, M2 V2


Practice – What is the concentration of a

solution prepared by diluting 45.0 mL of

8.25 M HNO3 to 135.0 mL?



Practice – What is the concentration of a solution prepared

by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL?

because the solution is diluted by a factor of 3, the

molarity should decrease by a factor of 3, and it does

M1V1 = M2V2

V1 = 45.0 mL, M1 = 8.25 M, V2 = 135.0 mL

M2, L

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

V1, M1, V2 M2


Practice – How would you prepare 200.0 mL of

0.25 M NaCl solution from a 2.0 M solution?


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Practice – How would you prepare 200.0 mL of 0.25 M

NaCl solution from a 2.0 M solution?

because the solution is diluted by a factor of 8, the volume

should increase by a factor of 8, and it does

M1V1 = M2V2

M1 = 2.0 M, M2 = 0.25 M, V2 = 200.0 mL

V1, L

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

M1, M2, V2 V1

Dilute 25 mL of 2.0 M solution up to 200.0 mL


What Happens When a Solute Dissolves? • There are attractive forces between the solute

particles holding them together

• There are also attractive forces between the solvent molecules

• When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules

• If the attractions between solute and solvent are strong enough, the solute will dissolve



Table Salt Dissolving in Water Each ion is attracted

to the surrounding

water molecules and

pulled off and away

from the crystal

When it enters the

solution, the ion is

surrounded by water

molecules, insulating

it from other ions

The result is a solution

with free moving

charged particles able

to conduct electricity


Solubility of Ionic Compounds

• Some ionic compounds, such as NaCl, dissolve

very well in water at room temperature

• Other ionic compounds, such as AgCl, dissolve

hardly at all in water at room temperature

• Compounds that dissolve in a solvent are said to be

soluble, where as those that do not are said to be

insoluble

NaCl is soluble in water, AgCl is insoluble in water

the degree of solubility depends on the temperature

even insoluble compounds dissolve, just not enough to be

meaningful



When Will a Salt Dissolve?

• Predicting whether a compound will dissolve in

water is not easy

• The best way to do it is to do some

experiments to test whether a compound will

dissolve in water, then develop some rules

based on those experimental results

we call this method the empirical method


Solubility Rules

Compounds that Are Generally

Soluble in Water


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Solubility Rules

Compounds that Are Generally

Insoluble in Water


Practice – Determine if each of the

following is soluble in water

KOH

AgBr

CaCl2

Pb(NO3)2

PbSO4

KOH is soluble because it contains K+

AgBr is insoluble; most bromides are soluble,

but AgBr is an exception

CaCl2 is soluble; most chlorides are soluble,

and CaCl2 is not an exception

Pb(NO3)2 is soluble because it contains NO3−

PbSO4 is insoluble; most sulfates are soluble,

but PbSO4 is an exception



Electrolytes and Nonelectrolytes

• Materials that dissolve

in water to form a

solution that will

conduct electricity are

called electrolytes

• Materials that dissolve

in water to form a

solution that will not

conduct electricity are

called nonelectrolytes


Molecular View of

Electrolytes and Nonelectrolytes • To conduct electricity, a material must have

charged particles that are able to flow

• Electrolyte solutions all contain ions dissolved in

the water

ionic compounds are electrolytes because they

dissociate into their ions when they dissolve

• Nonelectrolyte solutions contain whole molecules

dissolved in the water

generally, molecular compounds do not ionize when

they dissolve in water

the notable exception being molecular acids



Salt vs. Sugar Dissolved in Water


ionic compounds

dissociate into ions when

they dissolve

molecular compounds

do not dissociate when

they dissolve


Acids • Acids are molecular compounds that ionize when

they dissolve in water

the molecules are pulled apart by their attraction for the water

when acids ionize, they form H+ cations and also anions

• The percentage of molecules that ionize varies from one acid to another

• Acids that ionize virtually 100% are called strong acids

HCl(aq) H+(aq) + Cl−(aq)

• Acids that only ionize a small percentage are called weak acids

HF(aq) H+(aq) + F−(aq) 72 Tro: Chemistry: A Molecular Approach, 2/e

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Strong and Weak Electrolytes

• Strong electrolytes are materials that dissolve

completely as ions

ionic compounds and strong acids

their solutions conduct electricity well

• Weak electrolytes are materials that dissolve

mostly as molecules, but partially as ions

weak acids

their solutions conduct electricity, but not well

• When compounds containing a polyatomic ion

dissolve, the polyatomic ion stays together

HC2H3O2(aq) H+(aq) + C2H3O2−(aq)


Dissociation and Ionization

• When ionic compounds dissolve in water, the anions and

cations are separated from each other. This is called

dissociation.

Na2S(aq) 2 Na+(aq) + S2-(aq)

• When compounds containing polyatomic ions dissociate,

the polyatomic group stays together as one ion

Na2SO4(aq) 2 Na+(aq) + SO42−(aq)

• When strong acids dissolve in water, the molecule

ionizes into H+ and anions

H2SO4(aq) H+(aq) + HSO4−(aq)

HSO4−(aq) H+(aq) + SO4

2− (aq)



Practice – Write the equation for the process that

occurs when the following strong electrolytes

dissolve in water

CaCl2

HNO3

(NH4)2CO3

CaCl2(aq) Ca2+(aq) + 2 Cl−(aq)

HNO3(aq) H+(aq) + NO3−(aq)

(NH4)2CO3(aq) 2 NH4+(aq) + CO3

2−(aq)


Precipitation Reactions

• Precipitation reactions are

reactions in which a solid forms

when we mix two solutions

reactions between aqueous

solutions of ionic compounds

produce an ionic compound that

is insoluble in water

the insoluble product is called a

precipitate



2 KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2 KNO3(aq)


No Precipitate Formation =

No Reaction

KI(aq) + NaCl(aq) KCl(aq) + NaI(aq)

all ions still present, no reaction


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14


Process for Predicting the Products of

a Precipitation Reaction 1. Determine what ions each aqueous reactant has

2. Determine formulas of possible products

exchange ions

(+) ion from one reactant with (-) ion from other

balance charges of combined ions to get formula of each

product

3. Determine solubility of each product in water

use the solubility rules

if product is insoluble or slightly soluble, it will precipitate

4. If neither product will precipitate, write no reaction

after the arrow


Process for Predicting the Products of

a Precipitation Reaction

5. If any of the possible products are insoluble,

write their formulas as the products of the

reaction using (s) after the formula to indicate

solid. Write any soluble products with (aq)

after the formula to indicate aqueous.

6. Balance the equation

remember to only change coefficients, not

subscripts



Example 4.10: Write the equation for the

precipitation reaction between an aqueous

solution of potassium carbonate and an aqueous

solution of nickel(II) chloride 1. Write the formulas of the reactants

K2CO3(aq) + NiCl2(aq)

2. Determine the possible products

a) determine the ions present

(K+ + CO32−) + (Ni2+ + Cl−)

b) exchange the Ions

(K+ + CO32−) + (Ni2+ + Cl−) (K+ + Cl−) + (Ni2+ + CO3

2−)

c) write the formulas of the products

balance charges

K2CO3(aq) + NiCl2(aq) KCl + NiCO3




solution of potassium carbonate and an

aqueous solution of nickel(II) chloride

3. Determine the solubility of each product

KCl is soluble

NiCO3 is insoluble

4. If both products are soluble, write no

reaction

does not apply because NiCO3 is insoluble





solution of potassium carbonate and an

aqueous solution of nickel(II) chloride

5. Write (aq) next to soluble products and (s)

next to insoluble products

K2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)


K2CO3(aq) + NiCl2(aq) 2 KCl(aq) + NiCO3(s)

83 Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. 84

Practice – Predict the products and

balance the equation

KCl(aq) + AgNO3(aq) KNO3(aq) + AgCl(s)

Na2S(aq) + CaCl2(aq) NaCl(aq) + CaS(aq)

No reaction

(K+ + Cl−) + (Ag+ + NO3−) → (K+ + NO3

−) + (Ag+ + Cl−)

KCl(aq) + AgNO3(aq) → KNO3 + AgCl

(Na+ + S2−) + (Ca2+ + Cl−) → (Na+ + Cl−) + (Ca2+ + S2−)

Na2S(aq) + CaCl2(aq) → NaCl + CaS

KCl(aq) + AgNO3(aq)

Na2S(aq) + CaCl2(aq)


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Practice – Write an equation for the

reaction that takes place when an

aqueous solution of (NH4)2SO4 is mixed

with an aqueous solution of Pb(C2H3O2)2.

(NH4)2SO4(aq) + Pb(C2H3O2)2(aq)

NH4C2H3O2(aq) + PbSO4(s)

(NH4+ + SO4

2−) + (Pb2+ + C2H3O2−) → (NH4

+ + C2H3O2−) + (Pb2+ + SO4

2−)


(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) NH4C2H3O2 + PbSO4

85


2 NH4C2H3O2(aq) + PbSO4(s)


Ionic Equations

• Equations that describe the chemicals put into the

water and the product molecules are called molecular

equations

2 KOH(aq) + Mg(NO3)2(aq) 2 KNO3(aq) + Mg(OH)2(s)

• Equations that describe the material’s structure when

dissolved are called complete ionic equations

aqueous strong electrolytes are written as ions

soluble salts, strong acids, strong bases

insoluble substances, weak electrolytes, and nonelectrolytes

are written in molecule form

solids, liquids, and gases are not dissolved, therefore molecule form

2K+(aq) + 2OH−

(aq) + Mg2+(aq) + 2NO3

−(aq) 2K+

(aq) + 2NO3−

(aq) + Mg(OH)2(s)



Ionic Equations

• Ions that are both reactants and products are called spectator ions

2 K+(aq) + 2 OH−

(aq) + Mg2+(aq) + 2 NO3

−(aq) 2 K+

(aq) + 2 NO3−

(aq) + Mg(OH)2(s)

An ionic equation in which the spectator ions are removed is called a net ionic equation

2 OH−(aq) + Mg2+

(aq) Mg(OH)2(s)


Practice – Write the ionic and net

ionic equation for each

K2SO4(aq) + 2 AgNO3(aq) 2 KNO3(aq) + Ag2SO4(s)

2K+(aq) + SO42−(aq) + 2Ag+(aq) + 2NO3

−(aq)

2K+(aq) + 2NO3−(aq) + Ag2SO4(s)

2 Ag+(aq) + SO42−(aq) Ag2SO4(s)

Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l)

2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq)

2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l)

CO32−(aq) + 2 H+(aq) CO2(g) + H2O(l)

K2SO4(aq) + 2 AgNO3(aq) 2 KNO3(aq) + Ag2SO4(s)

Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l)



Acid-Base Reactions

• Also called neutralization reactions because the

acid and base neutralize each other’s properties

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

• The net ionic equation for an acid-base reaction is

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

as long as the salt that

forms is soluble in water


Acids and Bases in Solution

• Acids ionize in water to form H+ ions more precisely, the H from the acid molecule is donated

to a water molecule to form hydronium ion, H3O+

most chemists use H+ and H3O+ interchangeably

• Bases dissociate in water to form OH ions bases, such as NH3, that do not contain OH ions,

produce OH by pulling H off water molecules

• In the reaction of an acid with a base, the H+ from

the acid combines with the OH from the base to

make water

• The cation from the base combines with the anion

from the acid to make the salt

acid + base salt + water 90 Tro: Chemistry: A Molecular Approach, 2/e

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Common Acids

91 Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. 92

Common Bases



HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)


Example: Write the molecular, ionic, and net-

ionic equation for the reaction of aqueous

nitric acid with aqueous calcium hydroxide 1. Write the formulas of the reactants

HNO3(aq) + Ca(OH)2(aq)

2. Determine the possible products

a) determine the ions present when each reactant dissociates or ionizes

(H+ + NO3−) + (Ca2+ + OH−)

b) exchange the ions, H+ combines with OH− to make H2O(l)

(H+ + NO3−) + (Ca2+ + OH−) (Ca2+ + NO3

−) + H2O(l)

c write the formula of the salt

(H+ + NO3−) + (Ca2+ + OH−) Ca(NO3)2 + H2O(l)





nitric acid with aqueous calcium hydroxide

3. Determine the solubility of the salt

Ca(NO3)2 is soluble

4. Write an (s) after the insoluble products and an

(aq) after the soluble products

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


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




nitric acid with aqueous calcium hydroxide

6. Dissociate all aqueous strong electrolytes to get

complete ionic equation

not H2O

2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq)

Ca2+(aq) + 2 NO3−(aq) + H2O(l)

7. Eliminate spectator ions to get net-ionic equation

2 H+(aq) + 2 OH−(aq) 2 H2O(l)

H+(aq) + OH−(aq) H2O(l)


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2 HCl(aq) + Ba(OH)2(aq) 2 H2O(l) + BaCl2(aq)

H2SO4(aq) + Sr(OH)2(aq) 2 H2O(l) + SrSO4(s)

(H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−)

HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2

(H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO4

2−)

H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4

H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4

HCl(aq) + Ba(OH)2(aq)

H2SO4(aq) + Sr(OH)2(aq)


Gas-Evolving Reactions

• Some reactions form a gas directly from the ion

exchange

K2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g)

• Other reactions form a gas by the decomposition of

one of the ion exchange products into a gas and water

K2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq)

H2SO3 H2O(l) + SO2(g)



NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)


Compounds that Undergo

Gas-Evolving Reactions



Example 4.14: When an aqueous solution of

sodium carbonate is added to an aqueous

solution of nitric acid, a gas evolves

1. Write the formulas of the reactants Na2CO3(aq) + HNO3(aq)

2. Determine the possible products a) determine the ions present when each reactant

dissociates or ionizes

(Na+ + CO32−) + (H+ + NO3

−)

b) exchange the anions

(Na+ + CO32−) + (H+ + NO3

−) (Na+ + NO3−) + (H+ + CO3

2−)

c) write the formula of compounds

balance the charges

Na2CO3(aq) + HNO3(aq) NaNO3 + H2CO3





3. Check to see if either product is H2S - No

4. Check to see if either product decomposes –

Yes

H2CO3 decomposes into CO2(g) + H2O(l)

Na2CO3(aq) + HNO3(aq) NaNO3 + CO2(g) + H2O(l)


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5. Determine the solubility of other product

NaNO3 is soluble

6. Write an (s) after the insoluble products and an

(aq) after the soluble products

Na2CO3(aq) + 2 HNO3(aq) 2 NaNO3(aq) + CO2(g) + H2O(l)


Na2CO3(aq) + 2 HNO3(aq) 2 NaNO3(aq) + CO2(g) + H2O(l)




2 HCl(aq) + Na2SO3(aq) H2O(l) + SO2(g) + 2 NaCl(aq)

(H+ + Cl−) + (Na+ + SO32−) → (H+ + SO3

2−) + (Na+ + Cl−)

HCl(aq) + Na2SO3(aq) → H2SO3 + NaCl

(H+ + SO42−) + (Ca2+ + S2−) → (H+ + S2−) + (Ca2+ + SO4

2−)

H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4

H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4(s)

HCl(aq) + Na2SO3(aq)

H2SO4(aq) + CaS(aq)

HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + NaCl



Other Patterns in Reactions

• The precipitation, acid-base, and gas-evolving

reactions all involve exchanging the ions in the

solution

• Other kinds of reactions involve transferring

electrons from one atom to another – these are

called oxidation-reduction reactions

also known as redox reactions

many involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)


Combustion as Redox

2 H2(g) + O2(g) 2 H2O(g)



Redox without Combustion

2 Na(s) + Cl2(g) 2 NaCl(s)

2 Na 2 Na+ + 2 e

Cl2 + 2 e 2 Cl 107 Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.

Reactions of Metals with

Nonmetals

• Consider the following reactions:

4 Na(s) + O2(g) → 2 Na2O(s)

2 Na(s) + Cl2(g) → 2 NaCl(s)

• The reactions involve a metal reacting with a

nonmetal

• In addition, both reactions involve the conversion

of free elements into ions

4 Na(s) + O2(g) → 2 Na+2O

2– (s)

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)


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Oxidation and Reduction • To convert a free element into an ion, the

atoms must gain or lose electrons

of course, if one atom loses electrons, another must

accept them

• Reactions where electrons are transferred

from one atom to another are redox reactions

• Atoms that lose electrons are being oxidized,

atoms that gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

Na → Na+ + 1 e– oxidation

Cl2 + 2 e– → 2 Cl– reduction

Leo

Ger


Electron Bookkeeping

• For reactions that are not metal + nonmetal, or

do not involve O2, we need a method for

determining how the electrons are transferred

• Chemists assign a number to each element in

a reaction called an oxidation state that

allows them to determine the electron flow in

the reaction

even though they look like them, oxidation states

are not ion charges!

oxidation states are imaginary charges assigned

based on a set of rules

ion charges are real, measurable charges



Rules for Assigning Oxidation States

• Rules are in order of priority

1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)

2. monatomic ions have an oxidation state equal to their charge

Na = +1 and Cl = −1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0



3. (b) the sum of the oxidation states of all the

atoms in a polyatomic ion equals the charge

on the ion

N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1

4. (a) Group I metals have an oxidation state of

+1 in all their compounds

Na = +1 in NaCl

4. (b) Group II metals have an oxidation state of

+2 in all their compounds

Mg = +2 in MgCl2




5. in their compounds, nonmetals have oxidation

states according to the table below

nonmetals higher on the table take priority


Example: Determine the oxidation states of all

the atoms in a propanoate ion, C3H5O2–

• There are no free elements or free ions in

propanoate, so the first rule that applies is Rule 3b

(C3) + (H5) + (O2) = −1

• Because all the atoms are nonmetals, the next

rule we use is Rule 5, following the elements in

order:

H = +1

O = −2

(C3) + 5(+1) + 2(−2) = −1

(C3) = −2

C = −⅔

Note: unlike charges,

oxidation states can

be fractions!


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Practice – Assign an oxidation state

to each element in the following

• Br2

• K+

• LiF

• CO2

• SO42−

• Na2O2

Br = 0, (Rule 1)

K = +1, (Rule 2)

Li = +1, (Rule 4a) & F = −1, (Rule 5)

O = −2, (Rule 5) & C = +4, (Rule 3a)

O = −2, (Rule 5) & S = +6, (Rule 3b)

Na = +1, (Rule 4a) & O = −1 , (Rule 3a)


Oxidation and Reduction

Another Definition

• Oxidation occurs when an atom’s oxidation

state increases during a reaction

• Reduction occurs when an atom’s oxidation

state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O −4 +1 0 +4 –2 +1 −2

oxidation

reduction



Oxidation–Reduction

• Oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them

• The reactant that reduces an element in another

reactant is called the reducing agent

the reducing agent contains the element that is oxidized

• The reactant that oxidizes an element in another

reactant is called the oxidizing agent

the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent


Example: Assign oxidation states,

determine the element oxidized and reduced,

and determine the oxidizing agent and

reducing agent in the following reactions:

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O

0 −2 +7 +1 +3 −2 +4 +1 −2

Oxidation Reduction

Oxidizing

agent

Reducing

agent



Practice – Assign oxidation states,

determine the element oxidized and reduced,

and determine the oxidizing agent and

reducing agent in the following reactions:

Sn4+ + Ca → Sn2+ + Ca2+

F2 + S → SF4

+4 0 +2 +2

Ca is oxidized, Sn4+ is reduced

Ca is the reducing agent, Sn4+ is the oxidizing agent

0 0 +4−1

S is oxidized, F is reduced

S is the reducing agent, F2 is the oxidizing agent


Solution Stoichiometry

• Because molarity relates the moles of solute to

the liters of solution, it can be used to convert

between amount of reactants and/or products

in a chemical reaction


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Example 4.8: What volume of 0.150 M KCl is required to

completely react with 0.150 L of 0.175 M Pb(NO3)2 in the

reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)?

because you need 2x moles of KCl as Pb(NO3)2, and the

molarity of Pb(NO3)2 > KCl, the volume of KCl should be

more than 2x the volume of Pb(NO3)2

1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl

0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2

L KCl

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

L Pb(NO3)2 mol KCl L KCl mol Pb(NO3)2


Practice — Solution stoichiometry

• 43.8 mL of 0.107 M HCl is needed to

neutralize 37.6 mL of Ba(OH)2 solution.

What is the molarity of the base?

2 HCl + Ba(OH)2 BaCl2 + 2 H2O



Practice – 43.8 mL of 0.107 M HCl is needed to neutralize

37.6 mL of Ba(OH)2 solution. What is the molarity of the

base? 2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(aq)

the units are correct, the number makes sense because

there are fewer moles than liters

1 mL= 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH)2 : 2 mol HCl

43.8 mL of 0.107 M HCl, 37.6 mL Ba(OH)2

M Ba(OH)2

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

L HCl mol Ba(OH)2 mol HCl

L Ba(OH)2

M Ba(OH)2

0.0438 L of 0.107 M HCl, 0.0376 L Ba(OH)2

M Ba(OH)2


Titration • Often in the lab, a solution’s concentration

is determined by reacting it with another material and using stoichiometry – this process is called titration

• In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio.

the unknown solution is added slowly from an instrument called a burette

a long glass tube with precise volume markings that allows small additions of solution



Acid-Base Titrations

• The difficulty is determining when there has been

just enough titrant added to complete the reaction

the titrant is the solution in the burette

• In acid-base titrations, because both the reactant

and product solutions are colorless, a chemical is

added that changes color when the solution

undergoes large changes in acidity/alkalinity

the chemical is called an indicator

• At the endpoint of an acid-base titration, the

number of moles of H+ equals the number of

moles of OH

also known as the equivalence point 125 Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.

Titration


10/7/2014

22


Titration The titrant is the base

solution in the burette.

As the titrant is added to

the flask, the H+ reacts

with the OH– to form

water. But there is still

excess acid present so the

color does not change.

At the titration’s endpoint,

just enough base has

been added to neutralize

all the acid. At this point

the indicator changes

color.


Example 4.14:

The titration of 10.00 mL of HCl

solution of unknown concentration

requires 12.54 mL of 0.100 M

NaOH solution to reach the end

point. What is the concentration of

the unknown HCl solution?

• Write down the given quantity and its units

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH



• Write down the quantity to find, and/or its units

Find: concentration HCl, M

Example 4.14:







Information

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH


• Collect needed equations and conversion factors

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

1 mole HCl = 1 mole NaOH

0.100 M NaOH 0.100 mol NaOH 1 L sol’n

Example 4.14:







Information

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Find: M HCl



• Write a conceptual plan

mL

NaOH

L

NaOH

mol

NaOH

mol

HCl

Information

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Find: M HCl

Rel: 1 mol HCl = 1 mol NaOH

0.100 mol NaOH = 1 L

M = mol/L

Example 4.14:







mL

HCl

L

HCl



= 1.25 x 103 mol HCl

Information

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

Find: M HCl



M = mol/L

CP: mL NaOH → L NaOH →

mol NaOH → mol HCl;

mL HCl → L HCl & mol M

Example 4.14:


solution of unknown

concentration requires 12.54

mL of 0.100 M NaOH solution

to reach the end point. What is

the concentration of the

unknown HCl solution?


10/7/2014

23



Information

Given: 10.00 mL HCl

12.54 mL NaOH

Find: M HCl



M = mol/L




Example 4.14:


solution of unknown







Example 4.14:


solution of unknown







Information

Given: 10.00 mL HCl

12.54 mL NaOH

Find: M HCl



M = mol/L




• Check the solution

HCl solution = 0.125 M

The units of the answer, M, are correct.

The magnitude of the answer makes sense because

the neutralization takes less HCl solution than

NaOH solution, so the HCl should be more concentrated.


Practice − What is the concentration of

NaOH solution that requires 27.5 mL to

titrate 50.0 mL of 0.1015 M H2SO4?

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

50.0 mL 27.5 mL

0.1015 M ? M

Tro: Chemistry: A Molecular Approach, 2/e 135 Copyright © 2011 Pearson Education, Inc.

1 mL= 0.001L, 1 LH2SO4 = 0.1015mol, 2mol NaOH : 1mol H2SO4

Practice — What is the concentration of NaOH solution that

requires 27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4?

2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(aq)

the units are correct, the number makes because the volume of

NaOH is about ½ the H2SO4, but the stoichiometry says you need

twice the moles of NaOH as H2SO4

50.0 mL of 0.1015 M H2SO4, 27.5 mL NaOH

M NaOH

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

0.0500 L of 0.1015 M H2SO4, 0.0275 L NaOH

M NaOH

L H2SO4 mol H2SO4 mol NaOH

L NaOH

M NaOH


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