Modified from Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chapter 4 Chemical...

Modified from Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Chapter 4Chemical Quantities and Aqueous Reactions

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?

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?

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 reaction2 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 mathematically1 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 makeassuming you have enough crusts and tomato sauce

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 H2Obecause 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 H2Omol 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

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:

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

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 44.01mol 1

6126

6126

6126

2

6126

2

22

OHC g 5.82 OHC mol 1

OHC g 180.2

CO mol 6

OHC mol 1

CO g 44.01

CO mol 1CO g 7.83

2

6126

CO mol 6OHC mol 1

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:

Stoichiometry Road Mapa A b B

Moles A Moles B

mass A mass B

volume A (l) volume B(l)

Pur

eS

ubst

ance

Sol

utio

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 = molesL

More Making Pizzas• 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?

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.

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

• 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?

Limitingreactant

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:

Theoretical yield

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.

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 C88.2 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 reactanttheoretical yieldpercent yield

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


• Write a conceptual plan


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

kgTiO2

kgC }

smallestamount is

from limitingreactant

smallestmol Ti


• Collect needed relationships

1000 g = 1 kgMolar Mass TiO2 = 79.87 g/mol

Molar Mass Ti = 47.87 g/molMolar 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)


Find: lim. rct., theor. yld., % yld.CP: kg rct g rct mol rct mol Tipick smallest mol Ti TY kg Ti %Y Ti

• Apply the conceptual plan

InformationGiven: 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 Tipick 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


smallest moles of Tilimiting reactant


theoretical yield



• Check the solutions

limiting reactant = TiO2

theoretical yield = 52.9 kgpercent 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%



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)

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

1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2

9.05 g NH3, 45.2 g CuOg N2

Conceptual Plan:

Relationships:

Given:Find:

Choose smallest

because the percent yield is less than 100, the answer makes sense

Check:

Solution:

Theoreticalyield

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?

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

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 ConcentrationMolarity

• 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 solutionmolarity, M

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

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’nM

0.20 mol NH3, 0.2000 L solution

M

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 NaOHliters, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

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 Lmass CaCl2, g

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

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:

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)?

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:

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

solutionmoles solute in solution 1 = moles solute in solution 2

• The concentrations and volumes of the stock and new solutions are inversely proportional

M1 V∙ 1 = M2 V∙ 2

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 MV2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

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 mLM2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Practice – How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution?

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 mLV1, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Dilute 25 mL of 2.0 M solution up to 200.0 mL

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

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:

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 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

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 WaterEach 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

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 acidsHCl(aq) H+(aq) + Cl−(aq)

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

HF(aq) H+(aq) + F−(aq)

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)

Classes of Dissolved Materials

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) 2 H+(aq) + SO42−(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)

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 RulesCompounds that Are Generally

Soluble in Water

Solubility RulesCompounds 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

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

Process for Predicting the Products ofa Precipitation Reaction

1. Determine what ions each aqueous reactant has2. 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 ofa 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 reactantsK2CO3(aq) + NiCl2(aq)

2. Determine the possible productsa) determine the ions present

(K+ + CO32−) + (Ni2+ + Cl−)

b) exchange the Ions(K+ + CO3

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

c) write the formulas of the products• balance charges

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

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

3. Determine the solubility of each productKCl is soluble

NiCO3 is insoluble

4. If both products are soluble, write no reactiondoes not apply because NiCO3 is insoluble

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

5. Write (aq) next to soluble products and (s) next to insoluble productsK2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)

6. Balance the equationK2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)

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)

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)

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

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

Ionic Equations• Equations that describe the chemicals put into the water

and the product molecules are called molecular equations2 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 form2K+

(aq) + 2OH−(aq) + Mg2+

(aq) + 2NO3−

(aq) K+(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 properties2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)

• The net ionic equation for an acid-base reaction isH+(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

Common Acids

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 hydroxide1. 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)


aqueous calcium hydroxide

3. Determine the solubility of the saltCa(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)

5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)


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 equation2 H+(aq) + 2 OH−(aq) H2O(l)

H+(aq) + OH−(aq) H2O(l)


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)

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

Titration

TitrationThe titrant is the base solution in the burette.

As the titrant is added tothe 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


InformationGiven: 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 NaOH0.100 M NaOH 0.100 mol NaOH 1 L sol’n



12.54 mL of 0.100 M NaOHFind: M HCl


• Write a conceptual plan


12.54 mL of 0.100 M NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L

M = mol/L

mLNaOH

LNaOH

molNaOH

molHCl

mLHCl

LHCl




12.54 mL of 0.100 M NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L

M = mol/LCP: mL NaOH → L NaOH →

mol NaOH → mol HCl; mL HCl → L HCl & mol M

= 1.25 x 103 mol HCl




12.54 mL NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L





12.54 mL NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L



• Check the solutionHCl 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 thanNaOH 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 H2O50.0 mL 27.5 mL0.1015 M ? M

114

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 NaOHM NaOH

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 UndergoGas-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 reactantsNa2CO3(aq) + HNO3(aq)

2. Determine the possible productsa) determine the ions present when each reactant dissociates or

ionizes(Na+ + CO3

2−) + (H+ + NO3−)

b) exchange the anions(Na+ + CO3

2−) + (H+ + NO3−) (Na+ + NO3

−) + (H+ + CO32−)

c) write the formula of compounds balance the charges

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


acid, a gas evolves

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)


acid, a gas evolves

5. Determine the solubility of other productNaNO3 is soluble

6. Write an (s) after the insoluble products and an (aq) after the soluble productsNa2CO3(aq) + 2 HNO3(aq) NaNO3(aq) + CO2(g) + H2O(l)

7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) 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 Redox2 H2(g) + O2(g) 2 H2O(g)

Redox without Combustion2 Na(s) + Cl2(g) 2 NaCl(s)

2 Na 2 Na+ + 2 e

Cl2 + 2 e 2 Cl

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+

2O2– (s)

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

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– oxidationCl2 + 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 priority1. 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

Rules for Assigning Oxidation States3. (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

Rules for Assigning Oxidation States5. 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!

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 ReductionAnother 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

oxidationreduction

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

OxidationReduction

Oxidizingagent

Reducingagent

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 +2Ca is oxidized, Sn4+ is reduced

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

0 0 +4−1S is oxidized, F is reduced

S is the reducing agent, F2 is the oxidizing agent

Modified from Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chapter 4 Chemical...

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Transcript of Modified from Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chapter 4 Chemical...