Types of Chemical Reactions & Solution Stoichiometry

Chapter 4Chapter 4

Aqueous Solutions

Water is the dissolving Water is the dissolving medium, or medium, or solvent.solvent.

Some Properties of Water

Water is able to dissolve so many substances Water is able to dissolve so many substances because:because:

- Water is “bent” or Water is “bent” or V-shapedV-shaped..- The O-H bonds are The O-H bonds are covalentcovalent..- Water is a Water is a polarpolar molecule. molecule.- HydrationHydration occurs when salts dissolve in occurs when salts dissolve in

water.water.

04_40

H

O2

105

H

Water is a polar molecule because it is a bentmolecule. The hydrogen end is + while the oxygenend is -, Delta () is a partial charge--less than 1.

04_41

+

+

++

+

+

+

+

++

+

OH

H

O HH

+

+

Cation

Anion

Polar water molecules interact with the positiveand negative ions of a salt, assisting in the dissolving process. This process is called hydration.

A Solute

- dissolves in water (or other “solvent”)dissolves in water (or other “solvent”)

- changes phase changes phase (if different from the (if different from the solvent)solvent)

- is present in is present in lesserlesser amount (if the same amount (if the same phase as the solvent)phase as the solvent)

A Solvent

- retains its phase retains its phase (if different from the (if different from the solute)solute)

- is present inis present in greater greater amount (if the same amount (if the same phase as the solute)phase as the solute)

Solubility

The general rule for solubility is:The general rule for solubility is:

““Like dissolves like.”Like dissolves like.”

Polar water molecules can dissolve other polar Polar water molecules can dissolve other polar molecules such as alcohol and, also, ionic molecules such as alcohol and, also, ionic substances such as NaCl.substances such as NaCl.

Nonpolar molecules can dissolve other Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.substances. Gasoline can dissolve grease.

Miscibility

MiscibleMiscible -- two substances that will mix -- two substances that will mix together in any proportion to make a together in any proportion to make a solution. Alcohol and water are miscible solution. Alcohol and water are miscible because they are both polar and form because they are both polar and form hydrogen bonds.hydrogen bonds.

ImmiscibleImmiscible -- two substances that will not -- two substances that will not dissolve in each other. Oil and vinegar are dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and immiscible because oil is nonpolar and vinegar is polar.vinegar is polar.

Solubility

How does the rule “Like dissolves like.” apply to How does the rule “Like dissolves like.” apply to cleaning paint brushes used for latex paint as cleaning paint brushes used for latex paint as opposed to those used with oil-based paint?opposed to those used with oil-based paint?

CC66HH1414 HH2200

II22 C C66HH1414

HH2200 II22

NaNONaNO33 HH2200

Electrolytes & Nonelectrolytes

An electrolyte is a material that dissolves in An electrolyte is a material that dissolves in water to give a solution that conducts an water to give a solution that conducts an electric current.electric current.

A nonelectrolyte is a substance which, when A nonelectrolyte is a substance which, when dissolved in water, gives a nonconducting dissolved in water, gives a nonconducting solution.solution.

ElectrolytesStrongStrong - conduct current efficiently and are - conduct current efficiently and are

soluble salts, strong acids, and strong bases.soluble salts, strong acids, and strong bases.

NaCl, KNONaCl, KNO33, HNO, HNO33, NaOH, NaOH

Weak Weak - conduct only a small current and are - conduct only a small current and are

weak acids and weak bases.weak acids and weak bases.

HCHC22HH33OO22, aq. NH, aq. NH33, tap H, tap H22OO

NonNon - no current flows and are molecular - no current flows and are molecular

substancessubstances

pure Hpure H22O, sugar solution, glycerolO, sugar solution, glycerol

04_43 Power Source

(a) (b) (c)

+

+

+

+

+

+

Electrical conductivity of aqueous solutions. a) strong electrolyte b) weak electrolyte c) nonelectrolyte in solution.Svante Arrhenius first identified these electrical properties.

04_1529

= Ba2+

= Cl

BaCl2(s)dissolves

When BaCl2 dissolves, the Ba2+ and Cl- ions are randomlydispersed in the water. BaCl2 is a strong electrolyte.

Acids

Strong acids Strong acids -- dissociate completely (~100 %) to dissociate completely (~100 %) to produce Hproduce H++ in solution in solution

HCl, HHCl, H22SOSO44, HNO, HNO33, HBr, HI, & HClO, HBr, HI, & HClO44

Weak acids Weak acids - dissociate to a slight extent (~ 1 %) - dissociate to a slight extent (~ 1 %) to give Hto give H++ in solution in solution

HCHC22HH33OO22, HCOOH, HNO, HCOOH, HNO22, & H, & H22SOSO33

04_1530

= H+

+

+

+ +

+

+

+

+

+ +

= Cl

+

HCl is completely ionized and is a strong electrolyte.

BasesStrong bases Strong bases - react completely with water to - react completely with water to give OHgive OH ions. ions. sodium hydroxidesodium hydroxide

NaOHNaOH(s)(s) ---> Na ---> Na++(aq)(aq) + OH + OH--

(aq)(aq)

Weak bases Weak bases - react only slightly with water to - react only slightly with water to give OHgive OH ions. ions. ammoniaammonia

NHNH3(aq)3(aq) + HOH + HOH(l)(l) <---> NH <---> NH44++

(aq)(aq) + + OHOH--(aq)(aq)

04_1531

+

+

+

+

+

++

+

+

+

+

- = OH

= Na+

An aqueous solution of sodium hydroxide which isa strong bases dissociating almost 100 %.

04_1532

Acetic acid(CH3COOH) exists in water mostly as undissociatedmolecules. Only a small percent of the molecules are ionized.

Write the equation of the dissolving of the following compounds.

CaClCaCl22

HClHCl

Fe(NO3)Fe(NO3)33

KBrKBr

(NH(NH44))22CrCr22OO77

Molarity

Molarity (Molarity (MM) = moles of solute per volume of ) = moles of solute per volume of solution in liters:solution in liters:

M

M

molaritymoles of soluteliters of solution

HClmoles of HCl

liters of solution3

62

Molarity Calculations

Calculate the molarity of a solution prepared by Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.water to make 1.50 L of solution.

((

Molarity CalculationsCalculate the molarity of a solution prepared by Calculate the molarity of a solution prepared by

dissolving 1.56 g of gaseous HCl in enough dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.water to make 26.8 mL of solution.

(1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1000mL/1L) = 1.60M HCl(1000mL/1L) = 1.60M HCl

How many moles of Co(NOHow many moles of Co(NO33))22 are are

present in 25.00 mL of a 0.75 M present in 25.00 mL of a 0.75 M Co(NOCo(NO33))22 solution? solution?

Molarity Calculations

How many moles of nitrate ions are present in How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO25.00 mL of a 0.75 M Co(NO33))22 solution? solution?

(25.00mL)(1L/1000mL)(0.75mol Co(NO(25.00mL)(1L/1000mL)(0.75mol Co(NO33))22/1L)/1L)

(2 mol NO(2 mol NO33--/1 mol Co(NO/1 mol Co(NO33))22) = 3.8 x 10) = 3.8 x 10-2 -2 mol NOmol NO33

--

mol

Calculate the moles of each of the ions in Calculate the moles of each of the ions in 40.00 ml of the following solution.40.00 ml of the following solution.

.20 M Na.20 M Na22COCO33

P106 Q1,-,7, 8,9a-cP106 Q1,-,7, 8,9a-c

Calculate the moles of potassium ions in Calculate the moles of potassium ions in 50.00 ml of the following solution.50.00 ml of the following solution.

2 M K2 M K33PP22

Standard SolutionA standard solution is a solution whose concentration is A standard solution is a solution whose concentration is

accurately known. accurately known.

Standard solutions are made using a volumetric flask as Standard solutions are made using a volumetric flask as follows:follows:

• mass the solute accurately and add it to the mass the solute accurately and add it to the volumetric flaskvolumetric flask

• add a small quantity of distilled HOHadd a small quantity of distilled HOH• dissolve the solute by gently swirling the flaskdissolve the solute by gently swirling the flask• add more distilled HOH until the level of the solution add more distilled HOH until the level of the solution

reaches the mark on the neckreaches the mark on the neck• invert the capped volumetric 25X to thoroughly mix invert the capped volumetric 25X to thoroughly mix

the solution.the solution.

04_44

Volume marker(calibration mark)

Weighedamountof solute

Wash Bottle

(a) (b) (c) (d)

Steps involved in making a standard solution.

04_46

(a) (b) (c)

Rubber bulb

500 mL

Steps to dilute a stock solution.

Common Terms of Solution Concentration

StockStock - routinely used solutions prepared in - routinely used solutions prepared in concentrated form.concentrated form.

ConcentratedConcentrated - - relativelyrelatively large ratio of solute large ratio of solute to solvent. (to solvent. (5.0 5.0 MM NaCl NaCl))

DiluteDilute - - relativelyrelatively small ratio of solute to small ratio of solute to solvent. (solvent. (0.01 0.01 MM NaCl NaCl))

Make a solution demo Make a solution demo

Dilution of Stock Solutions

When diluting stock solutions, the moles of When diluting stock solutions, the moles of solute after dilution must equal the moles solute after dilution must equal the moles of solute before dilution.of solute before dilution.

Stock solutions are diluted using either a Stock solutions are diluted using either a measuring or a delivery pipet and a measuring or a delivery pipet and a volumetric flask.volumetric flask.

Dilution Calculations

What volume of 6 M sulfuric acid must be used to What volume of 6 M sulfuric acid must be used to prepare 1 L of a 3.0 M Hprepare 1 L of a 3.0 M H22SOSO44 solution? solution?

use dilution formulause dilution formula

What volume of 3 M sulfuric acid must be What volume of 3 M sulfuric acid must be used to prepare .5 L of a .25 M Hused to prepare .5 L of a .25 M H22SOSO44

solution?solution?

Types of Solution Reactions

- Precipitation reactionsPrecipitation reactionsAgNOAgNO33((aqaq) + NaCl() + NaCl(aqaq) ) AgCl( AgCl(ss) + NaNO) + NaNO33((aqaq))

- Acid-base reactionsAcid-base reactionsNaOH(NaOH(aqaq) + HCl() + HCl(aqaq) ) NaCl( NaCl(aqaq) + H) + H22O(O(ll))

- Oxidation-reduction reactionsOxidation-reduction reactionsFeFe22OO33((ss) + 2Al() + 2Al(ss) ) 2Fe( 2Fe(ll) + Al) + Al22OO33((ss))

1.      Which of the following substances would you expect to be insoluble in water?

  

Barium hydroxideBarium hydroxide Hydrochloric acidHydrochloric acid

Magnesium sulfateMagnesium sulfate Ammonium nitrateAmmonium nitrate

Silver chlorideSilver chloride Lithium Lithium carbonatecarbonate

Calcium carbonateCalcium carbonate Barium sulfateBarium sulfate

Ammonium acetateAmmonium acetate Lead I ChlorideLead I Chloride

Sodium hydroxideSodium hydroxide Ammonium nitrateAmmonium nitrate

Silver nitrateSilver nitrate

Solubility

Using the solubility rules, predict what will happen Using the solubility rules, predict what will happen when the following pairs of solutions are mixed.when the following pairs of solutions are mixed.

a) a) KOHKOH(aq)(aq) & Mg(NO & Mg(NO33))2(aq)2(aq)

b) Nab) Na22SOSO4(aq)4(aq) & Pb(NO & Pb(NO33))2(aq)2(aq)

c) KNOc) KNO3(aq)3(aq) & BaCl & BaCl2(aq)2(aq)

Mg(OH)2(s) forms

PbSO4(s) forms

No precipitate forms

Describing Reactions in Solution

1.1. Molecular equation Molecular equation (reactants and products as (reactants and products as compoundscompounds))

AgNOAgNO33((aqaq) + NaCl() + NaCl(aqaq) ) AgCl( AgCl(ss) + NaNO) + NaNO33((aqaq))

2.2. Complete ionic equation Complete ionic equation (all strong (all strong electrolytes shown as electrolytes shown as ionsions))

AgAg++((aqaq) + NO) + NO33((aqaq) + Na) + Na++((aqaq) + Cl) + Cl((aqaq) )

AgCl(AgCl(ss) + Na) + Na++((aqaq) + NO) + NO33((aqaq))

Describing Reactions in Solution (continued)

3.3. Net ionic equation Net ionic equation (show only (show only components that actually react)components that actually react)

AgAg++((aqaq) + Cl) + Cl((aqaq) ) AgCl( AgCl(ss))

NaNa++ and NO and NO33 are spectator ions. are spectator ions.

Write the balanced complete ionic and net ionic equations:

CuSO4(aq) + BaCl2(aq) →

P108 q 28P108 q 28

Sodium Sulfate and Lead II NitrateSodium Sulfate and Lead II Nitrate

Old limiting problem to compare

If 68.5 g of COIf 68.5 g of CO(g)(g) is reacted with 8.60 g of is reacted with 8.60 g of

HH2(g)2(g), what is the theoretical yield of , what is the theoretical yield of

methanol that can be produced?methanol that can be produced?

__H__H2(g)2(g) + __CO + __CO(g)(g) ---> __CH ---> __CH33OHOH(l)(l)

Precipitation CalculationsWhen aqueous solutions of NaWhen aqueous solutions of Na22SOSO44 & Pb(NO & Pb(NO33))22 are are

mixed. Calculate the mass of the percipitate mixed. Calculate the mass of the percipitate formed when 1.25 L of 0.0500 M Pb(NOformed when 1.25 L of 0.0500 M Pb(NO33))22 & &

2.00 L of 0.0250 M Na2.00 L of 0.0250 M Na22SOSO44 are mixed. are mixed.

1. 1.

04_STOICHIOMETRY FOR REACTIONS IN SOLUTION

STEP 1Identify the species present in the combined solution, and determinewhat reaction occurs.

STEP 2Write the balanced net ionic equation for the reaction.

STEP 3Calculate the moles of reactants.

STEP 4Determine which reactant is limiting.

STEP 5Calculate the moles of product or products, as required.

STEP 6Convert to grams or other units, as required.

Precipitation CalculationsContinued

3.3. (1.25L)(0.0500mol Pb(NO(1.25L)(0.0500mol Pb(NO33))22/1L) = /1L) =

0.0625 molPb(NO0.0625 molPb(NO33))22

(2.00L)(0.0250mol Na(2.00L)(0.0250mol Na22SOSO44/1L) = 0.0500 molNa/1L) = 0.0500 molNa22SOSO44

4.(0.0625 mol Pb(NO4.(0.0625 mol Pb(NO33))2 2 (1mol Na(1mol Na22SOSO44/1mol Pb(NO/1mol Pb(NO33))22 ) = ) =

0.0625 mol SO0.0625 mol SO442-2-

NaNa22SOSO4 4 is the limiting reactant.is the limiting reactant.

5.5. (0.0500mol Na(0.0500mol Na22SOSO44)(1mol PbSO)(1mol PbSO44/1mol Na/1mol Na22SOSO44))

(303.3g/1mol PbSO(303.3g/1mol PbSO44) = 15.2 g PbSO) = 15.2 g PbSO44

PRACTICE

What mass of precipitate should What mass of precipitate should result when 0.550 L of 0.500 M result when 0.550 L of 0.500 M aluminum nitrate solution is aluminum nitrate solution is mixed with 0.240 L of 1.50 M mixed with 0.240 L of 1.50 M sodium hydroxide solution?sodium hydroxide solution?

answeranswer

9.3g

What mass of precipitate should result when 0.350 L of 0.200 M aluminum nitrate solution is mixed with 0.540 L of .50 M sodium hydroxide solution?

What is the net ionic equation?

P108 29-32P108 29-32

What volume of 0.415 M silver nitrate will be required to

precipitate as silver bromide all the bromide

ion in 35.0 mL of 0.128 M calcium bromide?

answer

2 AgNO3(aq) + CaBr2(aq) Ca(NO3)2(aq) + 2 AgBr(s)

0.0350 L CaBr2 0.128 moles CaBr2 2 moles AgNO3 1 L AgNO3 = 0.0216 L AgNO3 1 L CaBr2 1 moles CaBr2 0.415 mole AgNO3

How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 25.0 mL of 0.350 M aluminum sulfate?

(93.8 mL barium nitrate)answer

Acid-Base Calculations

What volume of a 0.100M HCl solution is needed What volume of a 0.100M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?to neutralize 25.0 mL of 0.350 M NaOH?

1.1. ClCl--, Na, Na++, ,

2.2. 2. H2. H++(aq)(aq) + OH + OH--

(aq)(aq) ----> HOH ----> HOH(l)(l)

(25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solutionNaOH)(1L/0.100mol) = 87.5 mL HCl solution

What volume of a 0.500M HCl solution is What volume of a 0.500M HCl solution is needed to neutralize 32.0 mL of 0.250 M needed to neutralize 32.0 mL of 0.250 M LiOH?LiOH?

Key Titration Terms

TitrantTitrant - solution of known concentration used - solution of known concentration used in titrationin titration

AnalyteAnalyte - substance being analyzed - substance being analyzed

Equivalence point Equivalence point - enough titrant added to - enough titrant added to react exactly with the analytereact exactly with the analyte

Endpoint Endpoint - the indicator changes color so you - the indicator changes color so you can tell the equivalence point has been reached.can tell the equivalence point has been reached.

Ammonium sulfate is manufactured by reacting sulfuric acid with potassium hydroxide. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L potassium hydroxide solution if 50.0 mL of sulfuric acid is used?

Titration CalculationsTitration Titration

49,a,b 53,5449,a,b 53,54

practice for unit Textbook 181-183practice for unit Textbook 181-183

15a15a

17a17a

23a23a

29a-c29a-c

35ab35ab

39 try 39 try

45b 45b

Additional questions

36,a,b 414018,b1224

reminder ,Do redox in redox reminder ,Do redox in redox