Solution Concentration• Molarity = mol solute = M• L sol’n• Most commonly used unit of concentration• A 0.50 M sol'n = 0.50 mol solute

1.0 L sol'n• Use it as a conversion factor– Can change • 1 - moles (or grams) of solute to L of sol'n• 2 - L of sol'n to mol of solute

– Which can be converted to grams

Making Solutions

• What is the molarity of a solution of 29.25 g of NaCl in 500 mL of water solution?– Add water to the solid to the desired volume.– Solutions volumes are NOT additive

Types of problems

1. Deter M given the amount (mol or g) of solute and volume of solution

• Convert g mol and ÷ by L sol'n

2. Deter the amt of solute - in grams or moles -in a given vol of sol'n (L x M = moles)• 0.50 M means 0.50 mol/1L • use M as conver. fact. to convert L mol• Convert moles g if necessary

3. Determine the vol of sol’n containing a given amt of solute - in moles or grams • If given g convert to mol• Use M convert mol vol of sol'n in L

1. Calc the conc. (M) of a solution created by dissolving 0.60 mol of NaOH (molar mass = 40.0 g/mol) in enough water to make 1.75 L of solution.

2. Determine the number of grams of CaCO3 (mm = 100.1 g/mol) in 3.55 L of a 1.5 M solution

3. How many liters of a 2.50 M solution of H2SO4 (molar mass = 98.0 g/mol) could be created from 27.0 g of H2SO4?

Dilutions Problems

• Dilution means to add more solvent, usually water, and reduce the solution’s conc (M).– M x L = moles • Molarity x vol (L) = moles solute

– Mole solute before = mole solute after

– MbVb = MaVa

• M is not a conversion factor in these problems

Dilutions Problems• A 1.685 L solution of HCl is 0.055 M. If the solution is diluted

to 2.500 L what is the new molarity?

• How many mL of a 0.55 M solution could be made from 1.50 L of a 1.00 M stock solution?

• What mass of solid aluminum hydroxide (mm = 78.0 g/mol) is produced when 50.0 mL of 0.200 M Al(NO3)3 (mm = 213.0 g/mol) is added to 200.0 mL of 0.100 M KOH (mm = 56.1 g/mol)?

• • Al(NO3)3 (aq) + 3KOH (aq) Al(OH)3 (s) + 3KNO3 (aq)

• Solution Stoich• What mass, in grams, of AgCl will precipitate when

0.050 L of a 0.0500 M solution of AgNO3 reacts with 25.0 mL of .0330 M NaCl ?

AgNO3 (aq) + NaCl (aq) NaNO3 (aq) + AgCl (s)

• vol A mol A mol B g B

ElectrolytesElectrolyte

Substance that, when dissolved in water, produces an solution that conducts an electric current.

2 requirements

1-Charged particles2-Mobility

Ionic compounds and acids (only molecular electrolytes)

Electrolytes

Salts - made up of charged particlesNaCl (s) Na+ (aq) + Cl- (aq)

CaCl2 (s) Ca2+ (aq) + 2Cl- (aq)

Molecules are usually nonelectrolytesAcids = molecular electrolytes

HCl (g) H+ (aq) + Cl- (aq)

H2SO4 + (g) H+ (aq) + HSO4 - (aq)

Nonelectrolytes

Nonelectrolyte• compound that, when dissolved in water,

produces aq solution that does NOT conduct an electric current

• molecular compounds– C6H12O6 (s) C6H12O6 (aq)

no charged particles present

Strong versus Weak Electrolytes

• Strong electrolytes– Usually ionize completely or almost completely– Produce solutions that conduct a strong current

• Weak electrolytes– Ionize on the order of 1-10%– Produce solutions that conduct a weak current

Precipitation Reactions

• Why things dissolve• Precipitate = solid ionic compound formed

when water solutions of 2 diff ionic compounds are mixed

• Know rules fig 4.3 p 78• Know what ions combine to form a ppt

Net Ionic Equations

• NIE show – 1. Balance the formula equation– 2. Dissociate (pull apart) all (aq) compounds• Subscripts for R and P become coefficients• CaCl2(aq) Ca2+

(aq) + 2Cl-(aq)

– 3. Remove all spectator ions• spectator ions do nothing in the reaction and are excluded

from the equation

• NaOH(aq) + Cu(NO3)2(aq) --> NaNO3(aq) + Cu(OH)2(s)

First, balance the equation

2NaOH(aq) + Cu(NO3)2(aq) -->

2NaNO3(aq) + Cu(OH)2(s)

Write ionic eq (dissociate all aq compounds) and cancel the spectator ions

2Na+ + 2OH- + Cu2+ + 2NO3- -->

2Na+ + 2NO3- + Cu(OH)2(s)

Write NIE (write (+) ion first)CuCu2+2+

(aq)(aq) + 2OH+ 2OH--(aq)(aq) ---> ---> Cu Cu(OH)(OH)2(s)2(s)

This is all that really happened!!!This is all that really happened!!!

Net Ionic Equations

– BaCl2 and Na2SO4 are mixed. Write NIE for the reaction if a precipitate forms.• Ions present are Ba2+ Cl- Ag+ SO4

2-

– Only + and - ions join together, therefore– Possible ppts are (Ba2 + and SO4

2-) or (Na+ and Cl-)

» All chlorides are soluble (ex Ag+ and any Pb ion)» All sulfates are soluble (ex Ba2+)

– Barium and sulfate ions will form a ppt

Ba2+ + 2Cl- + 2Na+ + SO42- BaSO4 (s) + 2Na+ + 2Cl-

Ba+(aq) + SO42-(aq) BaSO4(s)

NIE from Reactants

• Write NIE for any reaction (or no rx) that occurs when (aq) solutions of

NaCl(aq) + AgNO3(aq) are mixedNa+ + Cl- + Ag+ + NO3

- --> • possible ppt = AgCl or NaNO3

– All nitrate are soluble – Most Ag compounds are insoluble therefore ppt is

– Ag+(aq) + Cl-(aq) AgCl(s)

Review

• Write NIE for any rx between aq sol’n of– AgNO3 (aq) + Ca(OH)2 (aq)

– AlCl3 (aq) + Pb(NO3) 2 (aq)

• How many grams of AgNO3(ag) are produced when 0.85L of 2.00 M HNO3 solution is added to 216 g of Ag according to the equation. Which reactant is the limiting reactant?

•

• 3 Ag(s) + 4 HNO3(aq) 3 AgNO3(aq) + NO(g) + 2 H2O(l)

Acids

• Commercially important• Contain H and a nonmetal or polyatomic ion• Taste sour – found in many foods • Rx with many other substances– w/ metals to produce H2 gas

– w/ carbonates to produce CO2 and a salt

• Molecular electrolytes– Produce H+ ions in water solution

• Neutralize bases

Bases

• Produce OH- in water solution • Feel slippery• Taste bitter• Commercially important• Present in many cleaning agents• Usually metal hydroxides, ammonia or

amines, and some anions• Neutralize acids

Acids – Proton donors

• Acid produce H+ ions– 2 types – strong acids and weak acids• Know examples in text pg 83• Strong acids – ionize (break apart) 100%

– HCl(aq) H+(aq) + Cl-(aq)– Acids lose only 1 H+ in aq sol’n– H2SO4(aq) H+(aq) + HSO4

- (aq)

• Weak acids - ionize only slightly – less than 5%– Use a 2 headed arrow to show both reactants and products

are present– HF(aq) H+(aq) + F-(aq)

Acids in Water

• Strong acids: Ionize 1 proton completelyinitial 1000 0 0

H2SO4(l) H+(aq) + HSO4

- (aq)

final 0 1000 1000

• Weak acids : Ionize 1 proton in eq system initial 1000 0 0 H2CO3 (aq) H+ (aq) + HCO3

- (aq)

final 995 5 5

Bases – Proton acceptor • Bases produce OH- ions in aq solution– Bases are proton grabbers

• Strong and weak bases (pg83)– Strong bases = grp 1 & 2 metal hydroxides – They dissociate all of their hydroxide ions

• NaOH (s) Na+ + OH-

• Ca(OH)2(s) Ca2+ + 2OH-

– Weak bases (H+ grabbers) do not contain OH- • They react with water (by grabbing a proton) leaving OH-

– H2O as a reactant and use means reversible rx

– NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

– F-(aq) + H2O(l) HF(aq) + OH-(aq)

Bases in water

• Strong Bases loses all their OH- to water • Ca(OH)2 (s) Ca2+

(aq) + 2OH- (aq)

• Weak bases break H2O apart and form OH- ions (grabs a proton)

• F-(aq) + H2O (l) HF (aq) + OH-

(aq)

Acids and Bases in H2O• Write NIE for reactions, in H2O, of– SA SB WA SB– SA WB WA WB

• Write SA & SB as the active ion–H+ or OH-

• Write WA & WB as the entire molecule–HF CH3COOH NH3 CH3NH2

• H+ ion jumps to the base

Acids & Bases in H2O

• SA= HCl (g) H+(aq) + Cl-

(aq) – Strong acids lose only 1 proton (i.e. H2SO4)

• WA= H3PO4 (g) H+(g) + H2PO4

-(g)

– Weak acid lose only 1 proton and set up an equilibrium system use a 2 headed arrow

• SB= Ca(OH)2 (s) Ca2+ (aq) + 2OH-

(aq) – All OH- ions are removed from the molecule

• WB= NH3(g) + H2O(l) NH4+

(aq) + OH-(aq)

– WB grab H+ from H2O to produce OH- ions

Know Rx for Strong and Weak Acids and Bases in Water

HCl(aq) H+ (aq) + Cl-

(aq)

H2CO3 (aq) H+ (aq) + HCO3-

(aq)

Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq)

NH3 (aq) + H2O (l) NH4+ (aq) + OH-

(aq)

Acids & Bases in Reactions

• Represent each of the following in reactions, or water solution:

– HNO3 H2SO4 HF HC2H3O2 H3PO4

– H+ H+ HF HC2H3O2 H3PO4

– KOH Ca(OH)2 NH3 F- CH3NH2

– OH- OH- NH3 F- CH3NH2

Acid Base Neutralization Reactions

• Acids and bases neutralize each other

• All H+ from the acid and all OH- from the base react to produce HOH (H2O)

• In NIE reactions: – Strong acids are represented by the H+ ion – Strong bases are represented by the OH- ion– Weak acid and bases are represented by the entire

molecule

Strong Acid + Strong Base

• Active ions are H+ and OH-

– They combine to form waterWrite the equation for the reaction between nitric

acid and potassium hydroxide– HNO3(aq) + KOH(aq) H2O(l) + KNO3(aq) • Break apart (aq) formulas and cancel spectators

H+ + NO3- + K+ + OH- H2O(l) + K+ + NO3

-

– H+ (aq) + OH-

(aq) H2O(l)

– In titrations use formula equations

Strong Acid Weak Base• Represent the strong acid as H+ and show the

entire molecular formula of the weak base– Combine the H+ with the formula for the base

• Write the eq for the reaction of nitric acid, HNO3,

and ammonia, NH3.– HNO3(aq) + NH3 (aq) NH4NO3

(aq)

• Break apart (aq) formulas (not the weak base) and cancel spectators• The weak base grabs H+ from the acid

– H+(aq) + NO3

-(aq) + NH3(aq) NH4

+(aq) + NO3

- (aq)

– H+ (aq) + NH3(aq) NH4

+(aq)

Weak Acid and Strong Base• Show the entire formula for the acid, and OH- for

the base. – The H+ from the acid and OH- from the base form water,

and the ion from the acid remains

• Write the equation for the reaction of carbonic acid and potassium hydroxide– H2CO3(aq) + 2KOH(aq) 2H2O(l) + K2CO3 (aq)

– H2CO3(aq) + 2K+(aq) + 2OH-

(aq) 2H2O(l) + 2K+(aq)

+ 2CO32-

(aq)

• Break apart (aq) species (not the WA) and cancel spectators– H2CO3(aq) + 2OH-

(aq) 2H2O(l) + CO32-

(aq)

A & B NIE Neutralization Equations

• SA SB Nitric acid + sodium hydroxide

–H+(aq) + OH-

(aq) H2O(l)

• SA WB Hydrochloric acid + ammonia

–H+(aq) + NH3(g) NH4

+(aq)

• WA SB Carbonic acid + potassium hydroxide

–H2CO3(aq) + OH-(aq) H2O(l) + HCO3-

(aq)

A & B equations

• Hydrobromic acid + potassium hydroxide

• Nitric acid + ammonia

• Phosphoric acid + lithium hydroxide

Acid Base Formula Eq• Neutralization Reactions are double

displacement reactions– HA + MOH HOH (water) + MA (a salt)– HA + NH3 NH4A (an ammonium salt)– HCl + LiOH – HCl + LiOH H2O + LiCl– H2CO3 + NH3 – H2CO3 + 2NH3 (NH4)2CO3

– HF + Al(OH)3 – 3HF + Al(OH)3 3H2O + AlF3

Acid Base Titration• Lab procedure used to determine the molarity of

a solution. You completely reacts 2 solution. – React a solution of known conc with a solution of

unknown conc– When equal moles of the 2 solutes have completely

reacted with one another, the reaction ends• End point (or equivalence point) of the titration is

determined by a color change in the solution• At the end point: moles solute A = moles solute B

Acid Base TitrationReact a sol'n of known M w/ a sol'n of unknown M– Solution of known conc is called standard sol’n– Determine the vol of each sol'n used in the titration

– you know the conc of one sol'n, but not the other– You’ll know the vol of both solutions

– At end point of the titration = number of moles of each solute have been reacted• mol soluteknwn = mol soluteunk

• You can calc the # mol of solute in the known solution from its vol and M – # mol soluteknwn = volknwn (L) x Mknwn = molesknwn = molunk

• M unk = mol unk/L unk sol'n

Acid Base Titration

• 1. Use L and M of the known (standard) solution to find the moles in the known solution

• 2. Use stoich factor to calc mol of unknown in it’s solution

• 3. Find M of the unknown • M = mol unk

• L unk

Beyond A/B Titration MathWrite a balanced formulaformula equation (NOT NIE)

- 1 - Calc moles of the known substance- 2 - Convert to moles of the unknown with the stoich factor- 3 - Divide by the L of the unknown to produce Molarity

2. Calc # moles of solute in the known sol'nmolkwn = Mkwn x vol (L) kwn

3. Calc the # moles of solute in the unk sol'nmolunkwn= molkwn x molunkwn (the stoich factor,

molkwn from the bal eq)

4. Divide the moles of solute in the unknown solution by the # of liters of solution used in the titration

Munkwn = molunkwn

Lunkwn

Titration Math a• A 15.0 mL sample of HCL is titrated to the eq pt with 25.0 mL of

0.500 M NaOH. Calc the M of the acid.

1. HCl(aq) + NaOH(aq) H2O(l)The stoich factor is 1:1

2. nNaOH = 0.025 LNaOH x 0.500 molNaOH = 0.0125 molNaOH

1 LNaOH

3. Use the stoich factor to deter moles H+

nHCl = 0.0125 molNaOH x 1 molHCl = 0.0125 molHCl 1 molNaOH

4. MH+ = 0.0125 molHCl = 0.833 M HCl

0.0150 LHCl

Titration Math b• A 15.0 mL sample of oxalic acid, H2C2O4 is titrated to the eq pt

with 25.0mL of 0.500 M NaOH. Calc the M of the acid.

1. H2C2O4 + 2NaOH 2H2O + Na2C2O4

» Stoic factor is 1 mol H2C2O4 = 2 mol NaOH

2. nOH- = 0.025 LNaOH x 0.500 molNaOH = 0.0125 molNaOH 1 L OH-

3. nox = 0.0125 molNaOH x 1 molox = 0.00625 molox

2 molNaOH

4. MH2C2O4 = 0.00625 molox = 0.417 M Ox acid

0.0150 Lox

Acids and Bases

• Definitions– Strong vs weak

• Equations of acids and bases in water– SA, WA, SB, WB

• Acid/Base neutralization reactions– SA/SB, WA/SB, SA/WB

• A/B Titration math

Oxidation Reduction Reactions

• Involve a transfer of e-– One reactant loses e-s and is oxidized– One reactant gains e-s and is reduced

• Oxidation and reduction occur together– e-s (charge) and atoms are conserved

• Oxidizing agent = subst reduced• Reducing agent = subst oxidized– Cu2+(aq) + Zn0(s) Cu0(s) + Zn2+(aq)

– Zn0(s) + 2H+(aq) Zn2+(aq) + + H2(g)

Oxidation Numbers• Way to keep track of e-s in redox rx• In molecules and PAI the oxid # are = to

number of e-s shared between atoms• Know rules for assigning oxid #s p89• Oxidation = increase in oxid #– Loss of electron(s)

• Reduction = decrease in oxid #– Gain of electron(s)

Half reactions• A redox rx can be divided into an oxidation and

a reduction ½ reactions– Cu2+(aq) + Zn0(s) Cu0(s) + Zn2+(aq)

– Oxidation ½ rx =• Zn0(s) Zn2+(aq) + 2 e-

– Oxidation number of Zn goes up– Zn is oxidized and is the reducing agent

» Zn losses e-s

– Reduction ½ rx = • Cu2+(aq) + 2e- Cu0(s)

– Oxidation number of Cu2+ goes down– Cu2+ is reduced and is the oxidizing agent

» Cu2+ gains f e-s

Balancing Redox Equations

• All redox equations must be balanced for both atoms and charge

• Know steps in the process • Balance all equation on the worksheet in

acidic and basic solutions• Redox reactions can be used for a titration

reaction– Follow steps for any titration

Balancing Redox Equations• Bal all atoms other than H and O

• Bal O using water

• Bal H using H+ ions

• Bal charges by placing e-’s on the side with more positive charges

– Charge, NOT oxidation number

• Bal both ½ rxs so e-s gained = e-s lost

• Combine ½ rx equations and cancel

• If basic, add OH- to each side to cancel out all H+ ions present – cancel H2O’s

Cr2O72- + NO2 Cr3+ + NO3

1-

Cr2O72- Cr3+ NO2 NO3

-

Cr2O72- 2Cr3+ NO2 NO3

-

Cr2O72- 2Cr3+ + 7H2O H20 + NO2 NO3

-

14H+ + Cr2O72- 2Cr3+ + 7H2O H20 + NO2 NO3

- + 2H+

14H+ + Cr2O7

2- + 6e- 2Cr3+ + 7H2O (x6) H20 + NO2 NO3- + 2H+

+ 1e-

14H+ + Cr2O72- + 6e- 2Cr3+ + 7H2O 6H20 + 6NO2 6NO3

- + 12H+ + 6e-

14H+ + Cr2O72- + 6H20 + 6NO2 2Cr3+ + 7H2O + 6NO3

- + 12H+

2H+ + Cr2O72- + 6NO2 2Cr3+ + H2O + 6NO3

-

Weak Acid + Weak Base

• Represent both acid and base as entire formula.• The acid loses 1 H+ to the base

• CH3COOH(l) + NH3 (g) NH4+

(aq) + CH3COO-(aq)

• H2CO3(aq) + CH3NH2(aq) HCO3-(aq) + CH3NH3

+(aq)