Transcript Slide 1

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
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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)
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0.50 M means 0.50 mol/1L
use M as conver. fact. to convert L
Convert moles
g if necessary
mol
3. Determine the vol of sol’n containing a given amt of
solute - in moles or grams
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•
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)?
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• 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
Electrolytes
Electrolyte
Substance that, when dissolved in water, produces an
solution that conducts an electric current.
2 requirements
1-Charged particles
2-Mobility
Ionic compounds and acids (only molecular
electrolytes)
Electrolytes
Salts - made up of charged particles
NaCl (s)  Na+ (aq) + Cl- (aq)
CaCl2 (s)  Ca2+ (aq) + 2Cl- (aq)
Molecules are usually nonelectrolytes
Acids = 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)
Cu2+(aq) + 2OH-(aq) ---> Cu(OH)2(s)
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+ SO42– Only + and - ions join together, therefore
– Possible ppts are (Ba2 + and SO42-) 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-
Ba+(aq) + SO42-(aq)
BaSO4 (s) + 2Na+ + 2Cl-
BaSO4(s)
NIE from Reactants
• Write NIE for any reaction (or no rx) that occurs
when (aq) solutions of
NaCl(aq) + AgNO3(aq) are mixed
Na+ + 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?
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• 3 Ag(s) + 4 HNO3(aq) 3 AgNO3(aq) + NO(g) + 2 H2O(l)
Acids
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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
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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 completely
1000
H2SO4(l)
final
0
initial
0
0
H+(aq) + HSO4- (aq)
1000
1000
• Weak acids : Ionize 1 proton in eq system
1000
H2CO3 (aq)
final 995
initial
0
0
H+ (aq) + HCO3- (aq)
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)
• Ca(OH)2(s)
Na+ + OHCa2+ + 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
– NH3(aq) + H2O(l)
– F-(aq) + H2O(l)
means reversible rx
NH4+(aq) + OH-(aq)
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 OHions (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
– SA WB
WA SB
WA WB
• Write SA & SB as the active ion
– H+ or OH-
• Write WA & WB as the entire molecule
– HF CH3COOH
NH3
• H+ ion jumps to the base
CH3NH2
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)
H2CO3 (aq)
H+ (aq) + Cl- (aq)
H+ (aq) + HCO3- (aq)
Ca(OH)2 (s)
NH3 (aq) + H2O (l)
Ca2+ (aq) + 2OH- (aq)
NH4+ (aq) + OH- (aq)
Acids & Bases in Reactions
• Represent each of the following in reactions,
or water solution:
– HNO3
– H+
– KOH
– OH-
H2SO4
H+
Ca(OH)2
OH-
HF
HC2H3O2
HF
HC2H3O2
NH3
NH3
FF-
H3PO4
H3PO4
CH3NH2
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 water
Write 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)
• SA WB
Hydrochloric acid + ammonia
– H+(aq) + NH3(g)
• WA SB
H2O(l)
NH4+(aq)
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 Titration
React 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 Math
Write a balanced formula 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'n
molkwn = Mkwn x vol (L) kwn
3. Calc the # moles of solute in the unk sol'n
molunkwn= 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
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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 OH3. 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)
– Oxidation ½ rx =
• Zn0(s)
Cu0(s) + Zn2+(aq)
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+ + NO31Cr2O72-  Cr3+
NO2  NO3-
Cr2O72-  2Cr3+
NO2  NO3-
Cr2O72-  2Cr3+ + 7H2O
14H+ + Cr2O72-  2Cr3+ + 7H2O
H20 + NO2  NO3H20 + NO2  NO3- + 2H+
14H+ + Cr2O72- + 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)
• H2CO3(aq) + CH3NH2(aq)
NH4+(aq) + CH3COO-(aq)
HCO3-(aq) + CH3NH3+(aq)