Transcript Intermolecular Forces

Chapter 14
14.1 Properties of Acids and Bases
Acids
– electrolytes – ionize to form H+ or H3O+(H+•
H2O) and anion
– taste sour – citric acid, carbonic acid, lactic
acid
– react w/ metals – form hydrogen gas and salt
– affect indicators – litmus = red,
phenolphthalein = clear
– highly corrosive – wear away by chemical
action
– neutralize bases  water and salt
• naming acids
– binary acids – hydrogen bonded to an anion
without oxygen
– molecular/un-ionized name
– HCl
•
•
•
•
start with hydrogen
use root of anion name
end in –ide
hydrogen chloride
– H2S
• hydrogen sulfide
– HCN
• hydrogen cyanide
• naming acids – binary cont.
– ionized or acid name
– HCl
•
•
•
•
start with hydrouse root of anion name
end in –ic acid
hydrochloric acid
– H2S
• hydrosulfuric acid
– HCN
• hydrocyanic acid
– oxyacids – H bonded to some anion
containing oxygen
– molecular/un-ionized name
– HNO3
• name hydrogen
• name anion
• hydrogen nitrate
– H2SO3
• hydrogen sulfite
– HClO4
• hydrogen perchlorate
– oxyacids cont.
– ionized/acid name
– HNO3
•
•
•
•
name anion
drop –ate and add –ic acid or
drop –ite and add –ous acid
nitrate  nitric acid
– H2SO3
• sulfite  sulfurous acid
– HClO4
• perchlorate  perchloric acid
• strong acids
– ionize completely, independent of
concentration
– HNO3 - nitric acid
HCl - hydrochloric acid
H2SO4 - sulfuric acid
HClO4 - perchloric acid
HBr - hydrobromic acid
HI - hydroiodic acid
100 HNO3  100 H+ + 100 NO3-
• weak acids
– partially ionize
– small percentage actually breaks apart,
independent of concentration
100 HC2H3O2  1 H+ + 1 C2H3O2- + 99 HC2H3O2
– properties not as strong
bases
– electrolytes – dissociate to form OH- and cation
– taste bitter – baking chocolate, soap
– slippery to touch – OH- react with oils in skin to
form soap
– affect indicators – litmus = blue,
phenolphthalein = pink/red (list indicators)
– neutralize acids  water + salt
– caustic - tending to eat away or consume,
especially to skin and tissue
• drain cleaners
• lime used on dead animals
– solubility varies
• slightly soluble/insoluble – Mg(OH)2
• very soluble – alkali
– hygroscopic – absorb water from air
– KOH
• alkaline/basic = OH- rich solutions
– neutralize acids – form water and salt
• strong bases
– totally dissociate to ions
– LiOH - lithium hydroxide
NaOH - sodium hydroxide
KOH - potassium hydroxide
RbOH - rubidium hydroxide
CsOH - cesium hydroxide
Mg(OH)2 - magnesium hydroxide
Ca(OH)2 - calcium hydroxide
Sr(OH)2 - strontium hydroxide
Ba(OH)2 - barium hydroxide
100 NaOH  100 Na+ + 100 OH-
• weak bases
– partially dissociate
– few OH100 NH3 + 2 H2O  2 NH4+ + 2 OH- + 98 NH3
Arrhenius acid
• any substance that produces H+ or H3O+
Arrhenius base
• any substance that produces OH-
– limited to aqueous solutions
HCl + H2O  H3O+ + ClHCl + NH3  NH4+ + Cl• acid in one case, not the other
• Bronsted-Lowry acid
– p+ donor
• H+ = proton
HCl + H2O  H3O+ + Cl– HCl donates a p+ to H2O
HCl + NH3  NH4+ + Cl– HCl donates a p+ to NH3
• Bronsted-Lowry base
– p+ acceptor
• H+ = proton
NH3 + H2O  NH4+ + OH– NH3 accepts a proton from H2O
HCl + OH-  H2O + Cl– HCl donates a p+ to OH-
• conjugate acid
– substance formed when a base accepts a p+
• conjugate base
– substance formed when an acid donates a p+
HCl + H2O  H3O+ + Cl-
• give the conjugate base for the following:
– HF
F-
– HNO3
NO3 -
– H2SO4
HSO4-
– H3O+
H2O
• give the conjugate acid for the following:
– OH-
H2 O
– NO3-
HNO3
– NH3
NH4+
– Cl-
HCl
• strength of conjugate acids/bases
– strong acid  weak conjugate base
– strong base  weak conjugate acid
HCl + H2O  H3O+ + Clstrong
acid
CH3COOH + H2O 
acid
weak
acid
strong
base
base
weak
base
H3O+ + CH3COOacid
base
page 485
acid
weak
weak
base

CH3COOH + H2O

H3O+ + CH3COOacid
strong
strong
base
• amphoteric
– substance that can be either an acid or a
base
HCl + H2O  H3O+ + Cl– water is a base
NH3 + H2O  NH4+ + OH– water is an acid
– when mixed with a stronger acid, water is a
base
– when mixed with a stronger base, water is an
acid
15.1 Aqueous solutions and the concept of pH
H2O + H2O
H3O+ + OH• autoprotolysis
• self ionization
• water pulls water apart and forms hydronium ions
and hydroxide ions
• not very common
• 2 water molecules/billion
• determines acidity and alkalinity of aqueous solutions
• concentration of H3O+ = 1 x 10-7 mol/L
[H3O+] = 1 x 10-7
• concentration of OH- = 1 x 10-7 mol/L
[OH-] = 1 x 10-7
• law of mass action
– describes the equilibrium conditions for a
reaction
– defined as a reaction constant or K
K = [prod]/[react]
for autoprotolysis,
H 2O + H 2 O
Kw = [H3O+][OH-]
[H2O][H2O]
H3O+ + OH-
– since [H2O] remains constant, [H2O] = 1
so Kw = [H3O+][OH-]
• in pure water [H3O+] = [OH-] = 1 x 10-7
Kw = [1 x 10-7 ][1 x 10-7 ]
Kw = 1 x 10-14
• constant for all water solutions
• because Kw = 1 x 10-14 ,
– if [H+] > 1 x 10-7 then [OH-] < 1 x 10-7
– if [OH-] > 1 x 10-7 then [H+] < 1 x 10-7
– if [H+] > 1 x 10-7 = acid
– if [OH-] > 1 x 10-7 = base
– if [H+] = 3.476 x 10-8  acid/base??
– if [OH-] = 9.75 x 10-7  acid/base??
– chemists devised an easier method
• pH – power of hydrogen or potentia
hydrogenii
– simplified method of determining acidity or
alkalinity
– pH scale - 0 to 14
• pH = - log[H+]
• if [H+] = 3.476 x 10-8  acid/base??
pH = -log[3.476 x 10-8]
pH = 7.459
• base
• [H+] = 1.75 x 10-3 , pH???
pH = -log[1.75 x 10-3]
pH = 2.76
• acid
• if [OH-] = 9.75 x 10-7  acid/base??
• pH = - log[H+]
– not [OH-]
Kw = [H+][OH-] = 1 x 10-14
Kw = [H+][OH-] = 1 x 10-14
[OH-]
[H+][OH-] = 1 x 10-14
[OH-] = 9.75 x 10-7
[H+ ] = 1.03 x 10-8
pH = -log[1.03 x 10-8]
pH = 7.99 base
= [H+]
• pOH = -log[OH-]
• pH + pOH = 14
• if [OH-] = 9.75 x 10-7  acid/base??
pOH = -log[9.75 x 10-7]
pOH = 6.01
pH = 14 – pOH
pH = 14 – 6.01
pH = 7.99
• What is the pH of a solution of the strong
acid HNO3 if it has a concentration of
0.375 M?
– strong acid = total ionization
HNO3  H+ +
0.375 mol/L 0.375 mol/L
[H+] = 0.375
pH = -log[0.375]
pH = 0.426
NO30.375 mol/L
• What is the pH of the strong base KOH if
the concentration is 0.0750 M?
• KOH 
K+ +
OH0.0750mol/L 0.0750 mol/L
0.0750 mol/L
[OH-] = 0.0750
Kw = [H+][OH-] = 1 x 10-14
[OH-] = 0.0750
[H+] = 1.33 x 10-13
pH = -log[1.33 x 10-13]
pH = 12.9
• what is the pH of the strong base of KOH if
the concentration is 0.0750M?
0.0750M KOH = 0.0750 M OHpOH = -log[OH-] or –log[0.0750]
pOH = 1.12
pH = 14-1.12
pH = 12.9
• What is [H+] of a solution with a pH of
9.47?
[H+] = inv log(-pH)
[H+] = inv log(-9.47)
[H+] = 3.39 x 10-10
15.2 Titrations
• neutralization Rx
– chem. Rx in which H+ and OH- react to form
water
– salt is prod. of cation(from base) and
anion(from acid)
ACID + BASE  SALT + WATER
HCl + NaOH  NaCl + H2O
strong
strong
neutral
neutral
• equivalence point
– point at which neutralization is complete
• titration
– controlled addition of a solution of known
conc. to a solution of unknown conc.
– concludes at equivalence point
– titrant - solution of known concentration used
to titrate
– analyte - solution of unknown concentration
titration video
• titration calculations
1) determine vol. of titrant and analyte used
•
25.5 mL of 0.750 M-HNO3 , 17.5 mL of ?? MKOH
2) calculate mol of titrant
mole = M x L
X mole HNO3 = 0.750 mol/L x 0.0255 L
X mole HNO3 = 0.0191 mol HNO3
3) write balanced neutralization equation
HNO3 + KOH  KNO3 + H2O
4) calculate mol of analyte neutralized
X mol KOH = 1 mol KOH x 0.0191 mol HNO3
1 mol HNO3
X mol KOH = 0.0191 mol KOH
5) calculate concentration of analyte
X mol KOH = 0.0191 mol KOH
L sol
0.0175 L sol
X M-KOH = 1.09 M
• What is the concentration of sulfuric acid if 75.0 mL of
0.375 M sodium hydroxide neutralized 47.6 mL of
acid?
1) 75.0 mL of 0.375 M-NaOH, 47.6 mL of ??M-H2SO4
2) X mol NaOH = 0.375 mol NaOH/L x 0.0750 L
X mol NaOH = 0.0281 mol NaOH
3) 2 NaOH + H2SO4  Na2SO4 + 2 H2O
4) X mol H2SO4 = 1 mol H2SO4 x 0.0281 mol NaOH
2 mol NaOH
X mol H2SO4 = 0.0141 mol H2SO4
5) X mol H2SO4 = 0.0141 mol H2SO4
L solution 0.0476 L solution
X M-H2SO4 = 0.296 M-H2SO4