Q-Acids and Bases
Prem Sattsangi
Copyright 2007
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Acids
Bases
Phosphoric(tri-basic)
H3PO4
Sulfuric (di-basic)
H2SO4
Nitric
(mono-basic)
HNO3
Acetic
HC2H3O2
Chloric
HClO3
Hydrochloric
HCl
• Ammonia/ammonium
hydroxide
• NH3/NH4OH
• Sodium hydroxide
• NaOH
• Calcium hydroxide
Ca(OH)2
• Aluminum hydroxide
• Al (OH)3
• Sodium carbonate
• Na2CO3
Definitions
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Arrhenius (1880s)
Acid: Produce H+ ions in solutions.
Base: Produce OH- ions in solutions.
Lowry Bronsted (1932)
Acid: Proton donor
Base: Proton acceptor
Lewis (1915)
Acid: Electron pair acceptor
Base: Electron pair donor
Commercial products
• Acidic
• Cola (Phosphoric acid)
• Jams and jellies (malic
and citric acid)
• Toilet bowl cleansers
and lime deposit
removers.
• Basic
• Detergents (sodium
polyphosphate,
Anionic surfactants)
• Drain cleaners, oven
cleaners (NaOH)
• Window, floor cleaners
(ammonia)
• Antacids (aluminum
hydroxide, magnesium
hydroxide, calcium
carbonate)
Neutralization reactions
Net ionic equation:
NaOH + HCl

H+ (aq) + OH- (aq)  H2O(l)
NaCl + H2O
CaCO3 + 2HCl 
CaCl2 + H2O + CO2
2NaOH + H2SO4 
Na2SO4 + 2H2O
CaCO3 + H2SO4 
CaSO4 + H2O + CO2
3NaOH +H3PO4 
Na3PO4 + 3H2O
3CaCO3+2H3PO4 
Ca3(PO4)2+3H2O + 3 CO2
Dilution of sulfuric acid Chemtrek, p.8-11
• Heat is evolved when a concentrated solution is
diluted. Explain.
• Cations and anions are farther separated in a
dilute solution.
• Which will have a higher heat of dilution?
(a) H2SO4 ~98% or (b) HCl ~38%
• (a) H2SO4 ~98%
• Explain.
• Amount of heat generated is directly
proportional to the amount of acid present.
Dilution of sulfuric acid Chemtrek, p.8-11
• Proper procedure for dilution is:
(a) Pour Concentrated acid into water.
(b) Pour water into concentrated acid.
• (a) Acid into water.
• Write equation for dilution of Conc. H2SO4
• H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) Step-1
HSO4-(aq) + H2O(l)  H3O+(aq) + SO42-(aq) Step-2
Handling of bases
• NH3(g) has a pungent JOVIAN odor which
quickly attacks eyes and lungs.
• Concentrated Ammonia (~30%) solution is
opened, after cooling in ice, inside a hood.
• NaOH is kept closed tight. Explain
• 1. NaOH(s) is hygroscopic.
• Give a reason for packaging NaOH in plastic
bottles as opposed to glass.
• NaOH reacts with glass to form
sodium silicate Na4SiO4(aq).
• NaOH also reacts with atmospheric CO2(g).
• Complete the equation:
NaOH(s) + CO2(g) 
NaHCO3
Questions p. 8-15 Chemtrek
• Q.8 For HCl(aq) and NaOH(aq) titration, write:
• The over-all equation: (Answer in slide 5)
• The net ionic equation:
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Q.10 Calculate MU.K.
Data: U.K. HCl 2 drs = 0.100 M NaOH 3 drs
MHCl x 2 DrsHCl = 0.100 MNaOH x 3DrsNaOH
M of U.K. HCl = 0.100M x 3Drs/2Drs
= 0.150M
• Q.11 Why is it unnecessary to use calibrated
micropipet in this section.
• Same micropipet. Same size drops. “Drs”
cancel out in calculation.
Questions p.8-18 Chemtrek
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Define, Standard solution.
A solution of known concentration.
Q. 12 What is meant by “Standardize” a NaOH solution?
Use a solution of known concentration to titrate and
find the Exact M of the NaOH solution.
• Q.15 What is the advantage of back titration in egg
shell titration?
• Reaction CaCO3 + 2HCl  CaCl2 + CO2 + H2O is very
slow at 25oC. It is markedly faster to dissolve the
CaCO3 in excess of HCl at higher temperature and then
titrate the un-reacted HCl(aq), as in back titration.
Prequiz - 1
• Q3. NaOH 2.5 mL, 1.00 M = ? NaOH (g)
• Strategy: n = M x V mol  g
• = 1.00 mol x (2.5 x 10-3L) x 40.0g
1L
1 mol NaOH
= 0.100 g NaOH
Prequiz - 2
Facts:
1. Egg shell, (a) Provides protective covering,
(b) is a source of Calcium, and
(c) is a respiratory membrane for the embryo.
2. Average weight of a chicken egg: 5 g (Ca = 2 g)
3. Shell formation takes 15 hrs.
4. Ca deposits at 133mg/hr (2000 mg/15 hrs)
5. Total Ca content of a chicken blood is 25 mg
6. Every 11 minutes, total Ca in blood is used up.
7. Deficit of “Ca” is made up by breakdown of
marrow bone growths. The released phosphate is
excreted in urine.
8. What disrupts these biochemical processes?
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Chlorinated hydrocarbons, e.g. PBB, PCB, DDT,
Dieldrin
%Ca in egg shell-1
Data:
MP calibration: 1 dr = 0.036 g or mL,
1g = 1mL (Density of water)
egg shell
= 0.0705 g,
• HCl 2.00M, 20 drs., needed NaOH 1.05 M, 11 drs,
• Moles of HCl added:
• 20 drs  mL  L x 2.00M = 1.44 x 10-3 mol HCl.
20drs x 0.036mL
1L
1dr
1000mL
2.00mL = 1.44x10-3mol HCl
L
%Ca in egg shell-1
• Moles of NaOH required to titrate unreacted HCl:
• 11 drs  mL  x 1.05M = 4.16 x 10-4 mol NaOH
• 4.16 x 10-4 mol NaOH = 4.16 x 10-4 mol HCl
• Moles of HCl that reacted:
• = Moles HCl added – moles of unreacted HCl
• = 1.44 x 10-3 - 4.16 x 10-4 = 1.02 x 10-3 mol
%Ca in egg shell-2
Previously calculated:
Moles of HCl that reacted with CaCO3 = 1.02x10-3mol
[CaCO3 + 2HCl  CaCl2 + H2O + CO2]
• CaCO3(g):
• =1.02x10-3molHCl x 1 mol CaCO3x 100 g CaCO3
2mol HCl
1 mol CaCO3
• = 5.12x10-2g CaCO3
%Ca in Egg shell: = 5.12x10-2g CaCO3 x 100
0.0705 g egg shell
=72.6%