Transcript Slide 1

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Binary acid (i.e. H & a non-metal)
◦
◦
◦
◦
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the
the
the
the
prefix hydro is used
root of the anion is used
suffix -ic is used
word acid is used as the second word of the name
Example
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◦
HCl = hydrochloric acid
HBr = hydrobromic acid
HI = hydroiodic acid
HF = hydrofluoric acid
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Polyatomic acids
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oxyacids: (acids with oxygen in the polyatomic anion)
change suffix of –ate with -ic
OR
change suffix -ite to -ous
These acids have the general formula HaXbOc where X = an
element other than Hydrogen or Oxygen
Examples
◦ HNO3
◦ HSO3
◦ H2SO4
(nitrate) (-ate  -ic)
(sulfite) (-ite  -ous)
(sulfate) (-ate  -ic)
Nitric acid
Sulfurous acid
Sulfuric acid
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Strong Acid:
◦ an acid that completely dissociates into ions.
◦ (100 molecules of HCl → 100 H+ ions)
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The six strong acids to be memorized
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HCl
HBr
HI
H2SO4
HNO3
HClO4
(Perchloric acid)
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Strong Bases
◦ a base that completely dissociates into ions.
◦ (100 formula units of NaOH → 100 OH- ions)
◦ When combined with the OH_ (hydroxide) ion, elements
found in group 1 (IA) and 2 (IIA) form strong bases
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Examples
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KOH
CsOH
Ba(OH)2
Ca(OH)2
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H+ (Hydrogen ion) indicates strong acid
◦ pH scale with a value of 2 or less
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OH- (Hydroxide ion) indicates strong base
◦ pH scale with a value of 12 or more
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Read p.271-273 Questions: P.273 #’s 29-33
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Current Assumption
◦ When strong acid and strong base are combined, all H+
and OH- ions join to form HOH (H2O)
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Neutralization reaction is a double displacement
Example
◦ For NaOH and HCl
 Predict the products of the reaction
 balance the equation
◦ NaOH(aq) + HCl(aq)  NaCl + HOH
◦ Use solubility rules to confirm whether each product will
be aqueous, solid or liquid
◦ NaOH(aq) + HCl(aq)  NaCl(aq) + HOH(l)
◦ Write the total ionic equation, showing all ions that are in
solution
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Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
Cancel the spectator ions and write the net ionic equation
OH-(aq) + H+(aq) → H2O(l)
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Example
◦ Write the molecular, ionic, net ionic equations for
 Sulfuric acid & potassium hydroxide
H2SO4(aq) + 2 KOH(aq)
→ K2SO4 (aq) + 2 HOH(l)
2 H+(aq)+ SO42-(aq)+ 2 K+(aq) + 2 OH-(aq) → 2 K+(aq) + SO42-(aq) +2 HOH(l)
2 H+(aq) + 2 OH-(aq) → + 2 HOH(l)
Simplify
H+(aq) + OH-(aq) → + HOH(l)
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Recall:
◦ Concentration is calculated as moles per litre
 mol/L
 M
◦ [NaOH] refers to the concentration of sodium hydroxide
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Equation
◦ [ ] =mol/L
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In the reaction of 35.0 mL of liquid drain cleaner containing NaOH,
50.08 mL of 0.409 mol/L HCl must be added to neutralize the base.
What is the concentration of the base in the cleaner?
◦ Write a balanced equation and the chart
NaOH(aq)
mol
0.0205mol
+
HCl(aq)
0.0205mol
Mm
m
V
[]
0.0350L
0.586M
0.05008L
0.409M
→
H2O(l)
+
NaCl(aq)
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Calculate the volume of 0.256 mol/L Ba(OH)2 that must be added to
neutralize 46.0 mL of 0.407 mol/L HClO4.
Ba(OH)2(aq)
mol
0.00935mol
+
2 HClO4(aq)
0.0187mol
Mm
m
V
0.0365L
0.0460L
[]
0.256M
0.407M
→
BaCl2(aq)
+
2H2O(l)
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P. 614
◦ #’s 30-31
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P. 616
◦ #’s 32-33, 36-37
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Aqueous Reactions Worksheet #1
◦ #’s 5-7
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Aqueous Reactions Worksheet #2