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Binary acid (i.e. H & a non-metal)◦ the prefix hydro is used◦ the root of the anion is used◦ the suffix -ic is used◦ the word acid is used as the second word of the name
Example◦ HCl = hydrochloric acid◦ HBr = hydrobromic acid◦ HI = hydroiodic acid◦ HF = hydrofluoric acid
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Polyatomic acids◦ 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 (nitrate) (-ate -ic)
◦ HSO3 (sulfite) (-ite -ous)
◦ H2SO4 (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)
The six strong acids to be memorized◦ 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
Examples◦ 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
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 hydroxideH2SO4(aq) + KOH(aq) → K2SO4 + HOH
2 H+(aq)+ SO4
2-(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)
(aq) (l)2 2
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Recall:◦ Concentration is calculated as moles per litre
mol/L M
◦ [NaOH] refers to the concentration of sodium hydroxide 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) + HCl(aq) → H2O(l) + NaCl(aq)
Mm
mol
V
m
[ ]
0.0350L 0.05008L
0.409M
0.0205mol0.0205mol
0.586M
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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) + 2 HClO4(aq) → BaCl2(aq) + 2H2O(l)
Mm
mol
V
m
[ ] 0.256M 0.407M
0.0460L
0.0187mol0.00935mol
0.0365L
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P. 614◦ #’s 30-31
P. 616◦ #’s 32-33, 36-37
Aqueous Reactions Worksheet #1◦ #’s 5-7
Aqueous Reactions Worksheet #2