Chem12 Unit 1 Aqueous Solutions: Neutralization Reactions and Titrations
1. Naming Acids
a) To name a binary acid, follow the steps:
i) The prefix “hydro” is used
ii) The root of the anion (negative ion) is used
iii) The suffix “ic” is used
iv) The word “acid” is used as the second word in the name
(a) e.g. HCl is hydrochloric acid
(b) e.g. HBr is hydrobromic acid
b) Naming polyatomic acids:
i) Many oxygen-rich polyatomic negative ions form acids that are named
by replacing the suffix –ate with –ic and the suffix –ite with –ous
ii) To name oxyacids (acids containing the element oxygen): Recognize
them by the general formula: HaXbOc
(a) X = an element other than hydrogen or oxygen
(b) If enough H+ ions are added to a (root)-ate polyatomic ion to
completely neutralize its charge, the (root)-ic acid is formed
iii) E.g. :
(a) If one H+ ion Is added to nitrate, NO3-, HNO3 is formed. This is
named nitric acid.
(b) If two H+ ions are added to sulfate, SO42-, H2SO4 is formed. This is
named sulfuric acid.
(Referenced: http://www.mpcfaculty.net/mark_bishop/acid_nomenclature_help.htm)
c) Strong Acids:
i) Defined as an acid that completely dissociates into ions
(a) ie. 100 molecules of HCl dissolved in water produces 100 ions of
H+ and 100 units of Cl(b) There are only 6 strong acids: (memorize these)
(i)
HCl = hydrochloric acid
(ii)
HBr = hydrobromic acid
(iii)
HI = hydroiodic acid
(iv)
H2SO4 = sulfuric acid
(v)
HNO3 = nitric acid
(vi)
HClO4 = perchloric acid
2. Naming Bases
a) To name a strong base:
i) Combine name of metal with the hydroxide ion, OHii) E.g. NaOH is sodium hydroxide
b) Strong Base:
i.) defined as a base that completely dissociates into ions
(a) i.e. 100 molecules of NaOH dissolved in water produces 100 ions
of Na+ and 100 units of OH(b) include any ionic compound that contains the hydroxide (OH-) ion
(c) When combined with the (OH-) ion, elements found in groups 1
(IA) and 2 (IIA) form strong bases.
3. Combining Strong Acids with Strong Bases:
When a strong acid and a strong base are combined, they react
“completely”. This means that all of the hydrogen ions (from the acid)
and all of the hydroxide ions (from the base) will react to form water.
(These are the only acids and bases that we will be dealing with in this
unit. We will come back to weak acids and weak bases in the acids and
bases unit later on.)
4. Writing a Balanced Equation for a Neutralization Reaction between
a Strong Acid and a Strong Base
1) Predict the products of the reaction and ensure that the equation is
balanced
2) Use the solubility rules to confirm whether each product will be (aq),
(s), or (l) (Note that water is a liquid since the aqueous solutions are
dissolved in water)
5. Practice Questions
Name each acid and base according to the naming rules previously
discussed, then predict the products, use the solubility rules table to
predict (aq), (s), or (l) , and balance the equation.
1) NaOH + HCl 
2) Ca(OH)2 + HCl 
3) NaOH + HNO3 
Titrations: Pre-Lab Preperation
Scenario:
Suppose you performed a titration using Ca(OH)2 and HCl. You know that the
concentration of HCl is 0.1 mol/L, but are trying to figure out the concentration of
the Ca(OH)2. You measure out 50 ml of Ca(OH)2 and place it in a beaker and dilute
it with 100 ml of water. You then put your 0.1 mol/L HCl into your burette and
run three trials to get a good estimate of the equivalence point of the titration.
Definitions:
The end point of a titration is when you observe a change in your indicator.
The equivalence point of a titration is when the pH reaches exactly 7.
These are the results of your titration:
Trial 1: 26.1 ml of HCl is used to reach the endpoint.
Trial 2: 26.3 ml of HCl is used to reach the endpoint.
Trial 3: 25.9 ml of HCl is used to reach the endpoint.
Questions:
1. What is the average volume of HCl used? This is your estimated
equivalence point.
2. How many moles of HCl are used?
3. How many moles of Ca(OH)2 were there in your original sample? (Hint: use
a balanced equation and a mole ratio!)
4. What is concentration of the original Ca(OH)2?
5. Did adding 100 ml of water to your original 50 ml of Ca(OH)2 have an effect
on the results of the concentration you found in question #4? Explain why
or why not.