Chem12 Unit 1 Aqueous Solutions: Neutralization Reactions and Titrations
Notes:
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:
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 Bases:
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 + HClO4 
3) NaOH + H2SO4 
Answer to practice questions:
1) NaOH(aq) + HCl(aq)  NaCl(aq) + H20(l)
sodium hydroxide and hydrochloric acid produces sodium chloride and water
2) Ca(OH)2(aq) + 2 HClO4 (aq)  CaCl2(aq) + 2 H20(l)
calcium hydroxide and perchloric acid produces calcium chloride and water
3) 2 NaOH(aq) + H2SO4 (aq) Na2SO4(aq) + 2 H20(l)
Sodium hydroxide and sulfuric acid produces sodium sulfate and water
Titrations: Pre-Lab Preperation
Background:
A titration is when we use a neutralization reaction along with an acid base indicator to find an
unknown concentration of an acid or a base. This is done using the following equation:
ACBVB = BCAVA
A = coefficient of acid from balanced equation
B = coefficient of base from balanced equation
CB = Concentration of base
VB = Volume of base
CA = Concentration of acid
VA =Volume of acid
Example #1
If 52.7 ml of 0.1 mol/L NaOH are required to completely neutralize 25.0 ml of HCl. What is the
concentration of the HCl?
NaOH(aq) + HCl(aq)  NaCl(aq) + H20(l)
ACBVB = BCAVA
(0.1mol/L)(0.0527 L) = CA(0.0250 L)
CA = 0.2 mol/L (rounded to 1 sig dig)
(Note: There are no coefficients in the balanced equation, so we can omit “A” and “B”.)
Example #2
If 42.5 ml of 0.1 mol/L Ca(OH)2 are required to completely neutralize 25.0 ml of HClO4. What is
the concentration of HClO4?
Ca(OH)2(aq) + 2 HClO4 (aq)  CaCl2(aq) + 2 H20(l)
ACBVB = BCAVA
(2)(0.1 mol/L)(0.0425 L) = (1)(CA)(0.0250 L)
CA = 0.3 mol/L (rounded to 1 sig dig)
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.
Also: Complete the “Titrations Practice Questions” worksheet.