Good mornin’
 Lot’s of things to pick up today:
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Quiz Scantron
Quiz version that you took
Lab Sheet
Final Study Guide
Dry Erase Board
Marker
 10 minutes to look over missed quiz questions &
answer the Catalyst
pH
Test Tube
•
•
•
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[H+] (mol/L)
pH
1
0.01
<4
2
0.001
<4
3
0.0001
4
4
0.00001
5
5
0.000001
6
How does the molarity of H+ change?
How does the pH change?
Why are the pH’s for test tubes 1 and 2 indicated as “<4”?
What is an Arrhenius acid? What is an Arrhenius base?
What is a Bronsted-Loewry acid? What is a Bronsted-Loewry base?
Example Brønsted
Acids and Bases:
NH3 + HOH 
+
NH4
+
OH
Here, H2O acts as a Brønsted acid by
donating a proton to NH3 which acts as a
Brønsted base.
Conjugate base and conjugate acid?
pH
 pH is a scale used to reflect the concentration of H+ ions!
 Acids and bases are aqueous solutions
 Water can act as an acid and a base:
HOH <--> H+ + OH-
Why is water neutral?
 Acids dissociate and increase the number of protons (H+)
in solution
 Bases dissociate and increase the number of hydroxide
(OH-) in solution either directly (Arrhenius) or indirectly
(Bronsted-Loewry)
Neutralization Reaction
 Remember that this is just an acid base reaction
 What would a product of a neutralization reaction be?
 Just like the stoichiometry you’ve been doing, in lab we
can use this fact to calculate the molarity of solution
whose concentration is unknown
Titration
standard solution
 Definition
◦ Analytical method in
which a standard
solution is used to
determine the
concentration of an
unknown solution.
unknown solution
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Buret
stopcock
Erlenmeyer flask
Titration Vocabulary
 Titrant
◦ The substance added to the analyte in a
titration (a standard solution)
 Analyte
◦ The substance being analyzed
 Equivalence point
◦ The point in a titration at which the quantity
of titrant is exactly sufficient for
stoichiometric reaction with the analyte.
If the concentration of the titrant is
Acid-Base
Titration
known, then
the unknown
concentration
of the analyte can be determined.
Titrant
Analyte
Buret
Reading
Why do chemists use titrations??
Quantitative analysis —
used to determine the
amounts or concentrations
of substances present in a
sample by using a
combination of chemical
reactions and
stoichiometric calculations
Acidic, basic, or neutral??
 The “perfect pink”
for a titration with
phenolphthalein
Indicator changes color to
indicate pH
change
pH
Example…
7
phenolphthalein
is colorless in
acid and pink in
basic solution
point at
which
exactly
enough
reactant
pink has been
added for
Endpoint =
the
solution to
be
neutralized
and no
Volume base added more
Equivalence point (endpoint)
 Point at which equal
amounts of H3O+ and
OH- are present in
solution.
 Determined by…
 indicator color change
 dramatic change in pH
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Titration
moles H3
+
O
= moles
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
OH
White Board Review!
 Solubility
 Molarity
 Dilutions
 Solutions Stoichiometry
 Acid and Base
Problem #1
 What mass of (NH4)2SO4 is required to make 1.25 L of
a 0.250 M solution of NH4+?
 Answer: 20.6 g (mm=132.14g/mol)
Problem #2
If 25 g of KCl is added
to 50 g of water at
40°C, the solution
would be:
1) Unsaturated
2) Saturated
3) Supersaturated
Problem #3
 Calculate the molarity of a solution prepared by
dissolving 4.1 g of solid KBr in enough water to make
1.10 L of solution
 Answer: 0.031 M (mm=119.02 g/mol)
Problem #4
One way to determine the amount of chloride ion in a a
water sample is to titrate the sample of standard AgNO3
solution to produce solid AgCl.
Ag+(aq) + Cl-(aq)  AgCl(s)
If a 25.0 mL water sample requires 27.2 mL of 0.104 M
AgNO3 in such a titration, what is the concentration of Cl- in the
sample?
Answer: 0.113 M
Problem #5
 What volume of of a a 5.00M Ca(NO3)2 solution is
needed to prepare 465 mL of a 0.250 M Ca(NO3)2
solution?
 Answer: 23.3 mL
Problem #6
 Calculate the mass of sodium iodide that must be
added to 425.0 mL of a 0.100 M lead (II) nitrate solution
to precipitate all of the lead (II) ions as lead (II) iodide.
Answer: 12.7 g NaI