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SCIENCE
AP* Chemistry
Finishing Strong!
Teacher Packet
AP* and Pre-AP* are trademarks of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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STUDENT PREP SESSIONS
Nothing takes the place of the qualified and creative teacher in the classroom every day. However,
formally scheduled prep sessions provide students opportunities to spend additional time on the allimportant task of preparing for upcoming examinations, measuring their current skills and knowledge
against those of other students and against the high standards set by the instructor and the materials.
Students are treated as mature individuals, capable of handling a rigorous, fast-paced review in six hours
of content instruction. Often, they realize how well-prepared they actually are and gain confidence in
their abilities. Sometimes their eyes are opened to the work that must still be done to successfully
manage the challenges of upcoming exams. Prep sessions can provide the instruction, materials, and
encouragement students need to reach their goals. Specifically, students who attend will have the
opportunity to
• Review topics taught in the classroom
• Gain tips and strategies for problem-solving
• Learn information from teachers other than their own
• Interact with other students who share an all-important goal
• Understand relationships and connections among many concepts
Teachers are encouraged to attend prep sessions with their students. They will have an opportunity to
• Share with other successful instructors and gain new strategies for test review
• Receive valuable materials created by master teachers
• Observe students other than their own for purposes of gauging classroom progress
TABLE OF CONTENTS
Acids, Bases, and Salts
Page 3
Buffers and Titrations
Page 21
The Lab Based Question
Page 42
Net Ionic Equations
Page 62
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*
AP CHEMISTRY
Acids, Bases, and Salts
Ka, Kb, and Ksp
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
®
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ACIDS, BASES AND SALTS
Ka, Kb, and Ksp
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions over acids, bases and salts.
Standards
The topic of acids, bases and salts is addressed in the topic outline of the College Board AP
Chemistry Course Description Guide as described below.
III Reactions
A. Reaction types
1. Acid-base reactions; concepts of Arrhenius, Brønsted-Lowry, and Lewis;
coordination complexes; amphoterism
2. Precipitation reactions
C. Equilibrium
2. Quantitative Treatment
b. Equilibrium constants for reactions in solution
(1) Constants for acids and bases; pK; pH
(2) Solubility product constants and their application to precipitation and the
dissolution of slightly soluble compounds
AP Chemistry Exam Connections
The topic of acids, bases, and salts is tested every year on the multiple choice and is often the
focus of question one on the free response portion of the exam. The list below identifies free
response questions that have been recently asked. These questions are available from the College
Board and can be downloaded free of charge from AP Central http://apcentral.collegeboard.com.
2007
2005
2003
2002
2001
Free Response Questions
Question 1 part a−b
2005 B Question 1 parts a−c
Question 1 parts a, b, d, and e 2002 B Question 1 parts a−b
Question 1 part a−b
Question 1 parts a, b, and e
Question 1 parts a−d
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ACIDS, BASES, AND SALTS
Ka, Kb, and Ksp
What I Absolutely Have to Know to Survive the AP Exam
The following might indicate the question deals with acids and bases and salts:
pH, pOH, [H3O+], [OH−], strong and weak, salt hydrolysis, solubility product, Ka, Kb, Ksp, acid or base dissociation
constant, percent ionized, Bronsted-Lowry, Lewis, Arrhenius, hydronium ion, etc…
Acids and Bases: By Definition!
Arrhenius
• Acids: Hydrogen ion, H+, donors
• Bases: Hydroxide ion, OH− donors
HCl → H+ + Cl−
NaOH → Na+ + OH−
Bronsted-Lowry
• Acids: proton donors ( lose H+)
• Bases: proton acceptors (gain H+)
NH4+ + H2O ⇌ H3O+ + NH3
HNO3 + H2O ↔ H3O+ + NO3−
Lewis
• Acids: electron pair acceptors
• Bases: electron pair donors
• Explains all traditional acids/bases + a host of coordination compounds. BF3 is famous!
H
F
B
F
+
N
F
H
F
H
F
H
B
N
F
H
H
Hydrogen Ion, Hydronium Ion, Which is it!!!?
+
Hydronium − H riding piggy-back on a water molecule; water is polar and the + charge of the naked proton is greatly
attracted to “Mickey's chin” (i.e. the oxygen atom)
• H3O+ “Anthony”
• H+
“Tony”
• Often used interchangeably in problems; if H3O+ is used be sure water is in the equation!
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Acids, Bases, and Salts
Bronsted−Lowry and Conjugate Acids and Bases: What a Pair!
Acid and conjugate base pairs differ by the presence of one H+ ion.
HC2H3O2 + H2O → H3O+ + C2H3O2−
−
• HC2H3O2 is the acid; thus C2H3O2 is its conjugate base (what remains after the H+ has been donated to the H2O
molecule)
• H2O behaves as a base in this reaction. The hydronium ion is its conjugate acid (what is formed after the H2O
accepts the H+ ion)
NH3 + H2O → NH4+ + OH−
+
• NH3 is the base; thus NH4 is it’s conjugate acid (what is formed after the NH3 accepts the H+ ion)
• H2O behaves as an acid in this reaction. The hydroxide ion is its conjugate base (what remains after the H+ has
been donated to NH3)
Understanding conjugate acid/base pairs is very important in understanding acid-base chemistry; this concepts allows for
the understanding of many complex situations (buffers, titrations, etc…)
Important Notes…
Amphiprotic/amphoteric--molecules or ions that can behave as EITHER acids or bases; water, some anions of weak acids,
etc… fit this bill.
Monoprotic − acids donating one H+
Diprotic − acids donating two H+
Polyprotic − acids donating 3+ H+
Regardless, always remember: ACIDS ONLY DONATE ONE PROTON AT A TIME!!!
Ionization: That’s What it’s All About!
Relative Strengths
A strong acid or base ionizes completely in aqueous solution (100% ionized [or very darn close to it!])
• The equilibrium position lies far, far to the right (products)…
• Since a strong acid/base dissociates into the ions, the concentration of the H+/OH− ion is equal to the original
concentration of the acid/base respectively.
• They are strong electrolytes.
• Do Not confuse concentration (M or mol/L) with strength!
• Strong Acids
• Hydrohalic acids: HCl, HBr, HI; Nitric: HNO3; Sulfuric: H2SO4; Perchloric: HClO4
• Oxyacids! More oxygen atoms present, the stronger the acid WITHIN that group. The H+ that is “donated” is
bonded to an oxygen atom. The oxygen atoms are highly electronegative and are pulling the bonded pair of
electrons AWAY from the site where the H+ is bonded, “polarizing it”, which makes it easier [i.e. requires less
energy] to remove − thus the stronger the acid!
O
H
O
Br
weaker than
H
O
Br
O
weaker than
H
O
Br
O
•
Strong Bases
• Group IA and IIA (1 and 2) metal hydroxides; be cautious as the pour solubility of Be(OH)2 and Mg(OH)2
limits the effectiveness of these 2 strong bases.
• IT’S A 2 for 1 SALE!! (with the Group 2 (IIA) ions), i.e. 0.10M Ca(OH)2 is 0.20M OH −
®
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Acids, Bases, and Salts
Ionization: That’s What it’s All About! con’t.
Where the fun begins…
A weak acid or base does not completely ionize (usually < 10%)
• They are weak electrolytes.
• The equilibrium position lies far to the left (reactants)…
• The [H+] is less than the acid concentration − thus to calculate this amount and the resulting pH you must
return to the world of EQUILIBRIUM Chemistry!
• The vast majority of acid/bases are weak. Remember, ionization not concentration!!!!
• Acids and Bases ionize one proton (or H+) at a time!
• For weak acid reactions:
HA + H2O → H3O+ + A−
[H 3O + ][A - ]
Ka =
= <1
[HA]
•
For weak base reactions:
B + H2O → HB+ + OH−
Kb =
[HB+ ][OH - ]
= <1
[B]
•
K a × K b = 1× 10−14 Very important when dealing with SALT pH
•
MONUMENTAL CONCEPT…
• If [HA]o is 1000×Ka or greater, you can use the initial concentration of the acid/base as the equilibrium
concentration −WITHOUT subtracting the amount of weak acid ionized (it is just mathematically
insignificant). Hallelujah!!! This just eliminated the Quadratic Equation (which is never needed on the AP
exam).
•
MUCHO time saver!!! Since acids ionize 1 H+ at a time the [H3O+] = [A−] thus all you need to know is…
Ka =
•
[x][x]
[x][x]
where x = [H3O+ ] or K b =
where x = [OH - ]
[M o ]
[M o ]
Percent Ionization
• Often will be asked to determine how ionized the weak acid or base is…
•
%=
[x]
× 100
[M o ]
where x = [H 3O + ]or[OH − ]
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Acids, Bases, and Salts
Salts and pH: It’s all about Hydrolysis
A salt is the PRODUCT of an acid base reaction; how the salt affects the pH of the solution depends on whether or not the
salt ions will hydrolyze water to any significant extent…
• Cation: M+ hydrolyzes H2O to produce MOH and H3O+ ions
• M+ + H2O → H3O+ + MOH
• If the MOH is a strong base this reaction WILL NOT occur (strong bases ionize 100%).
Therefore NO additional hydronium ion is formed in the solution and this salt ion has NO
effect on the pH of the solution!
• If the MOH is a weak base this reaction WILL occur; and the additional hydronium ion will
lower the pH of the solution!!
−
• Anion: A hydrolyzes H2O to produce HA and OH− ions
• A− + H2O → HA + OH−
• If the HA produced is a strong acid this reaction WILL NOT occur (strong acids ionize
100%). Therefore NO additional hydroxide ion is formed in the solution and this salt ion has
NO effect on the pH of the solution!
• If the HA is a weak acid this reaction WILL occur; and the additional hydroxide ion
will increase the pH of the solution!
• So… an easy way to tell, you ask…?
• Ask yourself, which acid and which base reacted….were they strong or weak?? Strong wins!
1. A salt such as NaNO3 gives a neutral solution—SA (HNO3) & SB (NaOH)
− Neither salt can hydrolyze water because it would result in the formation of a strong
acid and base
2. K2S should be basic—SB (KOH) & WA (H2S)
− Only the weak can effectively hydrolyze water so the S2− ion reacts with water to
produce H2S and OH− ions; therefore the salt solution is basic…
3. FeCl2 should be weakly acidic—since WB (Fe(OH)2) & SA (HCl)
− Only the weak can effectively hydrolyze water so the Fe2+ ion reacts with water to
produce Fe(OH)2 and H3O+ ions; therefore the salt solution is slightly acidic…
Calculating the pH of a salt solution
• A salt in solution is either behaving as a weak acid or a weak base.
• Remember, this means they do not ionize much. So treat them just like any other weak acid or base.
• The MAJOR DIFFERENCE; no Ka or Kb value will be provided. Its conjugate acid or base K will be
given instead.
• Example, the salt NaC2H3O2; the Ka for HC2H3O2 will be provided. The Kb has to be
calculated before any pH calculations can be attempted.
• K a × K b = 1× 10−14 Very important!
•
Then solve below to find the pH…
Kb =
®
[x][x]
where x = [OH - ]
[M o ]
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Acids, Bases, and Salts
Solubility Product: It’s Ksp!
When solids are first added to water no salt ions are present. As dissolution occurs, the ions dissolve in the water until
equilibrium is established and a saturated solution is formed. The extent to which a substance dissolves in the solvent is
the solubility. The equilibrium constant, Ksp, is the product of all ions in solution.
PbI2(s) ⇌ Pb2+(aq)+ 2 I−(aq) Ksp = [Pb2+][I−]2
•
•
Note: solids don’t appear in the K expression
Be aware of the coefficients
• the 1:2 ratio is important when determining concentrations of the ions present
• the 2 results in an expression of [I−]2
A precipitate is an insoluble compound formed when two soluble solutions are mixed. To determine whether a
precipitate will form the reaction quotient will need to be calculated.
• Q < K unsaturated solution is formed as the reaction has not reach the equilibrium position; no ppt
• Q = K a saturated solutions is made; no ppt
• Q > K a supersaturated solution is formed the concentration of ions are greater than allowed; a ppt will
form
Acids, Bases and Salts Cheat Sheet
Relationships
Equilibrium Expression
[H 3O + ][A - ]
= <1
[HA]
[x][x]
where x = [H 3O + ]
Ka =
[M o ]
[HB+ ][OH - ]
= <1
[B]
[x][x]
where x = [OH - ]
Kb =
[M o ]
Ka =
Kb =
%=
[x]
× 100
[M o ]
K a × K b = 1×10−14
where x = [H 3O + ]or[OH − ]
pH = −log[H+]
pOH = −log[OH−]
Acidic pH < 7
Basic pH > 7
pH + pOH = 14
Kw = [H3O+][OH−] = 1.0×10−14 @ 25°C
Neutral pH = 7
[H3O+] = [OH−]
Conjugate acids and bases
[H3O+] > [OH−]
[H3O+] < [OH−]
Ksp = [M+]x[A−]y
MA(s) ⇌ xM+(aq)+ yA−(aq)
Connections
Equilibrium
Buffers and Titrations
Precipitation and Qualitative Analysis
Bonding and Lewis Structures − justify oxyacid strengths
Potential Pitfalls
Coefficients in Ksp problems
Ka or Kb with salt pH
Weak acids and bases − be sure you know Ka or Kb;
solving a Kb problem gives [OH−] thus you are finding the Weak is about IONIZATION not CONCENTRATION
pOH!
®
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Acids, Bases, and Salts
Free Response
NH3(aq) + H2O(A) ⇌ NH4+(aq) + OH−(aq)
Kb = 1.80 × 10-5
1. Ammonia reacts with water as indicated in the reaction above.
(a) Write the equilibrium constant expression for the reaction represented above.
1 point is earned for the correct expression
[NH +4 ][OH − ]
Kb =
[NH 3 ]
(b) Calculate the pH of a 0.150 M solution of NH3
x2
where x = [OH − ]
M -x
x2
−5
1.80 x10 =
0.150
−3
1.64 x10 = x
− log x = pOH = 2.784
Kb =
1 point is earned for the correct set up and for
calculating the concentration of hydroxide ions
1 point is earned for the correct pH
pH = 11.216
(c) Determine the percent ionization of the weak base NH3.
%=
[x]
× 100
[M o ]
%=
1.64 × 10−3
×100 = 1.09%
[0.150]
1 point is earned for the correct percent
ionization
(d) Calculate the hydronium ion, H3O+, concentration in the above solution. Be sure to include units with
your answer.
[H 3O + ][OH − ] = 1.00 × 10−14
[H 3O + ]=
1.0 × 10−14 1.00 × 10−14
=
= 6.10 × 10−12 M
[OH − ]
1.64 × 10−3
1 point is earned for the concentration of
hydronium ions with units of mol L−1 or M
OR
[H 3O + ] = 10− pH
[H 3O + ] = 10−11.216 = 6.08 × 10−12 M
®
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Acids, Bases, and Salts
Ammonium ions react with water to form a solution with a pH of 4.827.
+
H
H
N
+
H
O
H
H
H
(e) Complete the reaction above by drawing the complete Lewis structures for both products of the reaction.
+
O
N
H
H
+ H
1 point is earned for each correct Lewis
structure.
H
H
H
(f) Determine the number of moles ammonium ions in 250 mL of the above solution.
[H 3O + ] = 10− pH = 10−4.827 = 1.49 × 10−5
Ka =
K w 1.00 × 10−14
=
= 5.56 × 10−10
K b 1.80 × 10−5
5.56 × 10−10 =
−5
−5
[1.49 × 10 ][1.49 × 10 ]
[M ]
M = 0.400
1 point is earned for calculating the
concentration of ions in the solution.
1 point is earned for calculating the Ka for
ammonium
1 point is earned for calculating the molar
concentration and the number of moles of
ammonium ions
x mol
0.250 L
x = 0.0100 mol
0.400M =
®
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Acids, Bases, and Salts
Free Response
2. Hydrocyanic acid, HCN, is a weak acid. The acid dissociation constant for HCN is 6.2 x 10−10. The equilibrium
expression for the reaction is shown below.
Ka =
[H 3O + ][CN − ]
= 6.2 × 10−10
[HCN]
(a) Write the equation for the dissociation of HCN in water.
1 point for the correct equation
HCN + H2O ⇌ H3O+ + CN−
(b) Calculate the hydronium ion, H3O+ concentration of a 0.250 M solution of HCN.
1 point for [H3O+] = [CN−] or x2
x2
where x = [H 3O + ]
M −x
x2
−10
6.2 × 10 =
0.25
−5
1.2 × 10 = x
Ka =
1 point for the correct hydronium ion concentration
When lithium cyanide, LiCN, is dissolved in water according to the equation below the resulting solution is basic.
LiCN(s) → Li+(aq) + CN−(aq)
(c) Write a net ionic equation that illustrates why the resulting solution is basic.
1 point for the correct reactants
CN− + H2O ⇌ HCN + OH−
1 point for the correct products
(d) Calculate the equilibrium constant for reaction in part c.
1 point for the correct Kb value
Kw
1× 10−14
=
= 1.6 × 10−5
Kb =
−10
K a 6.2 × 10
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Acids, Bases, and Salts
The overall dissociation reaction for carbonic acid, H2CO3, is shown below.
H2CO3 ⇌ CO32− + 2H+
Keq = 2.4×10−17
(e) Write a balanced chemical equation showing both the first and second dissociation reactions for carbonic
acid, H2CO3.
H2CO3 ⇌ HCO3− + H+
1 point for the first dissociation reaction
HCO3− ⇌ CO32− + H+
1 point for the second dissociation reaction
(f) The dissociation constant for the second dissociation, K a2 = 5.6 × 10−11 . Calculate the equilibrium
constant for the first dissociation, K a1 .
K eq = K a1 × K a2
2.4 × 10−17 = K a1 × (5.6 × 10−11 )
K a1 = 4.3 × 10
1 point for the correct K a1 value
−7
®
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Acids, Bases, and Salts
Free Response
3. Solid potassium lactate, KC3H5O3, is dissolved in enough water to make a 0.25 M solution. The solution has a pH of
12.25 at 25°C.
(a) Explain why the solution produced by dissolving KC3H5O3 is basic. Use an equation as a part of your
justification.
1 point for the correct hydrolysis reaction and
C3H5O3− + H2O ⇌ HC3H5O3 + OH−
explanation
The lactate ion hydrolyzes the water molecules,
producing hydroxide ions; which make the solution
basic. The potassium ion doesn’t hydrolyze the
water thus has no effect on the pH
(b) Calculate the hydroxide ion concentration of the solution described above.
pH = 12.25
pOH = 14.00 − 12.25 = 1.75
[OH−] = 10−1.75 = 1.8×10−2
1 point for the correct hydroxide concentration
(c) Calculate the value of the equilibrium constant for the reaction in part (a)
1 point for the correct Kb expression
[HC3 H 5O3 ][OH − ]
Kb =
[C3 H 5O −3 ]
Kb =
1 point for the correct value for Kb
(1.8 × 10−2 ) 2
= 1.4 ×10−3
−2
0.25 − (1.8 × 10 )
(d) Calculate the acid dissociation constant, Ka, for lactic acid, HC3H5O3.
Ka =
K w 1× 10−14
=
= 7.1×10−12
K b 1.4 × 10−3
1 point for the correct value for Ka
(e) The structural formula for lactic acid is shown below. Circle the hydrogen lost when HC3H5O3 behaves
as an acid.
O
H3C
C
CH
OH
OH
®
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Acids, Bases, and Salts
1 point for identifying the correct hydrogen
The circled hydrogen is a part of the carboxylic
acid group (COOH).
(f) A student needs to dilute a 0.25 M solution of KC3H5O3 to make 50.0 mL of 0.10 M potassium lactate
solution. Describe the steps needed to prepare the new solution. Be sure to include appropriate lab
equipment.
M1V1 = M2V2
(0.25 M)V1 = (50 mL)(0.10 M)
V1 = 20.0 mL KC3H5O3
1 point for correct volume of KC3H5O3 needed
Pipet 20.0 mL of 0.25 M KC3H5O3 into a 50 mL
volumetric flask. Add water and mix thoroughly.
Then fill the volumetric flask with water to the 50
mL mark and mix.
1 point for filling the flask to the 50.0 mL mark (adding
50.0 mL of water DOES NOT earn credit)
1 point for using a volumetric flask
®
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Acids, Bases, and Salts
Free Response
4. Answer the following questions about compounds of strontium.
In a saturated solution of strontium phosphate, Sr3(PO4)2, the concentration of Sr2+ is 7.5×10–7 M at 25°C.
(a) Write the balanced equation for the dissolution of solid Sr3(PO4)2 in water.
1 point for the correct equation
Sr3(PO4)2 ⇌ 3Sr2+ + 2PO43–
(b) Calculate the value of the solubility product constant.
1 point for correct [PO43–]
K sp = [Sr 2 + ]3 [PO 43− ]2
[Sr 2 + ] = 7.5 × 10−7
[PO 43− ] = 7.5 × 10−7 M Sr 2 + ×
2 mol PO 43−
= 5.0 × 10−7 M
2+
3mol Sr
K sp = [7.5 × 10−7 ]3 [5.0 × 10−7 ]2 = 1.1 × 10−31
1 point for correct Ksp
(c) Will a precipitate form when 20.0 mL of 0.0015 M strontium acetate and 30.0 mL of 0.0050 M lithium
phosphate are mixed? Justify your answer with appropriate calculations.
1 point for the correct [Sr2+] and [PO43–]
Qsp = [Sr 2 + ]3 [PO 43− ]2
[Sr 2+ ] = 0.020 L × 0.0015M =
3.0 × 10−5 molSr 2 +
= 6.0 × 10−4 M
0.050 L
[PO 43− ] = 0.030 L × 0.0050M =
1.5 × 10−4 molSr 2+
= 3.0 × 10−3 M
0.050 L
Qsp = [6.0 × 10−4 ]3 [3.0 × 10−3 ]2 = 1.9 × 10−15
1 point for the correct value of Qsp and an
explanation using Qsp and Ksp
Qsp > K sp ∴ a precipitate will form
The Ksp for strontium fluoride, SrF2, is 7.9×10–10 at 25°C.
(d) Calculate the molar solubility of a saturated solution of strontium fluoride.
1 point for the setting up and getting 4x3 in the Ksp
expression
K sp = [Sr 2 + ][F− ]2 = 7.9 × 10−10
[x][2x]2 = 7.9 × 10−10
4x 3 = 7.9 × 10−10
1 point for the correct molar solubility
x = 5.8 × 10−4 = molar solubility of SrF2
®
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Acids, Bases, and Salts
(e) A small sample of the saturated solution of SrF2 was allowed to evaporate. Would the concentration of
Sr2+ ions in the solution increase, decrease or remain the same? Justify your answer.
Remain the same. The [Sr2+] is independent of the volume
of the solution; as the volume is reduced so will be the
number of moles of strontium fluoride dissolved but the
[Sr2+] will not change.
1 point for stating the concentration Sr2+ will
remain the same with proper justification
(f) At 35°C, the amount of strontium fluoride present at the bottom of the flask increases. Is this dissolution
process endothermic or exothermic? Justify your answer.
When the temperature increases, more strontium fluoride is
present, thus the reaction shifted to favor the reactants.
Since the addition of heat (increasing temperature) drives
the reaction to the left the reaction must be exothermic.
®
1 point for indicating the reaction is
exothermic with justification
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Acids, Bases, and Salts
Multiple Choice
Questions 1−4 refer to the following answer choices. All solutions are 1.0 M.
(A) HCl
(B) LiCl
(C) HClO4
(D) NaC2H3O2
(E) BF3
1. Can be used as a Lewis acid
E
BF3 acts as an electron pair acceptor.
2. Would form a basic solution when dissolved in water
D
C2H3O2− hydrolyzes H2O which increases the amount of hydroxide ions in the solution
according to the following equation.
C2H3O2− + H2O ⇌ HC2H3O2 + OH−
3. Forms the solution with the lowest pH
C
HClO4 is the only strong acid in the list.
4. Can produce a buffer with a pH < 7 when combined with HC2H3O2
D
NaC2H3O2 is the salt of HC2H3O2. A weak acid together with it’s conjugate salt form a
buffer solution.
5. Which of the following can act as an amphoteric species?
I. SO42−
II. HC2O4−
III. NH4+
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
B
HC2O4− can both donate a hydrogen ion to form C2O42− and accept a hydrogen ion to form
H2C2O4.
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Acids, Bases, and Salts
6. Estimate the pH of a 1.0×10−5 M solution of the weak base aniline, C6H5NH2. Kb = 4.0 x 10−10
(A) 11
(B) Between 8 and 11
(C) 8
(D) Between 3 and 8
(E) 3
D
If the Kb value is approximately 10−10 then the pOH is approximately 10 and the pH around
4.
7. All the following statements are true about the reaction below EXCEPT:
650 kJ + HCO3−(aq) + HC2O4−(aq) ⇌ CO32−(aq) + H2C2O4(aq)
Kc < 1
(A) HC2O4− is a weaker base than CO32−.
(B) HCO3− and H2C2O4 are both acting as acids in the reaction.
(C) HCO3− is an amphoteric species.
(D) The reaction lies far to the right (favors products).
(E) Lowering the temperature of the solution will decrease the value of K.
A
The HC2O4− is NOT a weaker base than CO32− or else the CO32− ion would more readily
accept the proton.
8. A solution of a monoprotic weak acid, HA, has a pH of 5.00. Calculate the acid dissociation constant for the
weak acid if the solution has a molar concentration of 0.50 M.
(A) 2.0 x 10−3
(B) 1.0 x 10−5
(C) 2.0 x 10−6
(D) 2.0 x 10−9
(E) 2.0 x 10−10
E
[H + ][A − ]
Ka =
[HA]
[H + ]=10− pH = 10−5.00 = 1.0 × 10−5
(1.0 × 10−5 ) 2 1.0 × 10−10
Ka =
=
= 2 ×10−10
0.50
0.50
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Acids, Bases, and Salts
9. The solubility product constant of Ag2SO4 is 1.2×10−5 at 25°C. Which expression can be used to calculate the
solubility forAg2SO4?
1
⎡ 1.2 × 10−5 ⎤ 3
(A) ⎢
⎥
2
⎣
⎦
1
(B) ⎡⎣1.2 × 10−5 ⎤⎦ 2
1
(C) ⎡⎣1.2 × 10−5 ⎤⎦ 3
⎡ 1.2 × 10 ⎤
⎥
4
⎣
⎦
−5
(D) ⎢
1
2
⎡ (1.2 × 10−5 ) ⎤
⎥
(E) ⎢
4
⎣
⎦
1
3
Ag2SO4 ⇌ 2Ag+ + SO42−
K sp = [Ag + ]2 [SO 4 2 − ]
E
1.2 × 10−5 = [2x]2 [x]
1.2 × 10−5 = 4x 3
1
⎛ 1.2 × 10−5 ⎞ ⎛ 1.2 × 10−5 ⎞ 3
x=3⎜
⎟ or ⎜
⎟
4
4
⎝
⎠ ⎝
⎠
10. At a certain temperature, a saturated solution of silver phosphate has a [Ag+] of 3.0 ×10−5 M. What is the
solubility product constant, Ksp, for silver phosphate, Ag3PO4?
(A) 2.7 x 10−21
(B) 2.7 x 10−20
(C) 2.7 x 10−19
(D) 8.1 x 10−10
(E) 9.0 x 10−10
Ag3PO4 ⇌ 3Ag+ + PO43−
K sp = [Ag + ]3 [PO 43− ]
C
K sp = [3x]3 [x]
K sp = [3.0 × 10-5 ]3 [1.0 × 10-5 ]
K sp = (27 × 10-15 )(1.0 × 10-5 ) = 27 × 10-20 = 2.7 × 10-19
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*
AP CHEMISTRY
Buffers and Titrations
As Easy As It Ever Gets
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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BUFFERS AND TITRATIONS
As Easy As It Ever Gets
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions over acids, bases, particularly in the area of buffers and titrations.
Standards
The topic of acids, bases, buffers and titrations is addressed in the topic outline of the College
Board AP Chemistry Course Description Guide as described below.
III Reactions
C. Equilibrium
2. Quantitative Treatment
b. Equilibrium constants for reactions in solution
(1) Constants for acids and bases; pK; pH
(2) Solubility product constants and their application to precipitation and the
dissolution of slightly soluble compounds
(3) Common ion effect; buffers; hydrolysis
AP Chemistry Exam Connections
The topic of acids, bases, buffers and titrations is tested every year on the multiple choice and is
often the focus of question one on the free response portion of the exam. The list below identifies
free response questions that have been recently asked. These questions are available from the
College Board and can be downloaded free of charge from AP Central
http://apcentral.collegeboard.com.
2007
2005
2003
2002
1999
Free Response Questions
2006 B Question 1 parts a and b
Question 1 parts c−e
Question 1 part c
2005 B Question 1 parts c and d
2002 B Question 1 parts c and d
Question 1 part c−e
Question 1 parts c and d
Question 1 part e
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BUFFERS AND TITRATIONS
As Easy As It Ever Gets
What I Absolutely Have to Know to Survive the AP Exam
The following might indicate the question deals with buffers and/or titrations:
buffer solution, common ion, conjugate base/acid, equivalence point, ½ equivalence point, end-point, weak acid/base
“reacted with a strong..”, titrant, buret, analyte, etc…
Buffers: All about those “common ions”
IS: A buffer is a solution of a weak acid or base and its salt [which is its conjugate].
DOES: A buffer resists a change in pH.
HOW IT WORKS: Since a buffer consists of both an acid or base and its conjugate, which differ by an H+, both a weak
acid and a weak base are present in all buffer solutions.
In order to understand buffers and buffer problems we must be proficient with the following…
• MUST know the conjugate A/B concept to be successful at buffer problems. This means understanding that HA
has A− as it’s conjugate and that NaA is not only a salt, but that any soluble salt releases the common ion, A−
• pH = −log [H+]
• Ka × Kb = Kw
• Any titration involving a weak A/B, from the first drop to the last, before equivalence, is a buffer problem. If HA
is titrated with NaOH, as soon as the first drop splashes into the container, I’ve added A− ions to the solution and
the presence of both HA and A− constitutes a buffer solution.
Buffers: Calculating pH
Solving Buffer Problems: the one and only, handy dandy, all I’m ever going to need equation…
[ Acid ]
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
Where:
Ka – is the acid dissociation constant
[Acid] – is the concentration of the substance behaving as the weak acid in the buffered solution
ƒ Either the weak acid or the salt of the weak base
[Base] – is the concentration of the substance behaving as the weak base in the buffered solution
ƒ Either the weak base or the salt of the weak acid
To Solve:
ƒ MUST recognize who’s who in the buffered and then plug into the equation
ƒ If the buffer is a weak acid, HA, and its conjugate salt, A− then HA is the acid and A− is the base.
ƒ If the buffer is a weak base, WB, and its conjugate salt, B+, then WB is the base and B+ is the acid.
ƒ You still use the Ka; thus you must calculate it first using the Kb provided for the weak base
ƒ If concentrations are given plug and chug
ƒ If concentrations of separate solutions are given with volumes and then the two are added together you must
recalculate the “new concentrations” due to dilution.
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Buffers and Titrations
Buffers: Calculating pH
Con’t
Example:
Calculate the pH of a solution containing 0.75 M lactic acid, HC3H5O3 (Ka = 1.4×10−4), and 0.25 M sodium lactate,
NaC3H5O3.
ƒ Lactic acid is the weak ACID and lactate ion is its conjugate BASE
[ Acid ] = (1.4 × 10−4 ) [0.75] = 4.2 × 10−4
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[0.25]
pH = − log(4.2 × 10−4 ) = 3.38
Example:
A buffered solution contains 0.250 M NH3 (Kb = 1.80×10−5) and 0.400 M NH4Cl. Calculate the pH of this solution.
ƒ NH3 is the weak BASE and NH4+ is its conjugate ACID
ƒ Calculate Ka!
K a × K b = 1.00 × 10−14
1.00 × 10−14
= 5.56 × 10−10
1.8 × 10−5
[ Acid ] = (5.56 × 10−10 ) [0.400] = 8.89 × 10−10
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[0.250]
Ka =
pH = − log(8.89 × 10−10 ) = 9.051
Example:
Calculate the pH of a solution prepared by mixing 20.0 mL of 0.300 M acetic acid, HC2H3O2, with 20.0 mL of 0.350 M
NaC2H3O2. Ka for acetic acid is 1.80×10-5
ƒ Acetic acid is the weak ACID and acetate ion is its conjugate BASE
ƒ Recalculate molarities OR better yet just calculate the number of moles (volumes cancel)….
20.0 mL × 0.300M = 6.00 mM acid
20.0 mL × 0.350 M = 7.00 mM conjugate base
[ Acid ] = (1.80 × 10−5 ) [6.00] = 1.54 × 10−5
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[7.00]
pH = − log (1.54 × 10−5 ) = 4.812
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Buffers and Titrations
Buffer Attack: Who Defends?
The purpose of a buffer solution is to resist large swings in pH. If a small amount of strong acid is added to the buffer,
there is a base component ready and waiting to neutralize the “invader”. Same goes for the addition of a small amount of
strong base, there is an acid component ready and waiting to neutralize it…
If strong acid is added to the buffer, simply add the amount of acid added to the numerator (acid) AND subtract the same
quantity from the denominator (base). Why? If strong acid is added to a buffer solution the weak base component is selfsacrificing and neutralizes the acid − thus the amount present is reduced by the amount of strong acid present in the
invasion. The reaction between the weak base component of the buffer and the strong acid eliminates the strong acid and
produces more weak acid.
HCl + A− → HA + Cl−
Strong Acid Attack:
Example:
Calculate the pH of a solution containing 0.75 M lactic acid, HC3H5O3 (Ka = 1.4×10−4), and 0.25 M sodium lactate,
NaC3H5O3 after the addition of 0.10 M HCl.
[ Acid ] = (1.4 × 10−4 ) [0.75+0.10] = 7.9 × 10−4
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[0.25-0.10]
pH = − log(7.9 × 10−4 ) = 3.10
If strong base is added, simply add the base to the denominator and subtract from the numerator. Just the exact opposite
occurs; the acid component reacts with the base and converts the strong base to more weak base.
Strong Base Attack:
NaOH + HA → H2O + Na+ + A−
Example:
What is the pH of the final solution after the addition of 125 mL of 0.500 M NaOH to 250 mL of a solution that is 0.550
M acetic acid, HC2H3O2 and 0.450 M sodium acetate, NaC2H3O2?
ƒ Acetic acid is the weak ACID and acetate ion is its conjugate BASE; NaOH is the INVADER
ƒ Recalculate molarities OR better yet just calculate the number of moles (volumes cancel)….
125mL × 0.500 M = 62.5mM NaOH "the invader"
250 mL × 0.550 M = 137.5mM HA "the weak acid"
250 mL × 0.450M = 112.5mM A − "the conjugate base"
[ Acid ]
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[137.5 − 62.5] = 7.71× 10−6
⎡⎣ H 3O + ⎤⎦ = (1.80 × 10−5 )
[112.5+62.5]
pH = − log (7.71 × 10−6 ) = 5.113
CAUTION:
If the amount of invader exceeds the amount present in the buffer then the invader has overrun the buffer!
ƒ In this case calculate how much strong “invader” remains after all of the buffer is used
ƒ Recalculate the molar concentration of the remaining “invader”; this is a strong acid or base. Calculate the pH as
with any strong acid or base. pH = − log[H+]
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Buffers and Titrations
Preparing a Buffer: It’s all about Ka and the Ratio!
[ Acid ]
If asked to describe how to prepare a buffer of a certain pH… think ⎡⎣ H 3O + ⎤⎦ = K a
[Base]
ƒ Use 0.10 M to 1.0 M solutions of reagents & choose an acid whose Ka is near the [H3O+] concentration we want.
[ Acid ] to fine tune the pH.
Or its pKa should be as close to the pH desired as possible. Adjust the ratio of
[Base]
ƒ It is the relative # of moles (not M) of acid/CB or base/CA that is important since they are in the same solution and
share the same solution volume.
ƒ This allows companies to make concentrated versions of buffer and let the customer dilute--this will not affect
the # of moles present--just the dilution of those moles. So on the EXAM, never fall for the trick of “what
happens to the pH of the buffer solution after the addition of 250 mL of water. Yes the [Acid] and [Base] are
[ Acid ] ratio is the same and that is what matters in a buffer!
diluted but the
[Base]
IMPORTANT NOTE:
ƒ When equal concentrations of Acid and Base are present [which occurs at the ½ equivalence point of a titration]
the ratio of acid to base equals ONE and therefore, the pH = pKa.
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Buffers and Titrations
Acid-Base Titrations: It Depends on Where You Are!
There are 3 types of titrations you will be dealing with.
1. Strong Acid and Strong Base
ƒ Net ionic reaction: H+ + OH− → H2O
ƒ Only when the acid AND base are both strong is the pH at the equivalence point 7. [Any other
conditions and you get to do a nasty equilibrium problem.] It’s really a stoichiometry problem with
a limiting reactant. The “excess” is responsible for the pH
ƒ Just deal with what is left over. Be sure you are in moles if additive volumes are given and
recalculate the M after doing the stoichiometry work
ƒ Use pH = −log [H+] to determine the pH of the solution.
2. Weak Acid and Strong Base
ƒ The equivalence point > pH 7
ƒ 4 ZONES OF INTEREST ALONG A TITRATION CURVE for a Weak Acid and Strong Base
Titration
ZONE 1: The pH before the titration begins.
ƒ This is simply a weak acid problem.
Ka =
x2
M
where x = [H 3O + ]
ZONE 2: The pH during the titration but before the equivalence point.
ƒ Once the titration begins the weak acid has reacted with the strong base to produce salt and
water. The salt is the conjugate base of the weak acid. This is a BUFFER problem!!!
HA + OH− → H2O + A−
ƒ
Remember that the amount of strong base added = amount of weak acid reacted = amount
of salt made
ƒ
Plug in the amount of weak acid left and the amount of salt made into ⎡⎣ H 3O + ⎤⎦ = K a
[ Acid ]
[Base]
CAUTION! TIME SAVER: At the ½ equivalence point [Weak Acid] = [Conjuagte Base formed] thus
[H3O+] = Ka
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Buffers and Titrations
Acid-Base Titrations: It Depends on Where You Are!
Con’t
ZONE 3: The pH at the Equivalence Point
ƒ At the equivalence point of the titration ALL of the WEAK ACID has been reacted with
the STRONG BASE − the only species left are H2O and the Conjugate Base of the Weak
Acid (the SALT).
ƒ The pH is based only on the salts hydrolysis properties The M of the salt at equivalence
point must be calculated.
ƒ Remember that the amount of strong base added = amount of weak acid reacted = amount
of salt made
ƒ Thus the molarity of the salt in solution is:
moles of salt formed
M=
total L of solution at the equivalence pt.
ƒ Must use the Ka of the weak acid to calculate the Kb of the salt and then use
Kb =
x2
M
where x = [OH − ] ; find the pOH then convert to pH
ZONE 4: Beyond the Equivalence Point
ƒ It’s all about the excess!
ƒ Calculate the amount of excess strong base added beyond the equivalence point and then
recalculate its M. This is a strong base so pOH = − log [OH−]
3. Weak Base and Strong Acid
ƒ Just like the Weak Acid/Strong Base above, just flip flop the process…
ƒ The equivalence point < pH 7
IMPORTANT NOTES:
There is a distinction between the equivalence point and the end point. The end point is when the indicator
changes color. If you’ve made a careful choice of indicator, the equivalence point (the number of moles of acid
= number of moles of base) and the end point of the titration will be achieved at the same time.
ƒ When choosing an indicator, determine the pH at the equivalence pt. Then, pick the indicator that
has a pKa close to the pH at the equivalence point.
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Buffers and Titrations
Buffers and Titrations Cheat Sheet
Relationships
[ Acid ]
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
2
x
where x = [OH − ]
Kb =
M
pH = −log[H+]
Ka =
x2
M
where x = [H 3O + ]
K a × K b = 1×10−14
pOH = −log[OH−]
4 ZONES OF INTEREST ALONG A TITRATION CURVE for a Weak Acid and Strong Base
1. Initial pH − simply a weak acid/weak base problem K a =
x2
M
where x = [H 3O + ] or K b =
x2
M
where x = [OH − ]
[ Acid ] where the weak and its
2. During the titration − BUFFER, BUFFER, IT’S A BUFFER! ⎡⎣ H 3O + ⎤⎦ = K a
[Base]
conjugate salt are what we need to know about.
3. Equivalence Pt. − only salt and water left; the salt is either a weak base (if a weak acid was titrated) or a weak
x2
x2
where x = [H 3O + ] or K b =
where x = [OH − ]
acid (if a weak base was titrated); work it as such: K a =
M
M
BUT you have to convert Ka to Kb or vise versa!!!
4. Beyond the Equivalence Pt. − This is a stoichiometry problem; find out how much STRONG (acid or base) is left
after equivalence, recalculate its molarity (remember the volume increased during the titration), and use
pH = −log[H+] or pOH = −log[OH−]
Be aware of where the volume of STRONG added in the titration is ½ way to equivalence point!
[ Acid ] is 1 (same amount of weak acid (base) and its conjugate are
ƒ Here ⎡⎣ H 3O + ⎤⎦ = K a because the
[Base]
present. HUGE time saver and easy way to find the Ka or Kb of the weak acid or base.
In a titration problem ALWAYS know where the equivalence point is!!!!!!
[ Acid ] ; read carefully and know who is the weak acid (base)
BUFFERS: Only need to use the equation: ⎡⎣ H 3O + ⎤⎦ = K a
[Base]
ƒ
and who is the conjugate salt (it plays the other role!).
Don’t be tricked − NEVER select a strong acid or base as a part of a buffer solution; it HAS TO BE WEAK
When a buffered solution is “attacked” remember:
ƒ If the attacker is a strong acid; the base component of the buffer “sacrifices” and reacts with the invading acid,
converting it to water and more weak acid − thus resisting large decreases in the pH
ƒ If the attacker is a strong base; the acid component of the buffer “sacrifices” and reacts with the invading base,
converting it to water and more weak base − thus resisting large increases in the pH
Connections
Equilibrium
Stoichiometry
Potential Pitfalls and Things to Watch For
Knowing when you need Ka or Kb
Recognizing the weak organic bases
BE AWARE of when a weak acid (base) is present with its
conjugate salt − HELLO! YOU ARE IN THE BUFFER
ZONE!!!!
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Buffers and Titrations
Free Response
NH3(aq) + H2O(A) ⇌ NH4+(aq) + OH−(aq)
Kb = 1.80 × 10−5
1. Ammonia reacts with water as indicated in the reaction above.
(a) Write the equilibrium constant expression for the reaction represented above.
Kb =
1 point is earned for the correct expression
[NH +4 ][OH − ]
[NH 3 ]
(b) Calculate the pH of a 0.150 M solution of NH3
1 point is earned for the correct set up and for
calculating the concentration of hydroxide ions
x2
where x = [OH − ]
M
x2
1.80 x10−5 =
0.150
−3
1.64 x10 = x
Kb =
1 point is earned for the correct pH
− log x = pOH = 2.784
pH = 11.216
A 100 mL sample of the 0.150 M solution of NH3 is titrated with 0.100 M HCl
(c) Determine the volume of HCl required to reach the equivalence point.
(100 mL)(0.150 M ) = 15 mmol NH 3
1 point is earned for the volume of HCl
required to reach equivalence
( X mL)(0.100 M ) = 15 mmol HCl
X =150 mL HCl
(d) Calculate the pH of the solution at the equivalence point of the titration.
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Buffers and Titrations
At equivalence the pH depends on the salt NH +4
At equivalence: 15 mmol NH +4 ; 250 mL total volume
∴ [NH +4 ] =
Ka =
1 point is earned for the concentration of
ammonium ions
15mmol
= 0.060M
250mL
Kw
= 5.56 x10−10
Kb
1 point is earned for converting Ka to Kb.
x2
where x = [H 3O + ]
M
x2
5.56 x10−10 =
0.060
1 point is earned for the correct set up and pH
−6
5.77 x10 = x
calculation
− log x = pH = 5.239
To the above solution (at the equivalence point) 250 mL of 0.060 M Pb(NO3)2 are added. After mixing, the following
concentrations were determined: [Pb2+] = 0.030M; [NO3−] = 0.060M; [Cl−] = 0.030M
Ka =
(e) Will a precipitate form? Justify you answer. The Ksp for the insoluble salt in question is 1.6 × 10−5
PbCl2 → Pb2+ + 2Cl−
1 point is earned for the calculation of Qsp
Ksp = [Pb2+ ][Cl− ]2
Qsp = (0.030)(0.030)2 = 2.7 ×10−5
1 point is earned for comparing Qsp to Ksp and
determining that a ppt. will form
Qsp > Ksp ∴ ppt will form
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Buffers and Titrations
Free Response
CH3NH2 + H2O ⇌ CH3NH3+ + OH−
Kb = 4.40 x 10−4
2. Methylamine, CH3NH2, a weak base, is added to a flask to form 1.50 L of solution. The reaction shown
above occurs at 25°C.
(a) Write the expression for the base dissociate constant, Kb.
1 point for the correct expression. Charges
must be correct to receive credit.
[CH 3 NH 3+ ][OH − ]
Kb =
CH 3 NH 2
(b) The pH of the solution was determined to be 11.822. Calculate the molar concentration of
methylamine in this solution.
pH = 11.822
pOH = 2.178
1 point for the correct set up and for
calculating the concentration of hydroxide ions
[OH − ] = 10−2.178 = 6.63 × 10−3
x2
where x = [OH − ]
M
(6.63 × 10−3 ) 2
−4
4.40 × 10 =
M
M = 0.100M
Kb =
1 point is earned for the correct molar
concentration of methylamine
In a different experiment, 150.0 mL of a 0.150 M solution of methylamine was prepared. To this solution,
0.120 mole of methylammonium chloride was added.
(c) Calculate the pH after the addition of the methylammonium chloride, CH3NH3Cl.
0.150 L × 0.150M = 0.225mole methylamine
0.120 mole methylammonium chloride
K a × K b = 1.00 × 10−14
1.00 × 10−14
= 2.27 × 10−11
4.40 × 10−4
[ Acid ] = (2.27 × 10−11 ) [0.120] = 1.21× 10−11
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[0.225]
Ka =
1 point for the correct number of moles of
methylamine or the correct molarity of
methylammonium ion
1 point for the correct Ka for methylammonium
ion
1 point for the correct [H3O+]ions
pH = − log(1.21 × 10−11 ) = 10.917
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Buffers and Titrations
(d) To the 150 mL solution describe in part (c) 100 mL of 0.950 M HCl was added. Calculate the
resulting pH of the solution.
Buffer Attack
0.150 L × 0.150M = 0.225mole methylamine
1 point for the correct number of moles of HCl
added
0.120 mole methylammonium chloride
0.100 L × 0.950M = 0.095 mole HCl
K a = 2.27 × 10−11
[ Acid ] = (2.27 × 10−11 ) [0.120+0.095] = 3.75 × 10−11
⎡⎣ H 3O + ⎤⎦ = K a
[Base]
[0.225-0.095]
pH = − log(3.75 × 10−11 ) = 10.425
1 point for the correct pH
(e) Draw the Lewis structure for
i. the weak base methylamine, CH3NH2.
H
H
H
C
N
H
H
1 point for the correct Lewis structure; lone pair must be present on N
ii. methylammonium ion, CH3NH3+.
+
H
H
H
C
N
H
H
1 point for the correct Lewis structure; positive charge must be included for
credit.
H
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Buffers and Titrations
Free Response
3. A student performed a titration by dissolving a sample of a weak, monoprotic acid, HA, in 15.00 mL of water. The
solution was then titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH
solution had been added. The student’s data from the titration are shown in the table below.
Volume of 0.100 M
NaOH Added
0.00 mL
5.00 mL
10.00 mL
15.00 mL
20.00 mL
pH
2.22
3.44
3.92
8.13
(a) Calculate the number of moles of unknown acid originally present in the solution.
(0.100 M NaOH)(20.00 mL) = 2.00 mmoles OH−
Therefore there are 2.00 mmol of HA = 2.00 × 10−3 moles HA
1 point for the correct number of moles
of NaOH used to reach equivalence
point
1 point for the number of moles of
NaOH = HA
(b) Using the information provided above, calculate the value of the acid dissociation constant, Ka , for HA.
[H 3O + ] = K a
[Acid]
[Base]
1
equivalence (10.00 mL) [Acid]=[Base] ∴ [H 3O + ] = K a
2
pH = 3.44 ∴ [H 3O + ] = 10−3.44 = 3.63 × 10−4 = K a
at
®
1 point for correct set up Either using
[Acid]
[x ]2
[H 3O + ] = K a
or K a =
[M − x ]
[Base]
and calculating the initial M of HA
1 point for the correct calculation of Ka
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Buffers and Titrations
(c) Determine the pH of the solution after 5.00 mL of 0.100 M NaOH had been added, and record the value in the
student’s data table.
At the 5.00 mL point the solution formed has not yet reached the
equivalence point, thus is a buffer!
[Acid]
[H 3O + ] = K a
and
[Base]
moles of OH− added = moles of A− made = moles of HA used up
1 point for the correct calculation
number of moles of NaOH added
and relating that to number of
moles of HA used up and A−
formed
(5.00 mL)(0.100M) = 0.500 mmoles OH− added thus;
1 point for the correct pH
calculation
0.500 mmoles A− made and 0.500 mmoles HA used up
2.00 mmol HA originally − 0.500 mmol used up = 1.5 mmol HA left
[H 3O + ] = 3.63 × 10−4
[1.50]
= 1.09 × 10−3
[0.50]
pH = 2.96
(d) From the table below, select the most appropriate indicator for the titration. Justify your choice
Indicator
Methyl Red
Bromothymol Blue
Phenolphthalein
pH
5.5
7.1
8.7
Phenolphthalein because it changes color at the pH closest to the
equivalence point pH
®
1 point for the indicator and
justification
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Buffers and Titrations
(e) Using the axes provided below, sketch the titration curve that would result if the student had used 0.200 M NaOH
instead of the 0.100 M used above, to perform the titration. The equivalence point must be clearly marked.
10
pH
8
6
4
2
0
0
10
20
Volume NaOH added (mL)
Curve must have the following features:
Initial pH around 2.22
1 point for any 2 features
2 points for all 4 features
Vertical slope at 10.00 mL
Equivalence pH > 8.13 (not equal)
Shape of curve must resemble titration curve
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30
Buffers and Titrations
Free Response
HA (aq) + H2O(A) ⇌ H3O+(aq) + A− (aq)
Ka = 4.5×10−5
4. An unknown monoprotic acid HA, is a weak acid that ionizes according to the reaction above. A 50 mL sample of
the 0.020 M unknown weak acid solution described above was titrated with 0.050 M NaOH
(a) Write a balanced net ionic equation showing the reaction between the weak acid, HA and NaOH.
1 point for the correct equation. Must be in net ionic
form or no credit is awarded.
HA + OH− → H2O + A−
(b) For the titration, calculate the
(i)
equivalence point of the titration
(50 mL)(0.020 M) = 1.0 mmol HA = 1.0 mmol NaOH
1 point for correct equivalence point
1.0 mmol NaOH
= 20 mL to reach equivalence
0.050M
(ii)
pH of the solution in the titration flask after 5.0 mL of 0.050 M NaOH were added
Buffer Zone!
(5.0mL)(0.050 M ) = 0.25mmolOH − added
1 point for number of moles
(mmoles) of OH− added
0.25mmolOH − added = 0.25mmol A − formed = 0.25mmol HA reacted
[Acid]
[Base]
[1.0 mmol HA initial − 0.25 mmol reacted]
[H 3O + ] = 4.5 × 10−5
[0 mmol A − initial + 0.25 mmol formed ]
[0.75]
[H 3O + ] = 4.5 × 10−5
= 1.4 × 10−4
[0.25]
[H 3O + ] = K a
1 point for correct pH
pH = − log (1.4 × 10−4 ) = 3.85
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Buffers and Titrations
(iii) pH of the solution in the titration flask after 20.0 mL of 0.050 M NaOH were added
Equivalence Point!
1.0 mmol HA = 1.0 mmol NaOH = 1.0 mmol of A−
1.0 mmol A −
= 0.14 M A −
70 mL
1.0 × 10−14
Kb =
= 2.2 × 10−10
−5
4.5 × 10
x2
Kb =
M
x2
2.2 × 10−10 =
0.14
−6
x = 5.6 × 10
1 point for correct Kb conversion
pOH = − log (5.6 × 10−6 ) = 5.25
pH = 14 − 5.25 = 8.75
(iv)
1 point for [A−]
1 point for correct pH
pH of the solution in the titration flask after 30.0 mL of 0.050 M NaOH were added
Beyond Equivalence Point!
(30.0 mL)(0.050 M) = 1.5 mmol NaOH added
To get to equivalence 1.0 mmol NaOH reacted
1 point for [OH−] or NaOH in excess
0.50 mmol NaOH in excess = 0.50 mmol OH−
1 point for M of NaOH or OH− in
excess
0.50 mmol OH −
= 7.1 × 10−3 M OH −
70 mL
pOH = − log (7.1 × 10−3 ) = 2.15
pH = 14 − 2.15 = 11.85
1 point for correct pH
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Buffers and Titrations
Multiple Choice
1. If 50.0 mL of 0.25 M NH3 is titrated with 0.50 M HCl in a flask. Which of the following are present in
the flask at the equivalence point?
I. ammonia
II. hydrochloric acid
III. ammonium ions
(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
B
At the equivalence point of a titration, only the salt of the weak species and water remain.
For a reaction between NH3 and HCl the following occurs:
NH3 + HCl → NH4+ + Cl− + H2O
From the list, only ammonium ions are present.
2. A student titrates 25 mL of 0.50 M NH3 with 0.25 M HCl. Which of the following would be true about
the equivalence point of the titration?
(A) 25 mL of HCl is needed to reach the equivalence point.
(B) At the equivalence point, only the hydrochloric acid remains in the solution.
(C) At the equivalence point, the [OH−] is equal to the base ionization constant.
(D) There is no change in the pH of the solution after the equivalence point is passed.
(E) The pH at the equivalence point is less than 7.
E
At the equivalence point of a titration between a weak base and a strong acid, the salt that
remains will hydrolyze water and increase the concentration of hydronium ions (i.e. it will
be an acidic salt) therefore the pH of the solution is less than 7
3. Which pair of 1.0 M solutions below would form a buffer solution with a pH less than 6? (assume 1:1
ratios of each component)
(A) CH3NH2 and CH3NH3Cl
(B) HCl and NaOH
(C) HC2H3O2 and NaC2H3O2
(D) HF and HC2H3O2
(E) NH3 and CH3NH2
C
For a buffer solution with a pH less than 7 the components must be a weak acid and its salt.
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Buffers and Titrations
4. What is the pH of a buffer that is 0.30 M NH3 and 0.20 M NH4Cl? Kb for NH3 = 1.8 x 10−5
(A) 4.92
(B) 9.08
(C) 10.98
(D) 11.37
(E) 13.24
1.0 × 10−14
= ~ 0.5 × 10−9 or 5 × 10−10
1.8 × 10−5
[0.20] =~ 3.0 × 10−10
⎡⎣ H 3O + ⎤⎦ = 5 × 10−10
[0.30]
Ka =
B
pH = − log(3.0 × 10−10 ) = ~ 9.x
5. When used in a titration which ion below turns purple when the endpoint is reached?
(A) I−
(B) Cl−
(C) Cr2O72−
(D) CrO42−
(E) MnO4−
E
Permanganate ions produce purple solutions.
6. 200.0 mL of 0.25 M HC2H3O2 is titrated with 0.95 M Ba(OH)2. What volume of barium hydroxide is
needed to reach the equivalence point? Ka acetic acid = 1.8 x10−5
(A) 26 mL
(B) 40 mL
(C) 53 mL
(D) 1.0 x 102 mL
(E) 2.0 x 102 mL
A
(200)(0.25) = 50 mmol acid = 50 mmol base
50 mmol base
= ~ 25 mL
~1 M × 2OH −
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Buffers and Titrations
Questions 7 and 8 refer to the following titration data.
In order to reach the equivalence point (pH = 5.36), it takes 50.0 mL of 0.10 M HCl to titrate 100.0 mL of a
weak base. During the same titration, the halfway point (pH = 9.25) is reached with the addition of only
25.0 mL of 0.10 M HCl.
7. What is the Kb of the weak base?
(A) 6 x 10−10
(B) 2 x 10−8
(C) 4 x 10−6
(D) 2 x 10−5
(E) 5 x 10−2
1
equivalence [H 3O + ] = K a
2
pH = 9.25 ∴ [H 3O + ] = 10−9.25
at
A
[H 3O + ] =K a = ~ 10−10
8. Which indicator would be best to use in this titration?
(A) Phenolphthlein, Ka = 5.0 x 10−10
(B) Phenol red, Ka = 1.3 x 10−8
(C) Litmus, Ka = 1.0 x 10−7
(D) Bromophenol Blue, Ka = 7.9 x 10−6
(E) Thymol Blue, Ka =1.0 x 10−2
D
The equivalence point is at pH = 5.36; thus the Ka of the indicator needs to be close to 10−5
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*
AP CHEMISTRY
The Lab Based Question
Techniques, Types, and Tricks
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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THE LAB BASED QUESTION
Techniques, Types, and Tricks
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions over lab based problems, techniques, and situations.
Standards
The topic of laboratory chemistry is addressed in the topic outline of the College Board AP
Chemistry Course Description Guide as described below.
V. Laboratory
The differences between college chemistry and the usual secondary school
chemistry course are especially evident in the laboratory work. The AP Chemistry
Exam includes some questions based on experiences and skills students acquire in
the laboratory:
• making observations of chemical reactions and substances
• recording data
• calculating and interpreting results based on the quantitative data obtained
• communicating effectively the results of experimental work
AP Chemistry Exam Connections
The topic of laboratory chemistry is tested every year on the multiple choice and is the focus of
question one on the free response portion of the exam. The list below identifies free response
questions that have been recently asked. These questions are available from the College Board
and can be downloaded free of charge from AP Central http://apcentral.collegeboard.com.
2008
2007
2006
2005
2004
2003
2002
2001
2000
1999
Free Response Questions
2008 B Question 5
2007 B Question 5
2006 B Question 5
2005 B Question 5
2004 B Question 5
2003 B Question 5
2002 B Question 5
Question 2
Question 5
Question 5
Question 5
Question 5
Question 5
Question 5
Question 6
Question 5
Question 5
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THE LAB BASED QUESTION
Techniques, Types, and Tricks
What I Absolutely Have to Know to Survive the AP Exam
If asked to do the following, it might indicate the question deals with laboratory questions:
Design an experiment; list measurements needed; show setup of calculations needed, use sample data to do calculations;
interpret or draw graphs; explain the affect of error on the calculated value; use qualitative observations…
Parts of a Lab Question
You may not have done that exact experiment but you can use the knowledge gained doing other experiments to help
account for certain observations.
You may be asked to describe how to do an experiment or to design an experiment.
• Materials: If they give you a list of equipment, don’t think you have to use all of it.
• Procedure: Be sure to include important techniques like rinsing the buret with the solution before a titration or
heating to constant mass.
• Data needed: The data needed are values that can be measured like initial and final temperatures. Writing all the
mathematical equations needed to do the calculations will help you determine what data is needed.
• Calculations: A calculation is using what was measured like temperature change. Show the set up of the
mathematical equations required for the calculations. Use sample data when appropriate.
• Graphs: Be sure you label the axes and other important points on your graph.
• Error Analysis: State whether the quantity will be too high, too low, or no change. Use equations to help you
determine what change will occur and to support your answer.
Common Lab Procedure: Calorimetry
Calorimetry is used to determine the heat released or absorbed in a chemical reaction. A calorimeter can determine the
heat of a solution reaction at constant pressure.
Techniques:
• Use a double Styrofoam cup with a plastic top and hole for the thermometer
• Determine the change in temperature accurately
• Measure solution volumes precisely
• Start with a dry calorimeter
Information to know about calorimetry:
• Heat capacity (C) = the amount of heat needed to raise the temperature of an object by one degree Celsius or
Kelvin, J/°C or J/K.
• The heat capacity of 1 mol of a substance is called its molar heat capacity (Joules per mole per degree) J/mol⋅°C
or J/mol⋅K.
• Specific heat, c, also known as specific heat capacity, is defined as the amount of heat necessary to raise the
temperature of 1.00 g of a substance by one degree. Units are (joules per gram per degree), J/g⋅°C or J/g⋅K. You
often use the specific heat capacity in analyzing gathered data then convert to molar heat capacity.
Assumptions often made during calorimetry: (Be able to answer error analysis questions about each assumption below)
• The density of dilute solutions is the same for water. D= 1.0 g/mL
• The specific heat of the solutions is the same as that for water. c = 4.184 J/goC
• The solutions react in their stochiometric amounts.
• There is no loss of heat to the surroundings.
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The Lab Based Question
Equations:
• q = mcΔT (c = 4.184 J/g°C)
• q = ΔH = mcΔT , at constant pressure
• Use the density of the solution to convert from volume to mass for q = mcΔT
Common Lab Procedure: Titration
A titration is a laboratory procedure for quantitative analysis. In a titration two reagents are mixed, one with a known
concentration & known volume [or a solid with a known mass] and one with an unknown concentration. There is some
way to indicate when the two reagents have reacted completely (typically an indicator), and at the end of the titration the
unknown solution's concentration can be calculated since you have accurately determined the volume of that solution
required to complete the reaction.
Terms to know:
• Titrant – A solution of known concentration; it is often standardized
• Standardized solution – A solution in which the exact concentration is known.
• Indicator - A weak acid or base used in a titration to indicate the endpoint has been reached.
• Equivalence point – moles of acid = moles of base; point at which enough titrant has been added to completely
react with the solution being analyzed.
• End point – the point at which the indicator changes color; important to pick an indicator with a pKa very close to
the pH at the equivalence point.
Techniques:
Preparing the Buret:
1. Rinse a clean buret with distilled water and then the titrant (the solution that will be added to the flask).
2. Allow the titrant to drain through the buret so that the tip gets rinsed with titrant as well.
3. Discard the rinse solution. Fill it with the titrant. Remove air bubbles from the tip of the buret by draining several
milliliters of titrant.
Preparing the Sample:
4. Pipet the desired volume of the solution to be analyzed into an Erlenmeyer flask. Record the exact volume. If the
sample is a solid, weigh the desired mass, add the solid to an Erlenmeyer flask, and dissolve it in distilled water
(the amount of water does matter since it doesn’t change the moles of the solid). Be sure to record the exact mass
of sample used.
5. Add the titrant to the flask until the equivalence point is reached. Calculate the volume of titrant added.
6. Change in color of a chemical indicator is usually used to signal the endpoint of the titration. The endpoint for this
titration is reached when you reach a pale color that persists for several seconds.
Measured Data Required:
moles titrant = moles of substance analyzed @ equivalence point
ƒ mass of DRY substance analyzed OR accurately measured volume of solution analyzed [measure
with a pipet OR buret ]
ƒ initial volume of titrant (substance of known molarity) and final volume of titrant (required to
reach end point)
ƒ Molarity of titrant
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The Lab Based Question
Calculation Hints:
ƒ
ƒ
Substance analyzed is solution/liquid:
• M1V1 = M2V2 @ equivalence point
• Volume of titrant used to reach end point [difference between final and initial volumes]
• MtitrantVof titrant added = moles of titrant = moles of unknown
moles of unknown
• Molarity of unknown = Molarity of unknown =
Liters of unknown
Substance analyzed is solid:
• Same as process as above
• MtitrantVof titrant added = moles of titrant = moles of unknown
•
Molecular weight of the unknown =
mass of solid dissolved
moles of unknown
Graphs:
Common Lab Procedure: Gravimetric Analysis
One method for determining the amount of a given substance in solution is to form a precipitate that includes the
substance. The precipitate is then filtered and dried. This process is called gravimetric analysis.
Techniques/Procedure:
1. Weigh sample
2. Form precipitate
3. Filter precipitate (A buchner funnel and aspirator can be used)
4. Dry precipitate (Be sure to dry to constant mass)
5. Weigh precipitate
For example if we wanted to determine the amount of chloride ions present in a given solid, we would weigh the solid
sample, dissolve the sample in water, add an excess of silver nitrate solution to form the precipitate silver chloride. This
precipitant would be filtered, and dried to constant mass. From the mass of silver chloride formed, we can determine the
moles of silver chloride and the moles of chloride ion in the original sample.
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The Lab Based Question
Common Lab Procedure: Determining Molar Mass
molar mass =
•
mass of sample in grams
moles of sample
Organize answer around calculations—paying special attention to what quantities are measured versus
calculated!
A. Freezing point depression:
• ΔTf = mikf (nonelectrolyte i = 1 (most common); i is the van’t Hoff factor and is equal to the # of ions
released);
moles of solute
mass solute
• m=
or m =
molar
mass solute
kg of solvent
kg solvent
• Measured DATA:
o FP of pure solvent
o final FP of solution
o kg of solvent
o grams of solute
• Constants needed:
o kf = given OR determined by data (for H2O = 1.86)
• Calculations required:
o ∆T = freezing point pure solvent – freezing point of solution
o use ΔTf = mikf to solve for molality
o use molality equation to solve for molar mass
•
Graphs:
Temperature
20
15
10
5
Time
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The Lab Based Question
B. Titration Data
• moles titrant = moles of substance analyzed @ equivalence point
• Measured DATA required:
o mass of substance
o initial volume of titrant (substance of known molarity) and final volume of titrant (required to reach end
point)
o Molarity of titrant
• Calculations required:
o Molarity of titrant
o Substance analyzed is solid:
o MtitrantVtitrant = moles titrant
o
molar mass =
g of solid analzyed
moles of titrant used
c. Vaporization of a Volatile Liquid
• PV = nRT used to determine moles
• Measured DATA Required:
ƒ Pressure = atmospheric pressure unless collected over water
ƒ initial mass of flask
ƒ final mass of flask
ƒ Temperature of boiling water—don’t assume 100°C
ƒ Volume of gas = fill flask with water and measure the volume of water in a graduated cylinder
• Constants needed:
ƒ If collected over water, the water vapor pressure at the experimental temperature.
• Calculations Required:
ƒ If the gas was collected over water Pvapor = Patmospheric/baraometric - Pwater vapor at certain temperature
ƒ mass of sample = final mass of flask [includes vapors] – initial mass of flask
Common Lab Procedure: Colorimetric or spectrophotometric analysis
Colorimetric analysis is a quantitative analysis of a solution using color based on Beer’s Law. Colorimetric analysis can
be used to determine the concentration of an unknown solution, the rate constant of a reaction, the order of a reaction,
etc.
Beer’s Law is an expression than can be used to determine how much light passes through the solution. It also shows
that concentration and absorbance are directly related.
A = ε bc
•
•
•
•
A = absorbance (measured with a colorimeter)
ε = molar absorptivity (how much light will be absorbed by 1 cm of a 1 M solution of the chemical)
b = path length of the cuvette in cm
c = concentration in molarity
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The Lab Based Question
Lab Based Cheat Sheet
Relationships
Beer’s Law A = ε bc
Calorimetry q = ΔH = mcΔT at constant pressure
• c = specific heat
• A = absorbance
• Δt = change is temperature
• ε = molar absorptivity
• m = mass
• b = path length of the cuvette in cm
• q = heat
• c = concentration in molarity
Freezing point depression ΔTf = mikf
•
•
•
•
Ideal Gas Law PV = nRT
i = 1 for nonelectrolytes (most common)
∆T = freezing point pure solvent – freezing point
of solution
kf = molal freezing point constant
mass solute
m=
molar mass solute
kg solvent
molar mass =
%error =
mass of sample in grams
accepted
V = volume in liters
n = moles
T = temperature in Kelvin
P = pressure
If collected using water displacement:
Pvapor = Patmospheric/baraometric − Pwater vapor at certain temperature
@ Equivalence point
moles of sample
accepted - experimental
•
•
•
•
M1V1 = M2V2
× 100
Connections
Electrochemistry: draw diagram of galvanic or
electrolytic cell, qualitative observations that can be made
at the cathode and anode, etc.
Thermodynamics: heat of reaction, molar heat capacity,
etc.
Stoichiometry: empirical formula of a compound, percent
of an element in a compound, etc.
Acids and Bases: standardizing a solution, drawing a
titration curve, etc.
Qualitative Analysis: identifying a compound based on
observations and tests
Potential Pitfalls
Be aware there are quantities you measure [such an initial temperature and final temperature or initial pressure and
final pressure, etc.] and terms you calculate using what you measured such as Δ T or Δ P
Be sure to include important steps in the procedure like “heat to constant mass,” “rinse the buret with distilled water
and then the solution before titrating”, “dissolve the solid in about 100 mL of water and then add water to the 500 mL
mark on the volumetric flask,” etc.
Writing the mathematical equations may help determine how an error affects the results. Use the equations to justify
your error analysis.
If given a laboratory situation you have not specifically done, use the observations and the concepts you learned in
your labs throughout the year to reason your way through the lab question.
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The Lab Based Question
Free Response
1. Hydrochloric acid (about 5 mL) is placed at the bottom of the test tube and then carefully filled with distilled water so
not to disturb the HCl. A piece of magnesium is placed on top of the distilled water and then the test tube is inverted
into a beaker with distilled water. The magnesium is allowed to react completely. The gas is collected in the test tube
at room temperature. Before the test tube with the gas is removed, the water levels inside and outside the test tube are
the same. The chemicals and lab equipment used are listed below.
Hydrochloric acid
Table of water vapor pressures
Distilled water
Barometer
Graduated cylinder
600 mL beaker
Balance
Strip of magnesium metal
Test tube
Thermometer
(a) Write the balanced reaction that occurs when hydrochloric acid reacts with magnesium.
2Mg + 2HCl → MgCl2 + H2
1 point for the correct balanced
reaction
(b) What measurements are needed to calculate the molar volume of the gas at STP?
Mass of the magnesium
Temperature of water
Barometric pressure
Volume of water after the reaction
Total volume of test tube
2 points for all correct
measurements. 1 point for any three
measurements
(c) Show the setup for the calculations needed to determine:
(i)
moles of Mg =
the moles of gas produced
1 point for both correct calculations
mass Mg
molar mass of Mg
moles of H2 = moles of Mg ×
1 mol of H2
1 mol Mg
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The Lab Based Question
(ii) the molar volume of the gas at STP
Volume of H2(L) =
total volume of test tube(mL) - volume of liquid after reaction(mL)
1000
2 points for all the correct
calculations. 1 point for any two
correct calculations
PH2 = Pbarometric - Pwater vapor
VSTP =
PH2VH2TSTP
TH2PSTP
molar volume of H2 =
moles of H2 produced
volume of H2 at STP
(d) What test can be done to prove which gas is produced?
A glowing splint will “pop” when placed at the mouth of the
test tube since hydrogen gas is produced.
1 point for correct test
(e) What is the purpose of making the water level inside the test tube equal to the water level in the beaker?
To ensure that the barometric pressure is equal to the total
pressure in the test tube.
1 point for correct purpose
(f) The vapor pressure of water is not used in the calculation. How will the molar volume of the gas at STP be
affected? (higher, lower or the same) Explain.
If the water vapor pressure is not accounted for, the
pressure of the hydrogen gas will be higher making the
volume of H2 at STP higher. Therefore, the calculated
molar volume of H2 gas will be smaller.
®
1 point for correct justification
1 point for stating the molar volume will be
smaller
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The Lab Based Question
Free Response
2. A solution of calcium hydroxide is titrated with hydrochloric acid.
(a) Describe the steps needed to make 50.0 mL of a 0.050 M solution from a 2.00 M solution of hydrochloric acid
using a dropper, 5.0 mL pipet, 50.0 mL volumetric flask, and distilled water.
M1V1 = M2V2
1 point the correct volume of hydrochloric acid needed
(50.0 mL) × (0.050 M) = (2.00 M) × (V2)
1 point for the correct steps
V2 =
(50.0 mL) × (0.050 M)
(2.00 M)
= 1.25 mL
Put about 20 mL of distilled water into a 50.0 mL
volumetric flask. Measure out 1.25 mL of 2.00 M
HCl with the pipet. While swirling, add the 2.00
M HCl to the volumetric flask. Mix the solution.
Using the dropper, add enough distilled water to
fill the volumetric flask to the 50.0 mL mark and
mix.
(b) Write the balanced chemical equation for the dissociation of Ca(OH)2(s) in aqueous solution, and write the
equilibrium-constant expression for the dissolving Ca(OH)2.
Ca(OH)2 ⇌ Ca2+ + 2 OH−
1 point for correct chemical equation
Ksp = [Ca 2 + ] [OH - ]2
1 point for correct Ksp expression
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The Lab Based Question
A 15.0 mL saturated solution of Ca(OH)2 is titrated with a standardized 0.050 M solution of hydrochloric acid. It takes
10.50 mL of hydrochloric acid to reach the equivalence point where the pH is 8.50.
(c) Sketch the titration curve that shows the pH change as the volume of hydrochloric acid added increases from 0 to
16.0 mL. Label the equivalence point.
1 point for starting the titration curve at a basic pH and
for a gradual pH drop then a sharp drop in pH
1 point for correctly labeling the equivalence point
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The Lab Based Question
(d) Calculate the concentration of [OH]− from the titration.
1 point for the correct hydroxide concentration
mol H + = (0.0500 M) × (0.01050 L) =
+
mol H + = 5.24 × 10 - 4 = mol H = mol OH -
[OH − ] =
5.24 × 10 −4
= 3.49 ×10 − 2 M
0.0150 L
(e) Calculate the Ksp for calcium hydroxide.
R Ca(OH)2 (s) ↔ Ca 2 + + 2OH I
0
0
C
+x
+ 2x
E
+x
+ 2x
x=
1 point for the correct Ksp value consistent part d.
3.49 ×10 −2
= 1.75 × 10 − 2
2
Ksp = [Ca 2 + ] × [OH - ]2 = ( x) × (2 x) 2 = 4 x 3
Ksp = 4 x 3 = (4) × (1.75 × 10−2 )3 = 2.13 × 10−5
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The Lab Based Question
Free Response
3. A molecular compound is analyzed and found to be 76% C, 12.6 % H, and 11.2 % O.
(a) What is the empirical formula of the molecular compound?
1 point for moles of C, H, and O
moles C =
76.0 g C
= 6.33 mol ÷ 0.700 = 9.00
12.0 g C
moles H =
12.6 g H
= 12.5 mol ÷ 0.700 = 18.0
1.01 g H
moles O =
11.2 g O
= 0.700 mol ÷ 0.700 = 1.00
16.0 g O
1 point for the correct empirical formula
Empirical Formula: C9H18O
The freezing point of the pure solvent, lauric acid, is found to be 44.0 °C. A nonvolatile molecular compound is mixed
with lauric acid. The freezing point of a mixture of 8.11 grams of lauric acid and 1.12 grams of the organic compound is
found to be 42.1 °C. The freezing point molal constant for stearic acid is 4.9 °C/m, and the freezing point molal constant
for lauric acid is 3.9 °C/m.
(b) Calculate the molality of the mixture of lauric and and the unknown compound.
ΔTf = mikf
m=
m=
1 point for the correct molality
ΔTf
ikf
(44.0 C − 42.1 C)
o
o
(3.9 o C/m)(1)
m = 0.49 m
(c) Calculate the molar mass of the unknown compound.
1 point for the correct molar mass consistent with
part b.
mol solute
m=
kg solvent
mol solute = (m) × (kg solvent)
mol solute = (0.49 m) × (0.00811 kg )
mol solute = 0.0040 mol
mass
1.12 g
molar mass =
=
= 283 g/mol
moles 0.0040 mol
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The Lab Based Question
(d) Determine the molecular formula of the unknown compound.
1 point the correct molecular formula consistent with
part a and c.
MF mass 283
=
=2
EF mass 142
2(C9H18O) = C18H36O2
(e) If the same experiment is conducted but with an ionic compound instead of a molecular compound as the solute,
how would the molar mass of the compound be affected? (higher, lower or no change) Explain.
If an ionic compound is used the van Hoffmann
factor will be greater than 1. Therefore, the
molality will be lower making the moles of solute
to be lower. Thus, the molar mass will be higher.
m=
1 point for higher molar mass
1 point for justification
ΔTf
ikf
mol solute = (m) × (kg solvent)
molar mass =
mass of sample in grams
moles of sample
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The Lab Based Question
Multiple Choice
1. Which ion below would be colorless when in solution?
(A) Zn2+
(B) Fe3+
(C) Cu2+
(D) Mn2+
(E) Ni2+
In general, ions with unpaired electrons make colored solutions. The electrons in Zn2+ are
all paired. Therefore, the solution would be colorless. The other solutions all have at least 1
unpaired electron.
A
2. When solid LiCl is heated in a flame, what is the color of the flame?
(A)
(B)
(C)
(D)
(E)
Yellow
Orange
Red
Purple
Green
C
When LiCl is heated in a flame test, the color of the flame will be red.
3. A student heats a test tube containing a volatile liquid in a warm water bath and then condenses the gas using a
cold water bath. The data obtained is used to calculate the molar mass of the volatile liquid. Which measurements
below are needed to determine the molar mass of the volatile liquid?
I. Temperature of the gas
II. Mass of the condensate
III. Barometric pressure
IV. Vapor pressure of water
(A)
(B)
(C)
(D)
(E)
II only
I and III only
II and III only
I, II, and III only
I, II, III, and IV
molar mass =
B
mass of sample in grams
moles of sample
moles =
PV
RT
In order to calculate the molar mass of the volatile liquid, the moles of the sample is needed.
The measurements needed to calculate the moles of the sample are barometric pressure,
volume of test tube, and temperature. The gas is not collected over water; therefore, the
vapor pressure of water is not needed. The mass of the condensate is needed, but it is a
calculation rather than a measurement.
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The Lab Based Question
4. 10.00 mL of acetic acid is titrated with 0.100 M sodium hydroxide. If an indicator is used to signal the endpoint of
the titration, which indicator below is best to use?
(A)
(B)
(C)
(D)
(E)
Methyl red, Ka = 2.0 × 10 −4
Litmus, Ka = 3.2 × 10 −7
Thymolphthalein, Ka = 2.4 × 10 −10
Crystal violet, Ka = 7.9 × 10 −2
Bromothymol blue, Ka = 5.6 × 10 −8
C
When a weak acid is titrated with a strong base, the pH at the equivalence point is greater
than 7. The best indicator to use is the one that has pKa equal to the pH at the equivalence
point. The indicator that will change colors in the basic region is thymolphthalein.
5. Which solution would produce a precipitate with the addition of hydrochloric acid?
(A)
(B)
(C)
(D)
(E)
Ca(NO3)2
[Ag(NH3)2]NO3
Fe(C2H3O2)2
AlF3
Cr(ClO4)2
When an acid is added to a silver containing complex ion, silver chloride precipitate forms.
B
+
⎡⎣ Ag ( NH3) ⎤⎦ + H + +Cl− → AgCl+NH4 +
Calcium chloride, iron (II) chloride, aluminum chloride, and chromium (II) chloride are all
soluble products.
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The Lab Based Question
6. The concentration of a solution is determined using a spectrophotometer and Beer’s Law.
Beer’s Law:
A = ε bc
Which of the following statements are true about Beer’s Law?
I. The absorbance is directly related to the concentration.
II. The ε in Beer’s Law is the molar absorptivity constant.
III. As the absorbance of a solution increases, the percent transmittance decreases.
IV. The path length and the percent transmittance are indirectly related.
(A) I only
(B) I and III only
(C) I and II only
(D) I, II, and III only
(E) I, II, III, and IV
Beer’s Law A = ε bc
A = absorbance
ε = molar absorptivity
b = path length of the cuvette in cm
c = concentration in molarity
E
As the concentration goes up, the absorbance goes up. Thus, concentration and absorbance
are directly related. As the absorbance of light of a solution increase, the percent of light
transmitted would decrease. Thus, absorbance and transmittance are indirectly related. As
the path length of the cuvette increases, the absorbance increases and the transmittance
decreases. Therefore, all the statements are true.
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The Lab Based Question
7. An unknown gas in a vial is collected in a graduated tube over water.
Temperature of water
Atmospheric pressure
Mass of vial before gas collected
Mass of vial after gas collected
Volume of the gas collected
Water vapor pressure at 20ºC
20.0ºC
744.8 mm Hg
25.63 g
25.20 g
125.0 mL
17.54 mm Hg
Using the data above, which expression below can be used to calculate the molar mass of the gas?
(A)
(25.63 - 25.20) × (0.0821) × (293) × (760) × (1000)
(125) × (744.8)
(B)
(25.63 - 25.20) × (0.0821) × (293) × (1000)
(125) × (744.8 − 17.54)
(C)
(D)
(E)
(25.63 - 25.20) × (125) × (293) × (760)
(0.0821) × (744.8 − 17.54) × (1000)
(25.63 - 25.20) × (0.0821) × (293) × (760) × (1000)
(0.125) × (744.8 − 17.54)
(25.63 - 25.20) × (0.0821) × (293) × (1000)
(125) × (17.54)
PV = nRT
D
PV =
mass × R × T
MW
MW =
MW =
mass × R × T
Pgas × V
(25.63 - 25.20) × (0.0821) × (293) × (760) × (1000)
(125) × (744.8 − 17.54)
8. A student separates a mixture using distillation. Distillation can be used to separate the mixture because the two
substances have different
(A) melting points.
(B) reactivities.
(C) densities .
(D) boiling points.
(E) viscosities.
D
Distillation is a separation technique used to separate mixtures based on their boiling points.
The mixture is heated and the solution with the lower boiling point will vaporize first. The
vapor is then condensed as it is cooled in the condenser. The vapor is then captured into a
flask.
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The Lab Based Question
9. Which of the following is true about the oxidation-reduction reaction below?
2 I − + Cl 2 → I 2 + 2 Cl −
(A) I- is being reduced by Cl2.
(B) The reducing agent is the Cl2.
(C) The solution will be purple after the reaction.
(D) The half reaction that occurs at the anode is Cl2 + 2e − → 2Cl− .
(E) The voltage will increase if additional I2 is added.
C
As the reaction occurs, the Cl2 gas is used up and produces chloride ions and iodine. The
iodine makes the solution purple. The I- is being oxidized since the oxidation number is
changing from -1 to 0. The Cl2 is the oxidizing agent since it is being reduced from an
-
oxidation number of 0 to -1. The half reaction that occurs at the anode is 2I - → I 2 + 2e - .
The voltage will decrease if additional I2 is added because the reaction will shift to the left,
decreasing the voltage.
10. Which mixture of substances below will not react?
(A) Solutions of HI and CaCO3
(B) Solutions of (NH4)2SO4 and KOH
(C) Solutions of SnCl2 and KMnO4
(D) Solution of CaO and H2O
(E) Solution of CaCl2 and Br2 gas
E
CaCl2 and Br2 will not react because bromine is less reactive than chlorine.
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*
AP CHEMISTRY
Net Ionic Equations
Notes, Hints, and Practice
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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Net Ionic Equations
Notes, Hints, and Practice
Objective
To review the student on the types of reactions, processes and strategies necessary to successfully
answer questions over reaction chemistry and the net ionic equations free response problem on
the exam.
Standards
The topic of acids, bases and salts is addressed in the topic outline of the College Board AP
Chemistry Course Description Guide as described below.
III Reactions
A. Reaction types
1. Acid-base reactions
2. Precipitation reactions
3. Oxidation-reduction reactions
a. Oxidation number
b. The role of the electron in oxidation-reduction
IV. Descriptive Chemistry
Knowledge of specific facts of chemistry is essential for an understanding of principles and concepts.
These descriptive facts, including the chemistry involved in environmental and societal issues, should
not be isolated from the principles being studied but should be taught throughout the course to illustrate
and illuminate the principles. The following areas should be covered:
1. Chemical reactivity and products of chemical reactions
3. Introduction to organic chemistry: hydrocarbons and functional groups (structure, nomenclature,
chemical properties)
AP Chemistry Exam Connections
The topic of chemical reactions is tested every year on the multiple choice and is the focus of
question four (4) on the free response portion of the exam. The list below identifies free response
questions that have been recently asked. These questions are available from the College Board
and can be downloaded free of charge from AP Central http://apcentral.collegeboard.com.
2008
2007
2006*
2005*
2004*
2003*
2002*
2001*
2000*
1999*
Free Response Questions
2008 B Question 4
2007 B Question 4
2006 B* Question 4
2005 B* Question 4
2004 B* Question 4
2003 B* Question 4
2002 B* Question 4
Question 4
Question 4
Question 4
Question 4
Question 4
Question 4
Question 4
Question 4
Question 4
Question 4
* − these questions reflect a different format than the current AP exam
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Net Ionic Equations
Notes, Hints, and Practice
1.
2.
3.
4.
5.
6.
General Ideas Regarding AP Equations
Know your solubility rules inside and out.
Reactions are net ionic so no spectator ions are to be included.
All reactions happen so “no reactions” are not possible.
Equations must be balanced.
Do not include states of matter.
Strong acids and bases need to be written in ion form. There are a couple exceptions to this but no worries.
Double Replacement Reactions
When trying to decide reaction type, try to determine if it’s a double replacement FIRST. If you can form any of the
following substances, you know you have a double replacement reaction:
A precipitate (an insoluble solid)
Water
A gas
A weak acid or a weak base
Reminder: Double replacement reactions are NOT redox reactions. Keep your oxidation states the same.
Note: Ammonium chloride, sulfurous acid, and carbonic acid are unstable.
NH4OH → NH3 + H2O
H2SO3 → SO2 + H2O
H2CO3 → CO2 + H2O
To receive full credit, make sure to decompose those unstable compounds totally as indicated in the reactions above.
Redox Reactions
When faced with a redox reaction consider the following:
1. Remember that something must go up in oxidation state/charge and something must go down.
2. Never include group 1 metal ions in the reaction unless included in a “solid”.
3. Do not include the nitrate ion unless it is in the form of nitric acid reacting with a metal.
4. “Go To Metals First!” Meaning…..try and make redox happen between metals and metal ions first. If you can,
then that is where the redox happens. There are a few exceptions to this but none you need to worry about.
5. Polyatomic ions will decompose to gases. If you can’t make redox occur between metals and a polyatomic is
present, a gas that comes from that ion will be part of the redox coupling.
A couple must know rules:
MnO4- in an acidic solution produces Mn2+
Cr2O72- in an acidic solution produces Cr3+
Single Replacement Reactions
Single Replacement reactions are just redox. The same rules apply.
These are actually easy to identify due to the presence of a single element trying to displace another from a compound.
Note: Be very careful when there is a metal being added to water. The most common student mistake is to produce a
metal oxide and hydrogen gas. This is WRONG!!! When adding a reactive metal to water you produce a metal
HYDROXIDE (base) and hydrogen gas.
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Net Ionic Equations
Combustion Reactions
You know you have a combustion reaction when something is burned in air or burned in oxygen.
Combustion reactions are also REDOX equations so be sure to look for oxidation number changes.
No memorization is required for these equations. Oxygen will just smack into all the elements in present with oxygen
always going down in oxidation state from a zero to a minus 2.
Addition Reactions
Some addition reactions are easy to identify. Those easy ones include two elements that are coming together to form one
product. These are also another type of REDOX reaction where oxidation state changes occur.
However one of the most common addition reactions is the strangest one which includes an acidic or basic anhydride.
OXIDES ARE WEIRD!
Non metal oxides are “fake acids” – acids without water. When you add water to one of these you make an acid.
Metal oxides are “fake bases” – bases without water. When you add water to one of these you make a base.
Say to yourself “NO acid MO base” (NO is nonmetal oxide and MO is metal oxide)
Neutralization Reaction
Simply put a neutralization reaction is a specialized double replacement reaction where an acid and a base make a salt
and water. These are NOT redox reactions so keep your oxidation numbers the same.
Be careful about addition a fake acid to a fake base (nonmetal oxide to a metal oxide). Since water is not present you will
produce just the salt! This makes it an addition reaction as well.
Also if you’re adding a fake acid (nonmetal oxide) to a real base or a fake base (metal oxide) to a real acid you will still
produce water as a product. One of the ions from the salt will be a component derived from the metal or non metal oxide
involved.
Here are a few odd ones that might be helpful to memorize remember; OXIDES ARE WEIRD!
Metallic oxides plus CO2 make metallic carbonates
Metallic oxides plus SO2 make metallic sulfites
Decomposition Reactions
Decomposition reactions are EASY to recognize. One reactant is decomposing to multiple products. Simple compounds
containing just two types of elements easily split into two separate elements. (Be carefully not to produce ions in the
decomposition unless water is present)
Some of you more difficult decomposition reactions contain polyatomics and can get a bit harder. Rather than trying to
memorize a billion rules, just remember: Polyatomics can decompose to gases. Knowing that will get you through
many of these problems.
Remember those unstable compounds from double replacement? They are decomposition reactions as well!
NH4OH → NH3 + H2O
H2SO3 → SO2 + H2O
H2CO3 → CO2 + H2O
One last one you should know:
Hydrogen peroxide decomposes to make water and oxygen
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Net Ionic Equations
Anhydride Reactions
Anhydride means “without water” so you are adding water as a reactant in these problems. Pretty easy to identify!
Just as we saw in addition reactions:
Non metal oxides are “fake acids” – acids without water. When you add water to one of these you make an acid.
Metal oxides are “fake bases” – bases without water. When you add water to one of these you make a base.
Say to yourself “NO acid MO base” (NO is nonmetal oxide and MO is metal oxide)
OXIDES ARE WEIRD!
A couple additional ones you’ll want to learn:
Metallic hydrides plus water produce metallic hydroxides plus hydrogen gas (same as a metal plus water right?)
Phosphorus halides plus water produce a hydrohalic acid plus a ternary acid of phosphorus
Complex Ion Reactions
How do you know if you’re doing a complex ion problem? You need to have a metal in the presence of a very high
concentration of a ligand.
Metal: “Fish Metals” Ag, Cu, Cd, Zn and Al (These metals form a diving fish on the periodic table…check it out!
You’ll need to use your imagination because it is a stretch!) Theoretically any transition metal can complex but these are
your most common.
Ligand: High concentration of OH- (hydroxy) or NH3(amine) or SCN- (thiocyano). Again, there are other possible
ligands but these are the most common.
The number of the ligands that attach to the metal in the complex is usually double that of the most common oxidation
state of that metal OR using the fish mentioned above, the nose of the fish (Ag+) can attach two ligands and the rest of the
fish can attach 4 ligands.
The overall charge of the complex is based on the combination of charges from both the metal and the ligand (if the
ligand is even charged).
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Net Ionic Equations
Organic Nomenclature Review
Alkanes
Alkanes only contain single bonds (sigma bonding) and have the general formula CnH2n+2.
The prefixes for alkenes that you need to know in combination with the suffix –ANE are listed for below:
1-carbon
methane
4-carbons butane
7-carbons heptanes
10-carbons decane
2-carbons
ethane
5-carbons pentane
8-carbons octane
3-carbons
propane
6-carbons hexane
9-carbons nonane
Cycloalkanes
These hydrocarbons exist in ring structures with general formula CnH2n. You name them just as alkanes but add the
prefix CYCLOAlkenes
These are compounds containing at least one carbon to carbon double bond with the general formula CnH2n. They use
the same prefixes as alkanes but end in an –ENE suffix.
Alkynes
These are compounds containing at least one carbon to carbon triple bond with the general formula CnH2n-2. They use
the same prefixes as alkenes but end in a –YNE suffix.
Nomenclature of Alkanes, Alkenes, and Alkynes:
1. Use generic prefix according to the number of carbons then add the correct suffix.
2. Select the longest continuous carbon chain to dictate the number of carbons for the correct prefix “parent chain”.
3. Any branch off of the parent chain is called a substituent group and it is named according to the number of
carbon atoms it contains with a “-yl” ending rather than “-ane”.
4. Specify the position of the branch on the parent by putting a number in front with the lowest #s possible.
5. For multiple substituents that are identical, use di-, tri-, tetra- etc, and repeat the numbers.
6. Numbers are separated by commas and from letters by dashes (1,1-dichloroethane).
7. Arrange substituents by name, alphabetically, regardless of numbers or complexity.
8. For alkenes and alkynes put a number right before the parent chain name, specifying position of the double or
triple bond. Select numbering which always gives the lowest value.
Functional Groups and Naming:
R OH (hydroxyl group)
Alcohols
• Alcohols are names by the alkyl prefix + “-ol”
H3C
OH
Ex) ethanol
O
Ethers
•
R
R'
Ethers are named by listing substituents with a “-yl” ending + “ether”
Ex) diethyl ether H3C
O
CH3
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Net Ionic Equations
O
R
H (terminal carbonyl, -CHO)
Aldehydes
•
Aldehydes are named by the alkyl prefix + “-al” ending
O
H3C
Ex) ethanal
H
O
Ketones
•
R
(carbonyl group)
R'
Ketones are named by alkyl prefix + “-one” ending
H3C
Ex) propanone
CH3
O
O
R
OH
Organic acids
•
(-COOH)
Organic Acids are named by alkyl prefix + “-oic acid” ending
O
H3C
Ex) ethanoic acid
OH
O
R
Esters
•
O
R'
Esters are named from the alcohol name with a “-yl” ending, then the acid name with an “-oate” ending
O
H3C
Ex) methyl ethanoate
O
CH3
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Net Ionic Equations
Last Minute Attack Strategies
These probably won’t get you a 15 on Question 4, BUT they will help you if you are still having trouble with that and
they WILL HELP YOU BEAT THE NATIONAL AVERAGE. The national average on this question is usually between
7—8. If you can consistently score 11—12, that puts you WAY ahead of the game!
First, the minimum knowledge required to survive this question…
MUST KNOW:
1. Big Mamma: All nitrates are soluble.
2. Big Daddy: All IA metals and ammonium salts are soluble.
3. Halides (Cl−, Br−, I−): All are soluble except silver, mercury or lead.
4. Strong acids: hydrochloric, hydrobromic, hydroiodic, nitric, perchloric, sulfuric—WRITE THESE
DISSOCIATED except concentrated sulfuric, it really is 97% H2SO4 and 3% water so the molecules don’t
dissociate completely.
5. Strong bases: hydroxides [and oxides] of IA and IIA* metals—write these bases dissociated.
WRITE ALL WEAK ACIDS AND BASES AS MOLECULES—be on the look out for BF3 and its cousins BCl3, etc.
They are classic Lewis acids and when reacting with ammonia (a classic weak Lewis base), the product is F3BNH3
(just smash everything together) since nitrogen donated its unshared electron pair to boron in an act of extreme
generosity and formed a coordinate covalent bond. Lewis Acids Accept an electron pair.
* the little guys in the IIA’s have solubility issues, write Be and Mg UNdissociated—calcium can go either way, the
big guys are soluble. HF is not a strong acid since it’s the little guy as well in the halogen series. The little guys make
a stronger bond with H and do not dissociate in water. Also remember that the IA metals are named the alkali metals
and the IIA’s are the alkaline earth metals. What does “alkaline” mean? BASIC, so put them in water as metals, they
dissolve and you make OH-. Put IA metals in water and KABOOM. KABOOM = formation of explosive hydrogen
gas, H2.
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Attack Strategies
Before trying to figure out the “answers”, scan the words on each reaction for question 4 and do the following: [don’t
write any products until you’ve done all SIX things!]
1. Cross out the word nitrate any time it appears on the page.
2. Circle any word that implies solid or gas. (powdered, turnings, chunk, vapor, etc.)
3. Cross out any IA metal that you see UNLESS it is associated with a circled solid or gas word.
4. Underline halides then ask yourself if silver, mercury or lead is present—if not you can cross the halide off as
well such as with hydrochloric acid. The H+ is the reacting species. (Bring the halide back as a reacting ion
IF you need to oxidize something halide-1 Æhalogen2.)
5. Circle “burned in air” or “combines with oxygen” or anything that implies combustion and celebrate!
6. Circle the word concentrated. Get very excited if you see excess concentrated. It means you have entered
the land of complex coordinated ions (excess is not necessary, but often appears). Sounds scary, but VERY
easy. LOSE THE FEAR!
Now, WRITE THE REACTANT SETS FOR ALL OF THE ONES YOU’VE MARKED. Spend 8 minutes
writing products using those solubility rules and strong acid-base guidelines listed on the other side of this page.
To get the easy three points involved with step SIX above do the following:
Write the reactants. On the product side, open a set of brackets [ ]. Put the metal ion in the brackets first
then open a set of parentheses [M ion( )]. Next put a subscript on the parentheses that is twice the charge on
the metal—I’m not proud of this, but it will earn credit. For a +2 metal it becomes [M2+( )4]. Finally, plop
the ligand inside the parentheses and do the math to get the charge. If the ligand is ammonia or water, the
ligand is neutral, so our example carries a +2 overall charge, [M2+( )4]2+, if the ligand is hydroxide or a halide,
which are both negative one, then our example becomes [M2+(OH)4]2−. Other ligands are possible, like SCN-,
the thiocyanate ion and other polyatomic ions you should recognize.
Additional knowledge that contributes to survival:
• metal oxides + water → bases (ask yourself strong or weak? Dissociate the strong)
• nonmetal oxides + water → acids (ask yourself strong or weak? Dissociate the strong)
• metal carbonate heated → CO2 + metal oxide
• Redox, “acidified”? H+ is a reactant and water is a product.
• React a metal with oxygen → metal oxide
• React a nonmetal with oxygen [combustion] → make oxides of the nonmetal(s), NOT always CO2 & H2O!
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Free Response Examples
1. Sodium sulfide is added to a zinc nitrate solution.
(i)
(ii) What are the spectator ions in this reaction?
S2− + Zn2+→ ZnS
(ii) Na+ and NO3−
Don’t include the sodium or nitrate ions. Try double replacement first. An insoluble solid can be formed
so go with that!
2. Magnesium ribbon is added to an iron(III) chloride solution.
(i)
(ii)
What will be observed if a magnet is brought near the reaction vessel?
Mg + 2 Fe3+ → 3 Mg2+ + 2 Fe
(ii)
Iron particles will be attracted to the magnet
This cannot be a double replacement reaction so try a redox. Go to metals first to make redox happen!
3. Lithium solid is burned in air.
(i)
(ii) What is the change in oxidation number for lithium?
ii
Li + O2 → Li2O
(ii) Lithium changes from 0 to +1.
Burned in air alludes to combustion. Smack oxygen onto the lithim.
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4. Solutions of tin(II) chloride and iron(III) sulfate are mixed
(i)
(ii) What is the reducing agent? Explain.
Sn2+ + 2 Fe3+ → Sn4+ + 2 Fe2+
(ii) Sn2+ is the reducing agent because it loses 2 e−
This cannot be a double replacement so try redox. Go to metals first. Redox happens.
5. Solid aluminum hydroxide is added to a highly concentrated sodium hydroxide solution.
(i)
(ii) What is the name of the product for this reaction?
OH− + Al(OH)3 → [Al(OH)4]−
(ii) Tetrahydroxyaluminate ion
There is a fish metal and a ligand in high concentration. The body of the fish attached 4 ligands to the
metal. The overall charge is based on the +3 of the Al and the -1 of the hydroxide ions.
6. Solid calcium oxide is heated with sulfur trioxide gas.
(i)
(ii) What state of matter is the product?
CaO + SO3 → CaSO4
(ii) Solid
NO acid MO base. This is a neutralization reaction/addition reaction where you’re adding a fake acid to
a fake base produing the salt minus the water. OXIDES are WEIRD.
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7. Solutions of potassium fluoride and hydrochloric acid are mixed.
(i)
(ii) Would the resulting solution conduct electricity? Explain.
F− + H+ → HF
(ii) Yes. Even though HF is a molecular compound, spectator ions are still present.
This is a double replacement since you can form a weak acid.
8. Aluminum metal is added to a solution of copper(II) chloride.
(i)
(ii) What color changes will be observed during this reaction? Why?
2 Al + 3 Cu2+ → 2 Al3+ + 3 Cu
(ii) The blue color of the copper (II) will fade
This is not a double replacement so try redox. Go to metals first. Redox happens.
9. Water is added to solid sodium hydride.
(i)
(ii) Describe a test to positively identify the gaseous product.
H2O + NaH → Na+ + OH− + H2
(ii) A glowing splint will ignite causing a “bark”
Water being added is indicative of an anhydride reaction. Adding to a hydride produces a base
and hydrogen gas. Sodium hydroxide is a strong base so it needs to be dissociated.
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10. Ethanol is burned in oxygen completely.
(i)
(ii) Is the reaction endothermic or exothermic?
C2H5OH + O2 → CO2 + H2O
(ii) Exothermic because heat is released in the process.
Burned in oxygen means combustion. Smack oxygen into both the carbon and the hydrogen of the
alcohol to form carbon dioxide and water.
11. Hydrochloric acid is added to a potassium carbonate solution.
(i)
(ii) Describe a test to positively identify the gaseous product.
2 H+ + CO32− → H2O + CO2
(ii)
A glowing splint will be extinguished by the carbon dioxide.
This is a double replacement reaction since a weak acid, carbonic acid, is formed. That is an unstable
acid and decomposes further to water and carbon dioxide.
12. Iron (III) ions react with iodide ions.
(i)
(ii) How many electrons are transferred in this reaction?
2 Fe3+ + 2 I− → 2 Fe2+ + I2
(ii) 2 e− are transferred.
This does not form an insoluble substance so it can’t be a double replacement. Try redox.
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13. Aluminum pellets are added to a solution of silver nitrate.
(i)
(ii) What element is oxidized?
Al + 3 Ag+ → Al3+ + 3 Ag
(ii) Al
This is recognizable a single replacement due to the single element reacting with a solution. Treat like a
redox and go to metals first.
14. Solid lithium hydride is mixed with water.
(i)
(ii) If phenolphthalein is added to the final solution, what is the color of the solution?
LiH + H2O → Li+ + OH− + H2
(ii) Hot pink
As mentioned before, this is an anhydride reaction where a base and hydrogen gas is produced;
Recognized by the addition of water.
15. Chlorine gas is bubbled into a solution of potassium iodide.
(i)
(ii) What color is the solid product formed?
Cl2 +2 I− → 2 Cl− + I2
(ii) Purple
Potassium is a group I ion so leave it out. This can’t be a DR so attempt redox.
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16. Solid sodium is added to water.
(i)
(ii) Explain how inserting a glowing splint into the reaction test tube can positively identify the gas produced.
2 Na + 2 H2O → 2 Na+ + 2 OH− + H2
(ii) Hydrogen gas will pop upon ignition with a glowing splint.
This is a single replacement reaction. Watch out!!! You are not producing sodium oxide!!!
17. Solid potassium oxide is added to water.
(i)
(ii) Would the pH of the final solution be greater than 7, less than 7, or equal to 7?
K2O + H2O → 2 K+ + 2 OH−
(ii) Greater than 7
Oxides are Weird! This is a MO BASE. It forms a strong base that dissociates into ions.
18. A piece of solid bismuth is exposed to a large volume of oxygen.
(i)
(ii) Which substance is the oxidizing agent?
2 Bi + 3 O2 → 2 Bi2O3
(ii) O2 is the oxidizing agent.
COMBUSTION! Smack ‘em together.
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19. A concentrated solution of ammonia is added to a solution of zinc iodide.
i)
ii) name the ion produced
4 NH3 + Zn2+ → [Zn(NH3)4]2+
(ii) Tetraamine zinc(II) ion
Fish metal in a concentrated ligand!! Must be a complex ion with 4 ligands attached to the metal.
20. Magnesium turnings are burned in nitrogen gas.
(i)
(ii) Write a balanced half reaction for the substance being oxidized.
3 Mg + N2 → Mg3N2
(ii)
3 Mg0 → 3 Mg2+ + 6 e−
This is an addition reaction due to the fact there are two elements reacting. Add ‘em up!
21. A solution of potassium hydroxide is added to a solution of potassium dihydrogen phosphate until the same
number of moles of each compound has been added.
(i)
(ii) Is the resulting solution acidic, basic or neutral? Explain.
OH− + H2PO4− → HPO42− + H2O
(ii) The resulting solution will be basic because the salt produced is Na2HPO4 which is a basic.
This is an acid base rxn; (Bronsted Lowry). The statement of “equal moles” indicates the transfer of one
H+
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22. Solutions of silver nitrate and sodium chromate are mixed.
(i)
(ii) What are the spectator ions in this reaction?
2 Ag+ + CrO42− → Ag2CrO4
(ii) Na+ and NO3− are the spectator ions.
The chromate in the reaction makes many students think it is a redox reaction. This is NOT a redox.
Remember to try a double replacement first and attempt to make in insoluble solid.
\
23. The gases boron trifluoride and ammonia are combined
(i)
(ii) Which reactant is the Lewis acid?
BF3 + NH3 → F3BNH3
(ii) BF3 is the Lewis acid.
This is a Lewis acid base reaction where an electron pair is transferred creating a bond between the two
compounds. Also works with PF3. A Lewis acid is an electron pair acceptor and vice versa for a
Lewis base.
24. Solid calcium oxide and solid tetraphosphorus decaoxide are mixed and heated.
(i)
(ii) Is the product soluble in water? Explain.
6 CaO + P4O10 → 2 Ca3(PO4)2
(ii) No, only phosphates of IA and NH4+ are soluble, the rest are insoluble.
NO acid plus MO base gives a salt. This isn’t redox so make sure to maintain the same oxidation
number of phosphorus
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Multiple Choice
O
H3 C
C
H2
C
CH3
1) The compound above is classified as a/an
(A) Ester
(B) Ether
(C) Ketone
(D) Carboxylic acid
(E) Aldehyde
C
The carbonyl group of C=O, when within a chain and not on an end, is characteristic of a
ketone.
2) A yellow precipitate forms when a solution of KI is mixed with which of the following ions?
(A) SO42–(aq)
(B) Zn2+(aq)
(C) Pb2+(aq)
(D) Na1+(aq)
(E) CrO42–(aq)
C
PbI is a characteristically bright yellow colored solid
Questions 3-6 refer to the reactions represented below.
(A)
A piece of iron is added to a solution of iron(III) sulfate
(B)
Solid zinc sulfide is heated in an excess of oxygen
(C)
Manganese(II) nitrate solution is mixed with a sodium hydroxide solution
(D)
A suspension of zinc hydroxide is treated with concentrated sodium hydroxide solution
(E)
Solid lithium hydride is added to water
3) A reaction that produces a complex ions
D
A transition metal in the presence of a high concentration of a common ligand creates a
complex
4) A precipitation reaction
C
The precipitate of manganese II hydroxide is formed in this double replacement reaction.
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5) A combustion reaction
B
Anything heated or exposed to oxygen can be considered combustion
6) A reaction in which the same element undergoes both oxidation and reduction
A
Iron is both reduced and oxidized in this option from zero to +2 and +3 to +2
Questions 7–10 refer to the following solids.
(A) KMnO4
(B) NiSO4
6.
(C) CuSO4
(D) NaCl
(E) FeSCN+2
Makes a purple solution dissolved in water
A
7.
Is white and very soluble in water
D
8.
Table salt is colorless in solution and very soluble. The solid is also white.
Makes a blue solution when dissolved in water
C
9.
Permanganate ions are characteristically colored purple in solution
Copper II ions are characteristically blue in solution
Makes a green solution when dissolved in water
B
Nickel II ions are characteristically green in solution
10. Makes an orange coordination complex
E
Iron III thiocyanate is an orange colored complex due to the iron III ions. You may have
seen this complex in an equilibrium lab.
Note: Colored ions are paramagnetic (have unpaired electrons). Diamagnetic substances (all
paired electrons) are colorless in solution. When in doubt if a solution is colored or not,
look at the electron pairing.
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