Unit 13: Acids & Bases
Chemistry
Chapter 19
Welcome
To The
GowerHour
I. Bronsted-Lowry Acids and Bases:
A. Acid: Substance that _________
donates a ___________
H+, proton to another
substance.
accepts a __________
H+, proton from another
B. Base: Substance that ________
substance.
H+ is transferred from the ____
acid to the
C. Acid – Base reaction: ___
_____.
base
Ex.
HX + Y– ↔ HY + X–
Acid Base
Acid Base
D. Conjugate acid – base pair: The pair of acid and base that differ
by a ___.
H+
Ex.
ACID
BASE
–
HF
F
Conjugate Acid add H+
HNO2
NO2–
Conjugate Base subtract H+ HC2H3O2
C2H3O2–
Memorize: NH3 = Base
Memorize: NH4+ = Acid
E. Amphoteric substance: Substance that can either _______
donate or
Generally: H and
_______
accept a hydrogen ion.
– H+
+ H+
Ex. H2O
OH– ↔ H2O ↔ H3O+ ★Memorize
★ hydroxide
★ hydronium
+
+
+H
–H
– H+
+ H+
2–
–
SO4 ↔ HSO4 ↔ H2SO4
CO32– ↔ HCO3– ↔ H2CO3
F. Example: Label the acids and bases, draw lines to connect the
conjugate pairs.
(1) NH3 + H2O  NH4+ + OHBase
Acid
(CA)
(CB)
(2) HF + H2O  F- + H3O+
Acid
Base
(CB)
(CA)
G. Categorize each of the following as an Acid, Base, or Amphoteric
and below each, write the conjugate acid or base.
H2SO4
H2PO4NH3
NO2HSO4Acid
Ampho
Base
Base
Ampho
HSO4–
H3PO4
NH4+
HNO2
H2SO4
HPO42–
SO42–
II. Water Dissociation Constant (Ion product)
A. The acidic / basic properties of _________
aqueous solutions are
dependent upon the ____________
equilibrium that involves the solvent,
_______.
water
1. Reaction: H2O (l) ↔
H+ (aq) + OH– (aq)
Acid
2. Equilibrium expression:
Base
Keq = [H+][OH–] = Kw
1.0 x 10−14 @ 25 C. ___________
Reactants are favored.
3. Kw = ___________
H20  very little conc. of ions
B. In pure water: H2O (l) ↔ H+ (aq) + OH– (aq)
+][OH–]
ICE
Kw = 1.0 x 10−14 = [H
x
x
2
−14
x = 1.0 x 10 M
x = 1.0 x 10−7 M
[H+] = x = 1.0 x 10−7 M
[OH-] = x = 1.0 x 10−7 M
Kw = 1.0 x 10−14 = (1.0 x 10−7 M)(1.0 x 10−7 M)
C. In impure water (contains an acidic or basic substance):
Kw = 1.0 x 10−14 = [H+][OH–]  As [H+]  [OH–] 
If [H+] > 1.0 x 10-7 M, solution is _______.
acidic
basic or alkaline
If [H+] < 1.0 x 10-7 M, solution is _________________.
If [H+] = 1.0 x 10-7 M, solution is _________.
neutral
1.0 x 10−7 = pH 7 
pH < 7 = acid
pH > 7 = base
pH = 7 = neutral
III. pH and pOH
A. Because [H+] and [OH-] are generally very small numbers, a
_____
log based system of measuring acidity is used.
pH = – log [H+]
pOH = – log [OH–]
p
H
Practice: What is the pH of a solution with: [H+] = 10-5 M?
5
[H+] = 10-11 M? 11
unitless because we cannot take the
1. pH and pOH are _________,
logarithm of a unit.
2. For each pH change of 1, the [H+] changes by a factor of ____.
10
Richter scale
Ex. pH of 1 is 100 x’s more concentrated than pH of 3. (102 = 100)
104 = 10,000
pH = 1  pH = 5
3. [H+][OH-] = 1.0 x 10-14
Or
pH + pOH = 14
B. Example calculations
1. Calculate the pH of the following solutions, and indicate if the
solution is acidic or basic:
(a) [H+] = 1.0 x 10-11 M
pH = 11 basic or alkaline
(b) [H+] = 2.11 x 10-2 M
pH = – log [H+]
pH = – log (2.11 x 10-2 )
(c) [OH-] = 3.98 x 10-7 M
pOH = – log (3.98 x 10-7 )
pOH = 6.40 pH = 7.60
pH = 1.68
basic
acidic
2. Calculate the [H+] and [OH-] of the following and indicate if they
10x = antilog (opposite of log)
are acidic or basic:
10– 9.35 = [H+] 10x Key
(a) pH = 9.35 pH = – log [H+]
basic
9.35 = – log [H+]
[H+] = 4.47 x 10– 10 M
– 9.35 = log [H+]
[H+][OH-] = 1.0 x 10-14
Divide both side by log
[OH–] = 2.24 x 10– 5 M
(b) pH = 1.10 [H+] = 7.94 x 10– 2 M
acidic
[OH–] = 1.26 x 10– 13 M
(c) pOH = 2.98 [OH–] = 1.05 x 10– 3 M
[H+] = 9.55 x 10– 12 M
C. pH Scale:
NaOH
Lye
N
e
u
t
r
a
l
H2SO4
Alkaline/Basic
Acidic
0
1
2
3
4
5
Acid
Rain
6
7
High
quality
H2O
8
9
10
11
12
13
14
IV. Properties of Acids and Bases
A. Properties of Acids
Sour
Vinegar
1. _________
taste (ex. _________)
H2 (ex.___________)
Magnesium
2. Reacts with some metals to form ___
Conducts
electrolytes in solution)
3. __________electricity
(__________
4. React with _________
bases
B. Properties of Bases
Bitter
Soap
1. _______taste
(ex.________)
Slippery
Soap, NaOH
2. _____________
(ex._____________)
Conducts
electrolytes in solution)
3. ______________electricity
(___________
V. Strong vs. Weak
A. Strong Acids: Completely ______
ionize in solution (Not
_________reaction).
reversible
Ex.
HCl
H2O
+ Cl(aq)
H+(aq)
(1) HCl
(H3O+ = Hydronium ion)
H3O+ + Cl
(2) HCl + H2O
Chemists use H+ & H3O+ interchangeably. H+ is often
used for simplicity, but H3O+ more closely represents
reality.
*****MEMORIZE*****
HCl
H2SO4
hydrochloric
sulfuric
HNO3
nitric
HBr
hydrobromic
HClO4
perchloric
HI
hydroiodic
ionize in solution. (Group I and
B. Strong Bases: Completely _______
II hydroxides)
Ex.
NaOH
H2O
H2O
*****Memorize*****
LiOH
Ca(OH)2
+ OH
Na+ + OH
NaOH + H2O
Practice:
Ca(OH)2
Na+
+ H2O
Ca2+ + 2 OH
NaOH
KOH
Ba(OH)2
Sr(OH)2
C. Weak Acids: Ionize __________
partially (_________reaction).
reversible
Generally less than _____
1 % of the molecules ionize.
Ex. H2CO3
(1) H2CO3
H2O

(2) H2CO3 + H2O
H+ + HCO3

H3O+ + HCO3
Example of weak acids: HC2H3O2, H2CO3
If it is not a strong acid then it is a weak acid!!
D. Weak Bases: Ionize __________
partially (_________reaction).
reversible
Partial dissociation of an weak base in water:
NH4+
Ex. NH3
(1) NH3 + H2O

Example of weak bases: NH3, NH2CH2CH3
(aminoethane)
If it is not a strong base then it is a weak base!!
+ OH
VI. Calculations with weak acids and weak bases.
A. Weak Acids (Ka): As Ka decreases, less ____
H+ is formed in
solution, so acid is _______.
weaker
HA (aq)
H2O
↔
H+(aq) + A(aq)
[H  ][A - ]
K eq  K a 
[HA]
Ka = acid-dissociation
constant
OH is formed in
B. Weak Bases (Kb): As Kb decreases, less _____
weaker
solution, so base is _______.
B (aq) + H2O(l) ↔ BH+(aq)+ OH(aq) Kb = base-dissociation
constant
[BH  ][OH - ]
K eq  K b 
[B]
C. Examples
1. Example: Aspirin is a weak organic acid whose molecular
formula is HC9H7O4. An aqueous solution of aspirin is prepared
by dissolving 3.60 g/ L. The pH of this solution is found to be
2.60. Calculate Ka for aspirin.
HC9H7O4 ↔
H+
+ C9H7O4Initial (M)
2.00 x 10-2
0
0
Change (M)
Equilibrium (M)
3.60 g
L
x
1 mol
180.15 g
–x
+x
+x
0.0175
2.51 x 10-3
2.51 x 10-3
= 2.00 x 10-2 M = I
(2.51 x 103 ) 2
Ka 
 3.61 x 10-4
(0.0175)
pH = 2.60
[H+] = 2.51 x 10-3

-
[H ][C9 H 7 O 4 ]
Ka 
[HC9 H 7 O 4 ]
3.
Compare the pH of 0.100 M HCl and 0.100 M HCN
(Ka = 4.90 x 10-10)
HCN ↔ H+ + CN
HCl  H+ + Cl
[H+] = 0.100 M
I
0.100
0
0
pH = 1.00
C
-x
E 0.100 - x
+x
+x
x
x
2
2
x
[H  ][CN - ]
x
 4.90 x 10-10 
Ka 
 4.90 x 10-10 
0.100
(0.100 - x)
[HCN]
Assume x is small
x = 7.0 x 10–6 M
pH = 5.15
VII. Multiple Equilibria: When two reactions are added to give a
third (net reaction) the equilibrium constants are multiplied
K1 • K2 = Knet
(______________).
Ex. HC2H3O2
Ka = 1.8 x 10 – 5
↔ H+ + C2H3O2
C2H3O2- + H2O ↔ HC2H3O2 + OH
H2O
↔
H+
+
OH–
]
[H+] [C2H3O2] [HC
H
O
]
[OH
Kw = _______________
___________
2 3 2
•
[HC2H3O2]
[C2H3O2]
Kw = [H+][OH]
Equation:
Ka • Kb = K w
Kb = 5.6 x 10 – 10
Kw = 1.0 x 10−14
A. As Ka increases, Kb ________
decrease (_______
inverse relationship).
weaker its conjugate base.
B. The stronger the acid, the ________
Ex. If Ka = 1.0 x 1030, determine Kb.
Kb = 1.0 x 10 – 44
Ka = strong acid; Kb = weak base
VIII. Polyprotic Acids: Certain weak acids contain more than one
ionizable ___.
H+ These acids dissociate in multiple steps.
2 hydrogen ions.
A. Diprotic acid: Dissociates to form ___
H2O
Ka1 = 4.2 x 10 – 7
H+ + HCO3
Ex. H2CO3 ↔
H2O

Ka2 = 4.8 x 10 – 11
HCO3
H+ + CO32↔
3 hydrogen ions.
B. Triprotic acid: Dissociates to form ___
H2O
H+ + H2PO4
H+ + HPO42H+ + PO43-
–3
K
=
7.5
x
10
a1
H↔
2O
Ka2 = 6.2 x 10 – 8
↔
H2O
– 13
K
=
4.8
x
10
a3
↔
HCO3 , H2PO4
C. The _________
substance formed in one step (e.g. _______________)
dissociates in the next step.
equilibrium constant becomes
D. The dissociation constant (___________________)
smaller with each successive step: Ka1 > Ka2 > Ka3
E. The acids formed in successive steps become progressively
________.
weaker
Ex. H3PO4
H2PO4
HPO42-
F. Example: Write the dissociation reactions of sulfurous acid,
H2SO3.
H2SO3
HSO3
H2O
H+ + HSO3
H2O
H+ + SO32-
↔

↔
G. Example: Write the dissociation reactions of citric acid,
H3C5H5O7.
H2O
↔
H+ + H2C5H5O7 

H2O
H+ + HC5H5O7 2-
2-
H2O
H+ + C5H5O7 3-
H3C5H5O7
H2C5H5O7
HC5H5O7
↔
↔
IX. Acid / Base Properties of salt solutions:
A. A salt is an ionic
____ compound not containing ___
H+ or____.
OH
B. When a salt dissolves in water, the ___
ion are formed.
H2O
Ex. NaCl(s)
Na+ + Cl
H2O
Ex. K2CO3(s)
2 K+ + CO32C. Cations: Weak acids or “spectator” ions?
1. Cations derived from strong
___________
bases are _________
spectator ions.
(Do not react with water, therefore have no
_______
effect on pH.)
- pair
2. Other cations are slightly acidic
_____. (Lewis acid = e______acceptor).
D. Anions: Weak bases or “spectator” ions?
1. Anions derived from strong
__________are
________ ions.
acids spectator
2. Other anions are slightly basic
____.
Ex. HCl  H+ + Cl–
Spectator ions:
Cations: Li+ ; Na+ ; K+ ; Ca2+ ; Sr2+ ; Ba2+
Anions: NO3 ; SO42- ; ClO4 ; Cl ; Br ; I
E.
1.
2.
F.
(All others
weak acids)
(All others
weak bases)
Example: Consider water solutions of these four salts:
(a) NH4I, (b) Zn(NO3)2, (c) KClO4, (d) Na3PO4
Classify each salt solution as acidic, basic, or neutral. (Show the
dissociation reactions of each.)
H2O
a) NH4I
NH4+
+ I¯
Acidic
(w. acid)
b) Zn(NO3)2
c) KClO4
d) Na3PO4
H2O
H2O
Zn2+
(w. acid)
K+
(Spec)
H2O
3 Na+
(Spec)
(Spec)
+ 2 NO3¯
Acidic
(Spec)
+ ClO4¯
Neutral
+ PO43-
Basic
(Spec)
(w. base)
Which will undergo hydrolysis?
The salts that react w/water to change the pH!
X. Acid – Base Reactions
H2O
A. Strong acid + Strong base --> ____
salt + ____
Example:
Example:
HCl(aq) + NaOH(aq) -->
2 HNO3(aq) + Ca(OH)2(aq) --> Ca(NO3)2 (aq) + 2 H2O (l)
2 H+ + 2 NO3¯ + Ca2+ + 2 OH–
Net reaction:
NaCl (aq) + H2O (l)
Ca2+ + 2 NO3¯+ 2 H2O (l)
H+ + OH–
(neutralization)
H2 O
B.
Weak acid + Strong base
______+__________
water weak base
Note: Strong Base  weak base (Soln. will still be basic)
H2O + C2H3O2
Example: HC2H3O2 + OHH2O + F
Example: HF + OH-
C.
Strong acid + Weak base
__________
weak
acid
Note: Strong acid  weak acid (Soln. will still be acidic)
NH4+
Example: H+ + NH3
HClO
Example: H+ + ClO-
D. Weak acid + Weak base ↔
weak
acid weak base
_________+__________
Example: HC2H3O2 + NH3 ↔ NH4+ + C2H3O2
Example: HF + ClO- ↔ HClO + F
XI. Titration: A process in which one reagent is added to another
with which it reacts; an ________
indicator is used to determine the point
at which equal
_____ quantities of the two reagents have been added.
neutralization
A. Equivalence point: the point at which the _____________
reaction is complete. nA = nB
B. End point: The point at which the ________
indicator changes color.
pH’s
1. Indicators change at different ______.
2. In doing a titration, one must choose an _________
indicator where the
equivalence point and the _________
end point coincide.
3. Example Indicators:
Indicator
Color change
pH at end point
orange - red
5
Methyl Orange
___________
___________
Bromothymol Blue
Phenolphthalein
yellow
- blue
___________
clear - pink
___________
7
___________
9
___________
C. Acid – Base Indicators: Produce a _____
color change in an acid-base
reaction.
1. Example: Phenolphthalein, bromothymol blue
2. Natural indicators: purple cabbage, hydrangeas
3. Reversible reactions: (Phenolphthalein)
H-In 
H+
+
InCLEAR
PINK
left solution turns _____.
clear
Add acid, rxn shifts ____,
pink
Add base, rxn shifts right
_____, solution turns _____.
D. Titration Curves:
~ pH 7
1. Strong acid – Strong Base (Equivalence point = _______)
0.100 M HCl is titrated with 0.100 M NaOH
8.8
2. Weak Acid – Strong Base (Equivalence point = ~pH
_______)
0.100 M CH3COOH is titrated with 0.100 M NaOH
pH 5.3
3. Strong Acid – Weak Base (Equivalence point = ~_______)
0.100 M NH3 is titrated with 0.100 M HCl
E. Calculations with titrations:
1. STOICHIOMETRY
2. At equivalence point:
nA = nB
(mol acid = mol base)
M A VA M B VB

nA
nB
coefficients
Problems:
1. Example titration problem: 35.00 mL of 0.2500 M sodium
hydroxide is titrated with 0.4375 M HCl. (a) Write the balanced
chemical equation. (b) Determine the volume of HCl added at the
equivalence point (i) using stoichiometry and (ii) using the titration
equation.
(a)
HCl + NaOH
NaCl
+ H2 O
(b) (i)
35.00 ml NaOH
x
10-3 L
1 ml
x 0.2500 mol NaOH x
1 L NaOH
1 mol HCl
1 mol NaOH
x
1 L HCl
x 1 ml
0.4375 mol HCl
10-3 L
= 20.00 ml HCl
(b) (ii)
M A VA M B VB
M V

VA  B B
nA
nB
MA
(0.2500 M)(35.00 mL)
VA 
VA = 20.00 mL HCl
(0.4375 M)
2. Example: A 15.0 mL sample of a solution of H2SO4 with an
unknown molarity is titrated with 32.4 mL of 0.145 M NaOH to
the bromothymol endpoint. What is the molarity of the sulfuric
acid solution?
H2SO4
+ 2 NaOH
M A VA M B VB

nA
nB
Na2SO4
+ 2 H 2O
M B VB
MA 
VA n B
(0.145 M)(32.4 mL)
MA 
2(15.0 mL)
MA = 0.157 M H2SO4
3. A Ca(OH)2 solution was used to titrate 15.0 mL of a 0.125 M
H3PO4 solution. If 12.4 mL of Ca(OH)2 are used to reach the
endpoint, what is the concentration of the Ca(OH)2?
2 H3PO4 + 3 Ca(OH)2
M A VA M B VB

nA
nB
Ca3(PO4)2
+ 6 H 2O
M A VA n B
MB 
VB n A
3(0.125 M)(15.0 mL)
MB 
2(12.4 mL)
MB = 0.227 M Ca(OH)2
IQ 2
1.
Show the dissociation rxn of each of the following salts and
determine if each solution is acidic, basic, or neutral.
a) Ca(ClO3)2
H2O
Ca2+
(Spec)
a) FeCl2
H2O
Fe2+
(w. acid)
2.
+ 2 ClO3¯
Basic
(w. base)
+ 2 Cl¯
Acidic
(spec)
Predict the products and balance the following:
a. 2HNO3 + Sr(OH)2
Sr(NO3)2 + 2 H2O
b. HClO2 + OH–
H2O + ClO2¯
IQ 1
1.
Write formulas for two salts that:
(a) contain CO32- and are basic:
Li2CO3
CaCO3
(b) contain Li+ and are neutral:
LiBr
Li2SO4
2.
Predict the products and balance the following. Write the net
rxn for each.
Ca(NO3)2 + 2 H2O
a. 2HNO3 + Ca(OH)2
Ca2+ NO3H+ NO3- Ca2+ OHH2 O
Net Rxn:
H+ + OHb. HClO2 + LiOH
H+ ClO2- Li+ OHNet Rxn: HClO2 + OH-
H2O + LiClO2
Li+ ClO2H2O + ClO2-
Titration of a Strong Acid w/ a Strong Base
1.
2.
3.
4.
5.
6.
7.
8.
9.
Clean burette w/ NaOH; Drain into “waste beaker”.
Fill burette with NaOH to 0.00 mL (hundredths)(meniscus).
Record initial vol.
Measure out 10.00 mL (hundredths)(meniscus) of acid using the
graduated cylinder (Use the pipet). Record volume of acid; add
to the Erlenmeyer flask.
Add 2 drops of phenolphthalein to the acid. IMPORTANT!!!!
Add a little water to Erlenmeyer flask (Rinse acid off sides).
Start adding NaOH slowly; swirl flask as you add.
At the endpoint (when soln. remains light pink) stop adding
base and record volume of base added. (Check: rinse flask)
Dispose titrated solution in sink; rinse flask and repeat steps 3-8.
The final vol. of the base is your initial vol. for the next trial.
Reminders
1.
2.
3.
4.
5.
6.
Wear safety glasses/goggles.
Only use the pipette for acid!
Base in burette
Clean up after 3rd trial.
Must get a clean up stamp before leaving.
Leave excess acid and base in appropriate
beakers.
Titration of a Strong Acid w/ a Strong Base
HCl + NaOH
MA = ?
VA = 10.00 mL
nA = 1
M A VA M B VB

nA
nB
NaCl + H2O
MB = 0.0915 M
VB = ? mL
nB = 1
M B VB
MA 
VA
Titration of Vinegar Lab
1.
2.
3.
4.
5.
6.
7.
8.
Fill burette with NaOH to 0.00 mL (hundredths)(meniscus).
Record initial vol.
Measure out 1.00 mL (hundredths)(meniscus) of vinegar using
the graduated cylinder (Use the pipet). Record volume of
vinegar; add to the Erlenmeyer flask.
Add 2 drops of phenolphthalein to the vinegar.
IMPORTANT!!!!
Add a little water to Erlenmeyer flask (Rinse acid off sides).
Start adding NaOH slowly; swirl flask as you add.
At the endpoint (when soln. remains light pink) stop adding
base and record volume of base added.
Dispose titrated solution in sink; rinse flask and repeat steps 2-7.
The final vol. of the base is your initial vol. for the next trial.
Reminders
1.
2.
3.
4.
5.
6.
Wear safety glasses/goggles.
Only use the pipette for acid!
Base in burette
Clean up after 3rd trial.
Must get a clean up stamp before leaving.
Leave excess acid and base in appropriate
beakers.
Vinegar Calculations
1.
2.
HC2H3O2 + NaOH
H2O + C2H3O2–
Vinegar = Acetic acid (solute) in water (solvent).
Average number of moles of acetic acid:
M A  L vinegar  mol HC 2 H3O2
Average [ ]
of acid
3.
Average vol. of vinegar (i.e. 1 mL)
Mass (g) of acetic acid (HC2H3O2):
molar mass
mol HC 2 H 3O 2 x
 mass of acetic acid (g)
of HC 2 H 3O 2
4.
% mass of acetic acid:
g solute
%
x 100
g solution
g HC 2 H 3O 2

x 100
g vinegar
Density of vinegar = 1.02 g/mL
Vinegar Calculations
1.
2.
HC2H3O2 + NaOH
H2O + NaC2H3O2
Vinegar = Acetic acid (solute) in water (solvent).
Average number of moles of acetic acid:
mol HC 2 H 3O 2
MA 
x average vol in L
L vinegar
Average [ ]
of acid
3.
Average vol. of vinegar (i.e. 1 mL)
Mass (g) of acetic acid (HC2H3O2):
molar mass
mol HC 2 H 3O 2 x
 mass of acetic acid (g)
of HC 2 H 3O 2
4.
% mass of acetic acid:
g solute
%
x 100
g solution
g HC 2 H 3O 2

x 100
g vinegar
Density of vinegar = 1.02 g/mL
Neutralization Capacity of an Antacid
1.
Fill burette with 0.0953 M NaOH to 0.00 mL
(hundredths)(meniscus). Record initial vol.
2. Weigh ~ 0.10 g of Tums (CaCO3). Record and add to
Erlenmeyer flask.
3. Add water to E. flask and swirl to dissolve powder.
4. Measure out 5.00 mL (hundredths)(meniscus) of 0.336 M HCl
using the graduated cylinder (Use the pipet). Record volume of
acid; add to the E. flask containing antacid and swirl.
5. Add a little water to E. flask (Rinse acid off sides).
6. Add 2 drops of phenolphthalein to the acid. IMPORTANT!!!!
7. Start titrating; swirl E. flask as you add NaOH.
8. At the endpoint (when soln. remains light pink) stop adding
base and record volume of base added.
9. Dispose titrated solution in sink; rinse flask and repeat steps 2-7.
10. The final vol. of the base is your initial vol. for the next trial.
Reminders
1.
2.
3.
4.
5.
6.
Wear safety glasses/goggles.
Only use the pipette for acid!
Base in burette
Clean up after 3rd trial.
Must get a clean up stamp before leaving.
Leave excess acid and base in appropriate
beakers.
Antacid Calculations
Acid (HCl) (VA)
Antacid (CaCO3)(VA1)
NaOH (VA2)
VA1 = Volume of acid neutralized by the antacid.
VA2 = Volume of acid neutralized by the NaOH.
VA = Total volume of acid (5 mL).
VA = VA1 + VA2
Step 1: Solve for VA2
M A VA2 M B VB

nA
nB
Step 2: Solve for VA1
VA = VA1 + VA2
Antacid Calculations
Step 3: Solve for VA1/ g.
VA1
g
Mass of antacid used in the trial (i.e. 0.10)
Step 4: Solve for VA1/ Tablet
Antacid : 1 Tablet = 1.30 g
VA1 1.30 g
x
g
Tablet
VA1

Tablet
Step 5: Solve for VA1/ cent
Antacid : 72 Tablet = $5.49
VA1
Tablet
x
Tablet
cents

VA1
cents
Labs Due: Tuesday 05/15
Acid Nomenclature Review
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Acetic acid (w)
Oxalic acid (w)
Hydrocyanic acid (w)
Cyanic acid (w)
Sulfurous acid (w)
Sulfuric acid (s)
Hypochlorous acid (w)
Bromous acid (w)
Periodic acid (w)
Phosphoric acid (w)
11. Carbonic acid (w)
12. Perchloric acid (s)
13. Hypobromous acid (w)
14. Nitric acid (s)
15. Chloric acid (w)
16. Hydrofluoric acid (w)
17. Phosphorous acid (w)
18. Hydroiodic acid (s)
19. Nitrous acid (w)
20. Hydrochloric acid (s)