Period:
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Unit 13 HW Packet: Acids, Bases, and Neutralization
Part 1: Naming Binary Acids (hydro- prefix
a.
b.
c.
d.
e.
root
–ic suffix)
HCl
HBr
HF
H2S
H3P
Part 2: Naming Acids with Polyatomic Ions (Oxyacids)
If the polyatomic ion ends in –ate, the acid ends in –ic.
If the polyatomic ion ends in –ite, the acid ends in –ous.
a.
b.
c.
d.
e.
f.
g.
h.
H2SO4
HNO2
HClO3
H3PO4
H2CO3
HC2H3O2
HClO2
HNO3
Part 3: Naming Bases
a.
b.
c.
d.
e.
f.
g.
h.
KOH
Al(OH)3
Mg(OH)2
LiOH
Fe(OH)3
Zn(OH)2
NaOH
CsOH
Part 4: Writing Formulas for Acids and Bases
Remember the rules you just practiced and work backwards.
Acids always begin with “H” and Bases always end with “OH”.
You will also need to criss-cross the charges to determine subscripts.
a. Hydrosulfuric Acid
i.
Chloric Acid
b. Sulfuric Acid
j.
Chlorous Acid
c. Sulfurous Acid
k. Ammonium Hydroxide
d. Nitric Acid
l.
e. Nitrous Acid
m. Hydrophosphoric Acid
f. Silver Hydroxide
n. Phosphoric Acid
g. Iron (II) Hydroxide
o. Acetic Acid
h. Hydrochloric Acid
p. Magnesium Hydroxide
Aluminum Hydroxide
Part 5: Use these formulas to complete the chart below.
pH = - log [H3O+]
pH ===================
[H3O+] = antilog(-pH)
Acid/Base
HCl
NaOH
HBr
KOH
HI
LiOH
HC2H3O2
CsOH
HNO3
Mg(OH)2
H2SO4
Ca(OH)2
[H3O+]
1.0 x 10-4
-
2.6 x 10 3
9.5 x 10-11
[H3O+][OH-] = 1 x 10-14
[H3O ] ==================== [OH-]
+
[OH-]
pH
1.0 x 10-4
1.0 x 10
-3
1.00
5.8 x 10-4
3.6 x 10-10
4.70
11.90
0.20
2.89
Acid or Base?
Part 6: More pH Calculations
1. If the hydronium ion concentration is 3.4x10 -3M, calculate the pH.
2.
If the hydroxide ion concentration is 3.4x10 -3 M, calculate the pH.
3.
If the hydroxide ion concentration is 3.4x10 -3 M, calculate the hydronium ion concentration.
4.
If the pH = 5.47, calculate the hydronium ion concentration.
5.
If the pH = 5.47, calculate the hydroxide ion concentration.
Part 7: Bronsted-Lowry Acids and Bases
Label the Bronsted-Lowry acid, Bronsted-Lowry base, conjugate acid, and conjugate base.
1.
HCO3- + H2O
→
OH- + H2CO3
2.
HPO42- + H2O
→
H2PO4- + OH-
3.
CH3NH3+ + H2O
→
CH3NH2 + H3O+
4.
HClO4 + H2O
→
H3O+ + ClO4-
5.
HSO4- + H3O+
→
H2O + H2SO4
6.
NH3 + HSO3-
→
NH4+ + SO32-
Part 8: Neutralization Reactions
Predict the correct formula for the salt formed in the following equations.
Write the name of each acid under the formula for the acid.
ACID
+
BASE
SALT
+
WATER
HCl
+
NaOH
NaCl
+
H2O
H2SO4
+
Ca(OH)2
+
H2O
HNO3
+
KOH
+
H2O
H2CO3
+
NaOH
+
H2O
H3PO4
+
Ca(OH)2
+
H2O
HC2H3O2 +
KOH
+
H2O
HCl
+
NH4OH
+
H2O
H2SO4
+
Mg(OH)2
+
H2O
Hydrochloric acid
Part 9: Titration Calculations
Balance the equation. Substitute into the formula with units. Show your answer with units.
Use the following formula:
MAVA = MBVB
CB
CA
MA and MB:
Molarity of the acid/base
VA and VB:
Volume of the acid/base
CA and CB:
Coefficient of the acid/base from the balanced equation.
1. If it takes 25.0mL of HCl to neutralize 50.0 mL of 3.0M NaOH, what is the molarity of the acid?
HCl +
NaOH
NaCl +
H2O
2. If it takes 25.0mL of H2S to neutralize 50.0mL of 3.0M NaOH, what is the molarity of the acid?
H2S +
NaOH
Na2S +
H2O
3. If you start with 75.0mL of Ca(OH)2 and titrate it with 125mL of 1.5M hydrobromic acid, what is the
molarity of the base?
HBr +
Ca(OH)2
CaBr2 +
H2O
4. Given 80.0mL of 0.50M sulfuric acid, what vo lume of 0.75M potassium hydroxide is needed to
neutralize the acid?
H2SO4 +
KOH
K2SO4 +
H2O
5. Given 80.0mL of 0.50M nitric acid, what volume of 0.75M lithium hydroxide is needed to neutralize the
acid?
HNO3 +
6.
H2O
Mg(OH)2
Mg3(PO4)2 +
H2O
If you react 25.7mL of 0.0550M HCl with 38.0mL of NaOH, calculate the molarity of the base.
HCl +
8.
LiNO3 +
If you titrate with 0.0075M phosphoric acid, what volume of this acid is needed to neutralize 285mL of
0.025M magnesium hydroxide?
H3PO4 +
7.
LiOH
NaOH
NaCl +
H2O
If you react 25.7mL of 0.0550M H2SO4 with 38.0mL of NaOH, calculate the molarity of the base.
H2SO4 +
NaOH
Na2SO4 +
H2O
9. If you react 25.7mL of 0.0550M HCl with 38.0mL of Ca(OH) 2, calculate the molarity of the base.
HCl +
Ca(OH)2
CaCl2 +
H2O
10. If you react 25.7mL of 0.0550M H3PO4 with 38.0mL of Mg(OH)2, calculate the molarity of the base.
H3PO4 +
Mg(OH)2
Mg3(PO4)2 +
H2O