11.A – Nuclear Decay: Type and Predictions
Instructions: Predict the missing product or reactant in the following nuclear reactions. Determine the type of nuclear reaction (α emission, β emission, γ emission, positron
emission, artificial transmutation, fission, or fusion) described.
Emissions = single react as random and spontaneous
Artificial Transmutation = bombardment by “heavy” nuclei = 2 large react
Fission = bombardment by 10 n0 as a reactant to get multiple products
1.
Fusion = multiple reacts to 1 product
4219K
0-1e- β part
→
4220
+
23592
42He
2.
23994Pu
→
42 He
3.
11H
+
31H
→
4.
63Li
+
10 n0
n0bombardment→
5.
2713Al
+
42 He nuclei
6.
94Be
+
11 H
7.
3719K
→
U
9.
10.
8.
11.
Ca
β emission
U
α emission
1 product
fusion
31
H
+
42 He
fission
→
10
n0
+
3015 P
artificial transmutation
nuclei
→
63
Li
+
42 He
artificial transmutation
01e+
positron
+
3718
+
10 n0
n0bombardment→
23892U
+
42 He
146C
→
147 N
23692
10
n0 n0bombardment+
12.
2410 Ne
13.
21884
14.
α part
+
Po
23999Es
→
nuclei
18775 Re
2410
Ne
Ar
positron emission
14256 Ba
→
24194
+
0-1
→
18875 Re
+
2
+
9136 He
Pu
+
+
3 10 n0
fission
10 n0
artificial transmutation
e− β part
00γ
→
42He
α part
+
21482
+
42 He
nuclei
→
232101
β emission
+
11 H
fission
γ part
γ emission
Pb
α emission
Md
+
10 n0
artificial transmutation
11.B – Nuclear Decay: Writing and Predictions
Instructions: Write balanced nuclear equations for the following nuclear reactions.
1. Technetium-99 (9943 Tc) decays by beta emission to form ruthenium-99 (9944 Ru):
9943
Tc
2. Phosphorus-32 decays by beta emission to form sulfur-32:
23090
Th →
3. Fluorine-18 decays to oxygen-18 by positron emission:
189
F
→
188
4. Krypton-76 absorbs a beta particle to form bromine-76:
7636
Kr
→
7635
5. Aluminum-27 absorbs an alpha particle to form phosphorus-30 and emits a neutron:
2713
Al
→
42
6. Nitrogen-14 absorbs an alpha particle to form oxygen-17 and emits a proton:
147
N
→
+
+
0-1
He
+
22688
O
+
01
e+
+
0-1
e−
He
+
3015
He
→
178
+
42
He
+
0-1
e−
9944
42
42
Ru
Br
e−
Ra
P
+
10 n0
O
+
01
Instructions: Write balanced nuclear equations and predict the missing component of the following nuclear reactions.
7. Decay of polonium-218 by alpha (α) emission:
21884
Po
→
42
He
+
21482
Pb
8. The alpha (α) decay of thorium-230:
23090
Th
→
42
He
+
22688
Ra
9. Decay of carbon-14 by beta ( -) emission:
146
→
0-1
e−
+
147
10. The 3 gamma (γ) particle emission of 3H:
31
→
31
H
+
3 00 γ
11. The alpha decay (α) of radon-198:
19888
→
42
He
+
19486
12. The positron (β+) emission from 11C:
116
+
01
13. The beta (β -) decay of uranium-237:
23792
U
→
0-1
+
23793
14. The single alpha (α) and two gamma emission of 238U:
23892
U
→
23490
Th
+
2 00 γ
15. The positron (β+) emission from 201Hg:
20180
Hg
→
20179
Au
+
01
e+
16. The single alpha and single beta decay of plutonium-248:
24894
Pu
→
24493
Kp
+
42
He
C
H
Ra
C
→
115
B
e−
N
Rn
e+
Np
p+
11.C – Half-Life Calculations
Instructions: ON A SEPARATE SHEET OFPAPER (be sure to #), preform the following half-life calculations using reference Table N found above. NW = NC
Recommended Steps
(1) Variables
(2) Formulas
1. How long does it take a 100.00g sample of Au-l 98 to decay to 6.25g?
(1) R0 = 100.0g Rn = 6.25 g
t1/2 = 2.695 d/life
n=?
t=?
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t = n • t1/2
(3a) n = log ( 6.25g ÷ 100.0g ) ÷ log (0.5) (3b) t = 4.00 lives x 2.695 d/life
n = 4.00 lives
t = 10.8 d
2. How many half-lives will pass by the time a 60.0g sample of Co-60 decays to
7.59?
(1) R0 = 60.0 g
Rn = 7.59 g
n=?
(3a) n = log ( 7.59g ÷ 60.0g ) ÷ log (0.5)
n = 2.98 lives
3. What fraction of a sample of N-16 remains undecayed after 43.2 seconds?
(1) R0 = 1 g
Rn = ?
t1/2 = 7.13 s/life
n=?
t = 43.2 s
Assume you have 1, i.e. 100%
Find first
(2a) n = t ÷ t1/2
(2b) Rn = Ro
leave as fraction
2n
(3a) n = 43.2 s ÷ 7.13 s/life
(3b) Rn = 1 g ÷ 26.06
Rn =
N - 16
4. What is the half-life of a radioactive isotope if a 500.0g sample decays to 62.5g in
24.3 hours?
(1) R0 = 500.0 g Rn = 62.5 g
t1/2 = ?
n=?
t = 24.3 hrs
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t1/2 = t ÷ n
(3a) n = log ( 62.5g ÷ 500.0g ) ÷ log (0.5) (3b) t1/2 = 24.5 hrs ÷ 3.00 lives
n = 3.00 lives
t1/2 = 8.17 hrs/life
5. How old is a bone if it presently contains 0.3125g of C-14, but it was estimated to
have originally contained 80.000g of C-14?
(1) R0 = 80.000 g Rn = 0.3125 g
t1/2 = 5715 yrs/life
n=?
t=?
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t = t1/2 x n
(3a) n = log (0.3125 g ÷ 80.000g ) ÷ log (0.5) (3b) t = 5715 yrs/life x 8.00 lives
n = 8.00 lives
t1/2 = 4.140 x 104 yrs
6. If you are injected with 1.0000 mg of Tc-99, how long will it take for the sample to
decay to 1/64 of its original mass?
(1) R0 = 1.0000 mg Rn = 0.015625 mg t1/2 = 2.13x105 yrs/life n = ?
t=?
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t = t1/2 x n
(3a) n = log(1.0000mg÷0.015625mg)÷log(0.5) (3b) t = 2.13x105 yrs/life x 6.0000 lives
n = 6.0000 lives
t1/2 = 1.28 x 105 yrs
7. What is the half-live of a radioactive isotope if it takes 6.2 days for a 72g sample
to decay to 18g?
(1) R0 = 72 g
Rn = 18 g
t1/2 = ?
n=?
t = 6.2 d
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t1/2 = t ÷ n
(3a) n = log ( 18 g ÷ 72 g ) ÷ log (0.5)
n = 2.0 lives
8. How many half-lives of K-37 will pass after 6.15 seconds?
(1) R0 = X
Rn = X
t1/2 = 1.23 s/life
n=?
(2a) n = t ÷ t1/2
t = 6.15 s
(3a) t = 6.15 s ÷ 1.23 s/life
t = 5.00 lives
9. If a 700.00g sample of I-131 decays to 43.75g, how much time has passed?
(1) R0 = 700.00 g Rn = 43.75 g
t1/2 = 8.021 d/life
n=?
t=?
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t = t1/2 x n
(3a) n = log(43.75g ÷ 700.00g)÷log(0.5) (3b) t = 8.021 d/life x 4.000 lives
n = 4.000 lives
t1/2 = 32.08 d
(2) n = log (Rn ÷ R0) ÷ log (0.5)
n = 6.06 lives
(3) Numerical Step
(3b) t1/2 = 6.2 d ÷ 2.0 lives
t1/2 = 3.1 d/life
10. How long will it take a 3.5g sample of Fr-220 to decay so that only 1/4 of the
original amount of Fr-220 remains?
(1) R0 = 3.5 g
Rn = 0.88 g
t1/2 = 27.4 s/life
n=?
t=?
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t = t1/2 x n
(3a) n = log( 0.88 g ÷ 3.5 g)÷log(0.5)
n = 2.0 lives
(3b) t = 27.4 s/life x 2.0 lives
t1/2 = 55 s
11. If a radioactive sample of a pure material decays from 600g to 75g in 42.9 days,
what radioisotope could be in the sample?
(1) R0 = 600 g
Rn = 75 g t1/2 = ? n = ? t = 42.9 days
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t1/2 = t ÷ n
(3a) n = log ( 75 g ÷ 600 g ) ÷ log (0.5) (3b) t1/2 = 42.9 d ÷ 3 lives
n = 3 lives
t1/2 = 14.3
Referencing Table N then for the calculated t1/2 32 P
12. How many years would it take for a 1.000g sample of U-238 to decay to about
3.9 mg?
(1) R0 = 1000. mg Rn = 3.9 mg t1/2 = 4.47 x 109 yrs/life
n=?
t=?
Notice the unit conversion
Find first
(2a) n = log (Rn ÷ R0) ÷ log (0.5)
(2b) t = t1/2 x n
(3a) n = log(3.9 mg ÷ 1000.mg)÷log(0.5) (3b) t = 4.47x109 yrs/life x 8.0 lives
n = 8.0 lives
t1/2 = 3.6 x 1010 yrs
13. If 13.125g of K-42 remains undecayed after 62.0 hours, what was the original
sample size?
(1) R0 = ? Rn = 13.125 g t1/2 = 12.36 hrs/life
n=?
t = 62.0 hrs
Assume you have 1, i.e. 100%
Find first
(2a) n = t ÷ t1/2
(2b) R0 = Ro x 2n
(3a) n = 62.0 hrs ÷ 12.36 hrs/life
n = 5.02 lives
(3b) Rn = 13.125 g x 25.02
Rn = 426 g K-42
12.A – Defining and Identifying Acids and Bases
Instructions: This is a brief review of acid and base nomenclature practice. Thus, given the formula, provide the correct name. Spelling will of course count.
1.
NaOH
Sodium Hydroxide
2.
H2SO3
Sulfurous Acid
3.
H2S
Hydrosulfuric acid
4.
H3PO4
Phosphoric Acid
5.
NH3
Ammonia
6.
HCN
Cyanic Acid
7.
Ca(OH)2
Calcium Hydroxide
8.
Fe(OH)3
Iron (III) Hydroxide
9.
H3P
Hydrophosphoric Acid
10. NH4OH
Ammonium Hydroxide
Instructions: This continues the review of acid and base nomenclature. Thus, given the name, provide the correct formula. Watch your criss-crossing!!!!
11. hydrofluoric acid H+ F−
HF
12. hydroselenic acid H+ Se2−
H2Se
13. carbonic acid H+ CO32−
H2CO3
14. lithium hydroxide Li+ OH−
LiOH
15. nitrous acid H+ NO2−
HNO2
16. cobalt (II) hydroxide Co2+ OH−
Co(OH)2
17. sulfuric acid H+ SO42−
H2SO4
18. beryllium hydroxide Be2+ OH−
Be(OH)2
19. hydrobromic acid H+ Br −
HBr
20. hydrocarbonic acid H+ C4−
H4C
21. acetic acid H+ C2H3O2−
HC2H3O2
22. Barium hydroxide Ba2+ OH−
Ba(OH)2
Instructions: Based on the provided properties indicate whether each of the following describes an acid (A), a base (B), or both (AB).
(B) 24. Taste bitter
(A) 25. 1st element in formula is usually H
(A) 26. Lose a proton (B-L Theory)
(AB) 27. Conduct electricity
(B) 28. Feel slippery
(A) 29. Taste sour
(B) 34. A solution conducts electricity and feels slipper
(AB) 30. Change color of indicators
(A) 35. A solution conducts electricity and reacts with
metal.
(B) 31. Gain a proton (B-L Theory)
(A) 36. A solution is sour and turn blue litmus paper
red.
(B) 32. 2nd part of formula is usually OH
(A) 33. React with metals to form H2 gas
Instructions: From the formula alone, indicate if the following compounds are acid (A), a base (B), amphoteric (meaning both, AB), or a salt (S).
(A) 37.
CH3COOH
(S) 40.
NH4Cl
(B) 43.
KOH
(A) 38.
H3PO4
(S) 41.
Ca(NO3)2
(B) 44.
SO42-
(AB 39.
HCO3-
(A) 42.
H2SO4
(B) 45.
KOH
Instructions: ABOVE each of the chemical species of the following reactions, label the Bronsted-Lowry acids (A), bases (B), conjugate acids (CA), and conjugate bases
(CB):
A
B
CA
CB
46. HClO4(aq) + H2O(l) ⇄ H3O+(aq) + ClO4–(aq)
A
B
CA
CB
50. HSO3–(aq) + H2O(l) ⇄ H3O+(aq) + SO32–(aq)
A
B
CA
CB
47. H2SO3(aq) + H2O(l) ⇄ H3O+(aq) + HSO3–(aq)
B
A
CA
CB
51. NH3(g) + H2O(l) ⇄ NH4+(aq) + OH–(aq)
A
B
CA
CB
48. HC2H3O2(aq) + H2O(l) ⇄ H3O+(aq) + C2H3O2–(aq)
A
B
CB
CA
52. HF(aq) + HSO3–(aq) ⇄ F–(aq) + H2SO3(aq)
A
B
CA
CB
49. H2S(g) + H2O(l) ⇄ H3O+(aq) + HS–(aq)
A
B
CB
CA
53. HNO2(aq) + HS–(aq) ⇄ NO2–(aq) + H2S(aq)
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
54. What forms when an acid reacts with a base?
A salt water when the base is a hydroxide
55. What does it mean if a compound is said to be amphoteric?
Can BOTH donate anad accept a p+ (H+ ion)
Thus is both an acid and a base
56. Define conjugate acid.
The prod of an acid/base rxn that gained A SIGNLE H+, and losses it again w/the reverse reaction
i.e. The CA of SO32− is HSO3−
57. Define conjugate base.
The prod of an acid/base rxn that lost A SIGNLE H+, and gains it again w/the reverse reaction
i.e. The CB of H2SO3 is HSO3−
58. The stronger an acid is, the weaker its conjugate base. Basically, the more likelier it is for the species to lose a p+, the harder it is to regain it
59. What is the difference between a strong/weak acid/base?
Strong dissociate fully, weak dissociate to some degree less than full i.e. Strong (→) versus Weaker (⇌, equilibrium)
60. What is the difference between “strength” and “concentration”?
Strength = ABILITY to dissocate
Concentration = AMOUNT that dissociates/present per unit of volume
61. List the six (6) strong acids: 3 Monoatomics
HCl HBr HI
3 Polyatomics
H2SO4
HNO3
HClO4
Instructions: Use the rules for determining acid and base strength to decide if the following acids and bases are strong or weak. (Hint: when weak acids dissociate, just take
one hydrogen off the front of the formula. H3PO4 dissociates into H+ and H2PO4-.)
Acid or base?
Strong or Weak?
Particles in water solution:
Acid or base?
Strong or Weak?
Particles in water solution:
62. HCl
_A_
_S_
H+ (aq) Cl−(aq)
68. HCHO2
_A_
_W_
H+ (aq) CH2O−(aq)
63. HBr
_A_
_S_
H+ (aq) Br−(aq)
69. H3PO4
_A_
_W_
H+ (aq) H2PO4−(aq)
64. HNO3
_A_
_S_
H+ (aq) NO3−(aq)
70. NaOH
_A_
_S_
Na+ (aq) OH−(aq)
65. KOH
_A_
_S_
K+ (aq) OH−(aq)
71. H2SO4
_A_
_S_
H+ (aq) HSO4−(aq)
66. Ca(OH)2
_A_
_S_
Ca2+ (aq) 2 OH−(aq)
72. HF
_A_
_W_
H+ (aq) F−(aq)
67. NH4OH
_A_
_W_
NH4+ (aq) OH−(aq)
73. HC2H2O3
_A_
_W_
H+ (aq) C2H3O2−(aq)
Instructions: Complete the equation for the reaction of each of the following ionic molecules with water indicating the state of the products. Indicate whether the ion or
molecule is an acid/base or conjugate acid base according to the Bronsted-Lowry Theory.
A
74. HI(aq)
B
+ H2O(l)
→
CA
H3O+ (aq) +
CB
I−(aq)
B
77. CO32–(aq)
AB
+ H2O(l)
→
CA
HCO3+ (aq) +
CB
OH−(aq)
A
75. HF(aq)
B
+ H2O(l)
→
CA
H3O+ (aq) +
CB
F−(aq)
A
78. O2–(aq)
B
+ H2O(l)
→
CA
OH− (aq)
+
CB
OH−(aq)
B
A
76. C2H3O2–(aq) + H2O(l)
→
CA
HC2H3O2 (aq) +
CB
OH−(aq)
A
B
79. HC2H3O2(aq) + NaOH(aq)
→
CA
H2O (l)
+
CB
NaC2H5O2 (aq)
12.B – pH, pOH, [H+] and [OH-]
Instructions: Calculate the pH or [H+] showing your work and identify as acidic, basic, or neutral. NW = NC
1. [H+] = 1.0 x 10-6 M
pH = _6.0_
-log (1.0 x 10-6 M)
ABN
6. pH = 4.00
[H+] = _1.00 x 10—4 M_
10 −4.00
ABN
2. [H+] = 2.61 x 10-2 M
pH = _1.58_
-log (2.61 x 10-2 M)
ABN
7. pH = 5.89
[H+] = _1.29 x 10—6 M_ 10 −5.89
ABN
3. [H+] = 4.0 x 10-9 M
pH = _8.4_
-log (4.0 x 10-9 M)
ABN
8. pH = 7.00
[H+] = _1.00 x 10—7 M_ 10 −7.00
ABN
4. [H+] = 5.9 x 10-12 M
pH = _11_
-log (5.9 x 10-12 M)
ABN
9. pH = 12.25
[H+] = _5.623 x 10—13 M_ 10 −12.25
ABN
5. [H+] = 1.0 x 10-7 M
pH = _7.0_
-log (1.0 x 10-7 M)
ABN
10. pH = 9.11
[H+] = _7.76 x 10—10 M_ 10 −9.11
ABN
Instructions: Calculate the missing [H+] or [OH–] and identify as acidic, basic, or neutral. NW = NC and N3
11. [H+] = 4.2 x 10-6 M
[OH–] = _2.4 x 10—9 M_
-14
1.0 x 10 = [OH ] [4.2 x 10 -6 M]
ABN
12. [H+] = _2.3 x 10—9 M_
[OH–] = 4.3 x 10-5 M
-14
1.0 x 10 = [ 4.3 x 10-5 M ] [ H+ ]
ABN
Instructions: Complete the following chart. NW = NC show on a SEPARATE SHEET OF PAPER and N3
13.
pOH = 14.00 – 3.68
= 10.82
17.
[ H+ ] = 1.00x10-14 M ÷ 2.20 x10-8 M
pH = - log (4.55 x 10 -7 )
[ H+ ] = 10 -3.68
= 2.09 x 10-4 M
14.
15.
16.
[ OH− ] = 10 -10.32
= 4.786 x 10-11 M
[ H+ ] = 1.00x10-14 M ÷ 8.60x10-5 M
pOH = - log (8.60 x 10-5 )
= 1.16 x 10-10 M
= 4.07
pH = - log (1.16 x 10
= 9.94
-10
)
[OH-] = 1.00x10-14 M ÷ 1.80x10-9 M
pH = - log (1.80x10-9 M )
= 5.56 x 10-6 M
= 8.74
pOH = - log (5.56 x 10 -9 )
= 5.25
pH = 14.00 – 5.48
= 8.52
H+
10 - 5.48
[ ]=
[ OH− ] = 10 – 8.52
18.
19.
20.
10-6
= 3.31 x
M
= 3.02 x 10-9 M
= 4.55 x 10-7 M
= 6.34
pOH = - log (2.20 x10-8 M)
= 7.66
pOH = 14.00 – 10.84
= 3.16
[ H+ ] = 10 -10.84
[ OH− ] = 10 -3.16
= 1.445 x 10-11 M
= 6.92 x 10-4 M
[OH-] = 1.00x10-14 M ÷ 3.82x10-11 M
pH = - log (3.82x10-11 M )
= 2.62 x 10-4 M
= 10.4
pOH = - log (2.62 x 10 -4 )
= 3.58
pH = 14.00 – 2.85
= 11.15
H+
10 – 11.15
[ ]=
[ OH− ] = 10 – 2.85
= 7.08 x 10-12 M
= 1.41 x 10-3 M
12.C – Titration
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, and LEGIBLE regarding indicators.
1. What is a substance that is one color in an acid and another color in a base? _ Indicator _
2. What color does litmus paper turn in the presence of an acid? _ Red _ in the presence of a base? _ Blue _
3. What color does phenolphthalein turn in the presence of an acid? _ Colorless _ in the presence of a base? _ Pink _
4. A solution turns blue with bromcresol green yellow with bromthymol blue. What is the pH range? _ 5.4 → 6.0 _
5. A solution turns yellow with thymol blue and blue with bromthymol blue. What is the pH range? _ 7.6 → 8.0 _
Instructions: ON A SEPARATE SHEET OF PAPER, preform the following titration calculations. NW = NC N3….i.e. MAKE SURE YOU LABEL!
6. If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is
the concentration of the HCl?
1 OH- • 0.1M • 54 ml = 1 H+ • [ HCl ] • 125 ml
[ HCl ] = 0.04 M HCl
12. What is the concentration of a sodium hydroxide solution if 14.5 mL of it are
exactly neutralized by 30.0 mL of a 0.500 M hydrochloric acid solution?
1 H+ • 0.500 M • 30.0 ml = 1 OH+ • [ NaOH ] • 14.5 ml
[ NaOH ] = 1.03 M NaOH
7. If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the
concentration of the NaOH solution?
1 H+ • 0.1M • 25 ml = 1 OH+ • [ NaOH ] • 345 ml
[ NaOH ] = 0.007 M NaOH
13. Phosphoric acid is neutralized by potassium hydroxide according to the following
reaction: KOH (aq) + H3PO4 (aq)  K3PO4 (aq) + H2O (l)
What is the concentration of a phosphoric acid solution if 25.0 mL are exactly
neutralized by 20.0 mL of 2.000 M KOH solution?
1 OH- • 2.000 M • 20.0 ml = 3 H+ • [ H3PO4 ] • 25.0 ml
[ H3PO4 ] = 0.533 M H3PO4
8. If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric
acid solution (H2SO4), what is the concentration of the H2SO4 solution?
1 OH- • 0.5M • 50 ml = 2 H+ • [ H2SO4l ] • 125 ml
[ H2SO4 ] = 0.1 M H2SO4
9. What is the concentration of a hydrochloric acid solution if 35.00 mL of it are
exactly neutralized by 14.8 mL of a 0.500 M sodium hydroxide solution?
1 H+ • [ HCl ] • 35.00 ml = 1 OH- • 0.500 M • 14.8 ml
[ HCl ] = 0.211 M HCl
10. What volume of 0.500 M hydrochloric acid is required to exactly neutralize 40.00
mL 0.150 M NaOH?
1 H+ • 0.500M • VHCl = 1 OH+ • 0.150 M • 40.00 ml
VHCl = 120 ml of 0.500 M HCl
11. What volume of 2.500 M NaOH solution is required to neutralize 25.5 mL of
a 1.200 M HNO3 solution?
1 OH- • 2.500M • VNaOH = 1 H+ • 1.200 M • 25.5 ml
VNaOH = 12.2 ml of 2.500 M NaOH
14. Hydrochloric acid is neutralized by calcium hydroxide according to the following
reaction:
Ca(OH)2 (aq) + 2 HCl (aq)  CaCl2(aq) + 2 H2O (l)
What is the concentration of a calcium hydroxide solution if 15.0 mL are exactly
neutralized by 10.00 mL of 0.250 M HCl solution?
2 OH- • [Ca(OH)2 ] • 15.0 ml = 1 H+ • 0.250 M • 10.00 ml
[ Ca(OH)2 ] = 0.0833 M Ca(OH)2
15. Acid spills are often neutralized with sodium carbonate (baking soda). For
example:
Na2CO3 (s) + H2SO4 (aq)  Na2SO4 (aq) + CO2 (g) + H2O (l)
An instructor dropped a 2.50 L bottle of 18.0 M H2SO4 on a cement floor. How
much sodium carbonate would be required to neutralize it?
18.0 M H2SO4 x 2.50 L = 45.0 mol H2SO4
45.0 mol H2SO4
= 4770 g Na2CO3
1 mol Na2CO3
1 mol H2SO4
105.99 g Na2CO3
1 mol Na2CO3
13.A – Oxidation #’s
Instructions: Assign oxidation numbers to each species in the following. Use your Periodic Table and practice applying the “Rules for Predicting Oxidation Numbers”. After
you have assigned the number, review naming the substance using the correct IUPAC nomenclature. (Example: Cl2 Cl = 0, chlorine)
1.
Mn
Mn = 0
Manganese
10.
CO2
C = +4
2.
S8
S=0
Sulfur
11.
Ne
Ne = 0
3.
NO2
N = +4 O = – 2
Nitrogen Dioxide
12.
H2SO4
H = +1 S = +6
4.
CuNO3
Cu = +1 N = + 5 O = – 2
Copper (I) Nitrate
13.
NaH
Na = +1
5.
Cu(NO3)2
Cu = +2 N = + 5 O = – 2
Copper (II) Nitrate
14.
NaOH
Na = +1 H = +1 O = – 2
Sodium Hydroxide
6.
CuSO4
Cu = +2 S = + 6 O = – 2
Copper (II) Sulfate
15.
K2Cr2O7
K = +1 Cr = +6 O = – 2
Potassium Dichromate
7.
CO
C = +2
Carbon Monoxide
16.
HSO4-
H = +1 S = +6
O=–2
Bisulfate Ion
8.
OH-
H = +1 O = – 2
Hydroxide Ion
17.
HCO3-
H = +1 C = +4
O=–2
Bicarbonate Ion
9.
KMnO4
K = +1 Mn = +7 O = – 2
Potassium Permanganate
18.
C2O42-
C = +3
O=–2
O=–2
Carbon Dioxide
Neon
O=–2
H=–1
O=–2
Sulfuric Acid
Sodium Hydride
Oxolate Ion
Instructions: ON A SEPARATE SHEET OF PAPER, analyze the following redox reactions using the steps below:
19.
2Na0 + 2H+12O−2 → 2Na+1(O−2H+1) + H20
23.
Ca0 (s) + H+12O−2 → Ca+2 (O−2H+1)2 (aq) + H20 (g)
b)
Oxidized = Sodium
Reduced =Hydrogen
c)
Oxidizing Agent =
Reducing Agent=
as it goes from 0 to +1
as it goes from +1 to 0
Hydrogen
Sodium
2H+12S−2 + 3O20 → 2S+4O−22 + 2H+120−2
20.
b)
Oxidized = Sulfur
Reduced =Oxygen
c)
Oxidizing Agent =
Reducing Agent=
b)
Oxidized = Copper
Reduced =Silver
Oxygen
Sulfur
c)
Oxidizing Agent =
Reducing Agent=
as it goes from 0 to +2
as it goes from +1 to 0
Silver
Copper
Zn0 (s) + Cu+2 (N+5O−23)2 → Zn+2 (N+5O−23)2 + Cu0 (s)
22.
b)
Oxidized = Zinc
Reduced =Copper
c)
Oxidizing Agent =
Reducing Agent=
as it goes from 0 to +2
as it goes from +2 to 0
Copper
Zinc
Oxidized = Calcium
Reduced =Hydrogen
c)
Oxidizing Agent =
Reducing Agent=
as it goes from 0 to +2
as it goes from +1 to 0
Hydrogen
Calcium
S0 (s) + H+1N+3O−22 → H+12S+4O−23 + N+12O−2
24.
as it goes from -2 to +4
as it goes from 0 to -2
Cu0 + 2Ag+1(N+5O−23) → Cu+2 (N+5O−23)2 + 2Ag0
21.
b)
b)
Oxidized = Sulfur
Reduced =Nitrogen
c)
Oxidizing Agent =
Reducing Agent=
as it goes from 0 to +4
as it goes from +3 to +1
Nitrogen
Sulfur
Al0 (s) + H+1Cl−1 → Al+3Cl−13 (aq) + H20 (g)
25.
b)
Oxidized = Aluminum
Reduced =Hydrogen
c)
Oxidizing Agent =
Reducing Agent=
as it goes from 0 to +3
as it goes from +1 to 0
Hydrogen
Aluminum
13.B – Redox Reactions
Instructions: For each of the following, assign oxidation numbers, write the half reactions and decide which element is oxidized, which is reduced, which is the oxidizing
agent, and which is the reducing agent, and then combine the half reactions to write a balanced equation.
YOU CANNOT BREAK UP POLYATOMICS!!!!!!
1.
2.
3.

Ni0 +
Oxid =
Red=
Sn+4+4
nickel
tin
Oxid ½ :
Red ½ :
2 ( Ni0
Sn+4 + 4 e −
Combine:
2 Ni0 + Sn+4 + 4 e − → 2 Ni2+ + 4 e − + Sn0
Balanced :
2 Ni0 + Sn+4

Ni+2+2
+
Oxid Agent =
Red Agent=
→ Ni+2
→ Sn0
Sn0
tin
nickel
+ 2e−)
Ag+1 +1
mercury
silver
Oxid ½ :
Red ½ :
Hg0
2 (Ag+1 + 1 e −
Combine:
Hg0 + 2 Ag+1 + 2 e −→ Hg2+ + 2 e − + 2 Ag0
Balanced :
Hg0 + 2 Ag +1
H+1Cl+1O−2+ H+12O−2
iodine
chlorine
H+1N+5O−23 + H+13P+3O−23  N+2O−2 +
H+13P+5O−24 + H+12O−2
Oxid =
phosphorous
Oxid Agent =
nitrogen
Red=
nitrogen
Red Agent=
phosphorous
Oxid ½ :
Red ½ :
3 ( H2O + PO33→ PO43− + 2 e − + 2 H+ )
2(6 H+ + 3 e − + NO3− → N+2 + H2O )
Combine:
→
3 H2O + 3 PO33- + 12 = 6 H+ + 6 e − −
3 PO43− + 6 e − + 6 H+ + 2 N+2 + 6 = 3 H2O
Balanced :
3 PO33- + 6 H+
→ 3 PO43− + 2 N+2 + 3 H2O
Na+1I−1
Oxid =
Red=
H+12S+6O−24 
iodine
chlorine
I20 +
Na+12S+4O−23 + H+12O−2
Oxid Agent =
chlorine
Red Agent=
iodine
→ 2 Ni2+ + Sn0
Hg0 +
Oxid =
Red=
I20 +
Oxid =
Red=
4.
Ag0
+
Oxid Agent =
Red Agent=
Hg+2+2
silver
mercury
5.
→ Hg+2 + 2 e −
→ Ag0 )
→ Hg2+ + 2 Ag 0

H+1I+5O−23 +
H+1Cl−1
chlorine
Oxid Agent =
Red Agent=
iodine
Oxid ½ :
Red ½ :
2 (3 H2O + I0
→ IO3− + 5 e − + 6 H+ )
5(2H+ + 2 e − + ClO − → Cl− + H2O )
Combine:
→
6 H2O + 2 I0 + 10 H+ + 10 e − + 5 ClO −
2 IO3 − + 10 e − + 12 = 2 H+ + 5 Cl − + 5 H2O
Balanced :
H2O + 2 I0 + 5 ClO − → 2 IO3 − + 2 H+ + 5 Cl −
6.
+
Oxid ½ :
Red ½ :
2 I −1
→ I2 0 + 2 e −
+
−
2−
2 H + 2 e + SO4 → SO32− + H2O
Combine:
→
2 I −1 + 2 H+ + 2 e − + SO42−
I2 0 + 2 e − + SO32− + H2O
Balanced :
2 I −1 + 2 H+ + SO42− → I2 0 + SO32− + H2O
H+12S−2 +
Oxid =
Red=
O2 0
sulfur
oxygen
Oxid ½ :
Red ½ :
2 ( S −2
3 (O20 + 4 e −
Combine:
→
2 S −2 + 12 e − + 3 O20
2 S +4 + 12 e − + +2 O −2
Balanced :
2 S −2 + 3 O2

S+4O2−2 +
Oxid Agent =
Red Agent=
→ S +4 + 6 e − )
→ 2 O −2 )
→ 2 S +4 + 6 O −2
H20
oxygen
sulfur