Chemistry 114
Fourth Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. (12 points) Calculate the pH of the following solutions
A. 1.45x10-5M HCl
[H+]=1.45x10-5
pH=-log(1.45x10-5)
=-(-4.84)
=4.84
B. 1.45x10-5 M Fe(OH)3
1.45x10-5M Fe(OH)3 x (3 moles OH-/1 mole Fe(OH)3)
=4.35x10-5 OHpOH = -log(4.35x10-5) = 4.36
pH + pOH =14; pH= 14-4.36
pH=9.64
C. 1.45x10-5M Nitrous Acid (KA = 5.6x10-4)
Ka=[A-][H+]/[HA]
5.6x10-4 = X2/(1.45x10-5 -X) Assume ~ X2/1.45x10-5
X2 = 5.6x10-4 x 1.45x10-5
X = sqrt(8.12x10-9)
X =9.01x10-5, pH =-log(9.01x10-5) = 4.05
D. For a bonus point, What is the formula of nitrous acid
HNO2
2. (12 Points) Identify the following compounds as : SA - Strong Acid, WA- Weak Acid, NNeutral, WB - Weak Base, or SB - Strong base.
HClO4 ___SA_______
NH3 ____WB______
NH4NO3 ___WA_______
SeO2 ___A (S or W?)_______
BaO __SB________
Mn(NO3)5__A(S or W?)________
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3A. (5 points) Sketch the titration curve for a weak acid being titrated with a strong base.
On this curve identify the following features: X-Axis, Y- Axis, Buffer region, Stoichiometric
point (or equivalence point), point where pH = pKa.
Should look like figure 21.9 from your text
(1 point) Is the pH at the equivalence point <,=,or > 7?
B. (5 points)Sketch the titration curve for a weak base being titrated with a strong acid. On
this curve identify the following features: X-Axis, Y- Axis, Buffer region, Stoichiometric point
(or equivalence point), point where pH = pKa.
Should look like figure 21.11 from your text
(1 point) Is the pH at the equivalence point <, =, or > 7?
4.Define the following terms:
Conjugate base
What remains of an acid after it donates its proton.
Second law of thermodynamics
The entropy of the universe is always increasing.
Thermal disorder
Entropy derived from the distribution of energy states at a given temperature
Entropy-favored reaction
A reaction where ÄS is positiveÄ
van’t Hoff Equation
Hydrocarbon
A compound that contains only hydrogen and carbon.
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5. (12 points) Does S increase, decrease, or remain the same in the following reactions:
H2O (l) 6H2O (s) ___decrease_______
4Al(s) + 3O2(g) 6 2Al2O3 (s) __decrease________
2C2H6(g) + 5O2(g)64CO2(g) + 6H2O(g)_increase_________
C2H6(g)6C2H4(g) + H2(g) __increase________
2H2(g) +O2(g) 62H2O(g)__decrease________
Cu2+(aq) + 6NH3(aq) 6Cu(NH3)6(aq) _decrease_________
6. (12 points) When will the following process be spontaneous?
A.
Br2(l) 6Br2 (g) (ÄH is positive)
Never
At low T
At High T
Always
(Circle one)
Why?ÄS+,ÄH+ Will be favorable at High T where favorable ÄS term is maximal
B.
Never
CH4(g) + 2O2(g) 6CO2(g) + H2O(l) (ÄH is negative)
At low T
At High T
Always
(Circle one)
Why?__ÄS-,ÄH- Will be favorable at Low where unfavorable S term is a minimum
C.
Never
3O2(g) 6 2O3 (g) (ÄH is positive)
At low T
At High T
Always
(Circle one)
Why? Both ÄH and ÄS are unfavorable at all T
D.
Never
NaOH(s) 6Na+(aq) + OH-(aq) (ÄH is positive)
At low T
At High T
Always
Both ÄH and ÄS are unfavorable at all T
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(Circle one)
7A. (4 points) ÄGo for the reaction CH3COO-(aq) + H2O(l) 6 CH3COOH(aq) + OH-(aq) is
27.1 kJ/mol. What is the K for this reaction at 25oC ?
ÄG0=-RTlnK
27100 J = -8.315 x 298 x ln(K)
-27100/(8.314x298) = lnK
-10.94=lnK
K=e-10.94 = 1.77x10-5
B. (4 points) The reaction CH3COO-(aq) + H3O+ (aq) 6CH3COOH(aq) + H2O(l) has a K of
5700. What is the ÄGo for this reaction at 250C ?
ÄG0=-RTlnK
X = -8.315 x 298 x ln(5700)
X=-8.314x298x8.65
= -21,400J or -21.4 kJ
C. (4 points) The reaction A(aq) + B(aq) 6C(aq) has a ÄGo of +10 kJ/mol. What is the ÄG
for this reaction when the concentrations of A, B, and C are all 5M and T = 25oC ?
ÄGRXN = ÄG0 + RTlnQ
=10,000 + 8.314(298)ln(5/5A5)
=10,000 + 2478ln(.2)
=10,000 + -3990
=6,010 J or 6.01 kJ
8. There are nine different structural isomers for Heptane (C7H16). You get 6 points for
drawing 6 different structural isomers for this compound and 6 more points for naming
them correctly.
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