Name:_____________
(4 points)
Chemistry 114
Fourth Hour Exam
Remember- Show all work for partial credit. Each problem is worth 12 points.
1.
A. What is the pH of .0000576M HNO3
Strong Acid, 100% dissociation, [H+] = .0000576
-log(.0000576) = 4.24
B. A solution of RbOH has a pH of 9.56, what is the concentration of RbOH?
14=pH + pOH, pOH= 14 - 9.56 = 4.44
[OH-] = 10-4.44 = 3.63x10-5 = [RbOH]
C. What is the pH of .076M Ca(OH)2?
Ca(OH)2 W Ca2+ + 2 OH.076M Ca(OH)2 x [2OH/1Ca(OH)2] = .152M OHpOH = -log(.152) = .818
14 = pH + pOH ; 14-.818 = pH = 13.182
D. A solution of HClO4 has a pOH of 15, what is the concentration of HClO4?
14 = pH + pOH; 14-15 = pH;
[H+] = 10-(-1) = 10 = [HClO4]
pH=-1
2. An aqueous solution is 13.9% HNO3 by mass and has a density of 1.052g/mL What
is the pH of this solution?
Let’s start with 1 liter of solution, or 1000 mLs
1000 mLs x 1.052 g/mLs = 1052 grams of solution
The solution is only 13.9% acid so
1052 x (13.9/100) = 136.76 grams of acid
HNO3 has a molar mass of 1.008+ 14.01 = 3(16.000) = 63.018 g/mol
136.76 g/63.018 = 2.17M
[H+] = 2.17; pH=-log(2.17) = -.336
2
3. A. (9 points) Hydrazoic acid (HN3) has a Ka of 3.0x10-5. What is the pH of a .001M
solution of Hydrazoic acid.
Weak acid problem
I
C
E
HA W H+ + A.001 0
0
-X
+X
+X
.001-X
X X
Ka = 3x10-5 = X2/(.001-X)
Ka is small enough that we can approximate that .001 ~ .001-X
Ka = 3x10-5 = X2/(.001)
X = [H+] = [A-] = sqrt(.001 x 3x10-5) = 1.73x10-4
pH = -log(1.73x10-4 = 3.76
If you used the quadratic I gave you some bonus points and the answer was
3.80 about an 8% error
3B. (3 points) What is the KB of the Hydrazoic anion, N3-?
This is the conjugate base of a weak acid so Kw = Ka @Kb
Kw/Ka = Kb = 1x10-14/3x10-5 = 3.33x10-10
4. Classify the following compounds as strong acids (SA), Weak Acids (WA) Neutral
(N), Weak bases (WB), or strong bases (SB)
_WA____
HNO2 (Ka = 5.6x10-4)
_WB____
CH3NH2 (Kb = 4.6x10-4)
_WA____
CH3NH3ClO4
_WB____
KNO2
__B___
MgO
Could be either strong or weak
__A___
NO2
Could be either strong or weak
3
5. Define the following terms:
Lewis Acid
An electron pair acceptor
Conjugate acid
What remains after a weak base has accepted a proton
pOH
-log[OH-]
Stoichiometric point
In a titration it is the point where the moles of titrant exactly equals the moles of
analyte. In the case of acids and bases, moles of acid = moles of base.
amphoteric
A substance that can act as either an acid or a base.
spontaneous process
A process that will occur without the addition of energy from an outside source.
entropy driven reaction
A chemical reaction that is spontaneous because ÄS is positive.
Second law of thermodynamics
The entropy of the universe is always increasing.
6. Circle the compound with the higher S in the following sets AND EXPLAIN WHY
H2O (l)
H2O(s)
WHY: Liquids more random than solids
H2O @ 200C
H2O @ 45oC WHY: At higher temp, more motion so
more randomness
NaCl(s)
NaCl(aq)
WHY: Aqueous ions more random than
a solid
F2(g)
Cl2(g)
WHY: Greater mass so more energy
levels at a given temperature.
C2H2(l)
C2H6(l)
WHY: More atoms so more motion and
more randomness
WHY: The square compound has
restricted motion so it is less random
4
7. The ÄHVap for water is 40.65 kJ/mol @ 1000C. What is the ÄSVap at this
temperature
ÄS = q/T
At constant pressure ÄHÄ=q
ÄS = ÄH/T
T must be in K
ÄS = 40.65 kJ/mol/ 373K
= .109 kJ/mol@K
Or 109 J/mol@K
8. A( 6 points) . Calculate ÄGrxn for the reaction C2H4(g) + H2O(g)6C2H5OH(g) given the
following data:
ÄGf0 (kJ/mol)
C2H4(g)
+68.4
ÄGrxn = sum of ÄGfo products - ÄGfo reactants
C2H5OH(g) -167.9
C2H5OH(l)
-174.8
= -167.9 -[68.4 + (-228.6)]
= -167.9 -(-160.2)
H2O(g)
-228.6
H2O(l)
-237.1
= -7.7 kJ/mol
B (6 points). Given the ÄGrxn found in part A, determine the K for this reaction at 25o C
(NOTE: If you did not calculate a ÄGrxn in part A, TAKE A GUESS so I can give
you credit for this calculation)
ÄG = -RTlnK
-7.7 kJ/mol = -8.314 J/mol x 298K x lnK
-7700 J/mol / (-8.314 x 298) = lnK
3.11 = lnK
e3.11 =K
K= 22.42