Chemistry 114
Fourth Hour Exam
Name:____________
1. (10 points) What is the pH and pOH of a .00057M solution of HCl?
HCl will completely dissociate so there will be .00057 of [H+] in solution
pH = -log[H+]
= -log(.00057)
=3.24
pH + pOH = 14
3.24 + pOH = 14
pOH = 14-3.24 = 10.76
2. (10 points) What is the pH and pOH of a 3.7x10-3 M solution of HF, K a = 7.2x10-4 ?
Since I gave you a K a you know this is a weak acid and will not dissociate completely, thus
you must use the K a expression:
Ka=[H+][A-]/[HA]
X=[H+]=[A -]
[A-] = 3.7x10-3 -X
our usual assumption is that X in this last term is small so [A -] • 3.7x10-3
Thus 7.2x10-4 = X 2 / 3.7x10-3
X=sqrt(7.2x10-4 × 3.7x10-3 )
X = .00163
pH = 2.79
pOH = 14-2.79 = 11.21
One thing we talked about in class was that as a weak acid gets more dilute, it ionizes
more, hence the assumption we made about [A -] is probably wrong. If you do this correctly:
Thus 7.2x10-4 = X 2 / (3.7x10-3 -X)
Using either the quadratic or the solver function on the calculator you will fined that
X = .00131
pH = 2.88
pOH = 14-2.79 = 11.12
A significant difference. Note I accepted the first answer for full value, but I gave bonus
points to anybody who did this problem the long way
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3. (10 points) Arrange the following 0.1M solutions in order from most acidic to most basic:
Al(NO3)3, NaNO3, NH4NO3, NaCH3COO, HClO4, HClO3, K2O
Acidic
Strong A
HCLO4
Weak A
(Any order)
HClO3, Al(NO3)3,NH4NO3
Neutral
Weak Base
Basic
Strong base
NaNO3
NaCH3COO
K2 O
4. (10 points) I am going to make a buffer using compound with the trivial name Tris. The
buffer is going to contain.015M Tris in the acid form and .035M Tris in the conjugate base
form. What is the pH of this solution if the pKa of Tris is 8.075?
Henderson Hasselbach
pH = pKa + log ([A -]/[HA])
HA = acid form, A - = conjugate base
= 8.075 + log (.035/.015)
= 8.44
5. (2 points) A. Make a rough sketch of a titration curve of a weak base being titrated with a
strong acid.
B. (2 points) On this curve indicate where the equivalence point is.
See arrow
C. (2 points) Define what is meant by the equivalence point.
The point where you have added an amount of acid equivalent to the base that you
started with
D. (2 points) Is the pH at the equivalence point <7, =7 or >7?
<7 (have converted weak base to conjugate acid so acidic)
E. (2 points) How would you use this curve to determine the pKB of the base? Indicate the
point you would use and the equation that you would use at this point.
In middle of buffer region (½ way to equivalence point) pOH = pKB
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6A. (8 points) Calculate the K SP of Bi2S3, given that its solubility is 1.0x10-15 mol/L.
If the solubility of Bi2S3 is 1.0x10-15 that means the 1.0x10-15 moles of this salt dissolves in 1
liter of solution. The reaction will be
Bi2S3 (s) W 2Bi3+(aq) + 3 S 2-(aq)
So [Bi3+(aq)] = 2x10-15 and [S 2-(aq)] = 3x10-15]
Ksp = [Bi3+]2[S2-]3
[2x10-15]2[3x10-15]3
=1.08x10-73
B(2 points) Would ths solubility of Bi2S3 increase, decrease, or remain constant if tried to
dissolve this salt in .1M Bi(NO3)3
This would add the common ion Bi3+, a product in the solubility reaction, and hence
would drive the reaction toward reactants (solid) and the solubility would decrease
7 (10 points) Define
Entropy
A thermodynamic measure of randomness
Free energy
G = H + TS, or the energy available in a system to do work, or measure of spontaneity
in a process.
The second law of thermodynamics
The entropy of the universe is always increasing
A galvanic cell
A system of chemical reactions in which a spontaneous chemical reaction is used to
generate an electrical current.
An anode
The negative pole of a galvanic cell or the pole of a galvanic cell in which the oxidation
reaction occurs
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8. (10 points) I have a chemical reaction with a )Hrxn of +18 kJ/mol, and a )Srxn of
-50J/K @mol
A . What is the )Grxn at 25K
)G = )H - T)S
= 18000- 25(-50)
= 19.25 kJ/mol
B What is the )Grxn at 25oC
= 18000- 298(-50)
= 32.9 kJ/mol
C. What is the )Grxn at 250oC
= 18000- 523(-50)
= 44.15 kJ/mol
9. (10 points) If the Ka for acetic acid (CH3COOH) is 1.8x10-5 . What is the )G for the
reaction CH3COOHWCH3COO- + H+ at 25oC ?
)G = -RTln(K)
= 8.3145(298)ln(1.8x10-5 )
= +27.1 kJ/mol
10. (10 points) Diagram a galvanic cell. Indicate on this diagram the anode, cathode,
direction of electron flow, the place where the oxidation reaction occurs and the place where
the reduction reaction occurs
Diagram is difficult to make flow of electrons should be from left to right the left side has
the oxidation reaction and the - or cathode pole, the right side has the reduction reaction and
is the + or anode pole.
Question to contemplate: If the second law of thermodynamics is true, what will the universe
look like at the end of time?
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