Final review problems
Or… breathe easy, you’re almost
done!
Practice Problem 1: Half-Life
From the first exam: The half-life of
cobalt-60 is 5.3 yr. How much of a
1.000 mg sample of cobalt-60 is left
after a 15.9 yr period?
Problem 1: Solution
Most of you got this right on the exam, but
through a very difficult pathway. There
is an easier solution!
One half-life is 5.3 yr. The time that
elapsed is 15.9 yr = three half-lives.
More problem 1: solution
After 1 half-life, 50% of the original
amount remains.
After 2 half-lives, 25% of the original
amount remains.
After 3 half-lives, 12.5 % of the original
amount remains.
More problem 1: solution
Also note that on this exam, I didn’t give
you the first-order integrated rate
equation -- suggesting that there may
be an easier pathway!
12.5% of 1.000 mg = 0.1250 mg remains
after 15.9 yr.
Problem 2: pH of weak acid
Calculate the pH of a 0.100 M aqueous solution of
hypochlorous acid (HOCl, Ka = 3.5 * 10-8).
Assume that the principal acid/base reaction is that with
HOCl, not with water! So,
Ka = 3.5 * 10-8 = [H+][-OCl]/[HOCl]
and I’ll set up an ICE table on the next page.
Problem 2: Solution
HOCl
H+
+ -OCl
------------------------------------------------------------I(M)
0.100
0
0
C(M)
-x
+x
+x
E(M) 0.100 - x
x
x
Ka = 3.5 * 10-8 = (x)(x)/0.100 - x
Problem 2: Solution
Assume x is small and continue to solve. In this case,
(0.100 - x) ~ 0.100.
Ka = 3.5 * 10-8 ~ x2/0.100
and x = 5.9 * 10-5, which is much less than 5% of our
initial amount, so the assumption that “x is small” is
valid.
[H+] = x = 5.9 * 10-5
pH = -log [H+] = 4.23
Problem 3: Adding strong
acid to a buffer
Calculate the change in pH when 0.010 mol HCl (a
strong acid) is added to 1.0 L of a solution containing
0.050 M acetic acid and 0.050 M acetate. (Ka of
acetic acid is 1.8 * 10-5)
First, calculate the initial pH of the buffer prior to adding
the HCl, using the Henderson-Hasselbalch equation:
pH = pKa + log ([acetate]/[acetic acid])
= (- log 1.8 * 10-5) + log 1 = 4.74
Problem 3: Solution
Now, what happens when we add acid to our buffer?
The acid is consumed by the base component of the
buffer.
H+
+ acetate
acetic acid
-----------------------------------------------------------------------Before
0.010M
0.050 M
0.050 M
After
0
0.040 M
0.060 M
Problem 3: Solution
Use this information to calculate your new pH, to see
how far it has deviated from the pH of the original
buffer solution.
pH = 4.74 + log (0.040/0.060) = 4.56
The change in pH would therefore be 0.18 pH unit.
Problem 4: Electrolysis
Stoichiometry
How long must a current of 5.00 A be applied to a
solution of Ag+ to produce 10.5 g silver metal?
Thinking of the overall process before you start…
Given g of silver --> mol of silver --> mol of electrons
--> coulombs of charge required --> time required
Remember this process, and you can solve any related
problem!
Problem 4: Solution
Convert mass of silver into mol:
10.5 g Ag (1 mol Ag/107.9 g Ag) = 9.73 * 10-2 mol Ag
Since Ag forms a +1 ion, this particular electron transfer
reaction transfers 1 mol electrons per mol Ag -- so,
9.73 * 10-2 mol electrons is also transferred, here.
Problem 4: Solution
Use the Faraday Constant to convert this into coulombs
of charge:
9.73 * 10-2 mol e-(96,500 C/mol e-)
= 9.39 * 103 C
Lastly, the current of 5.00 C/s must produce 9.39 * 103
C of charge. Time required =9.39 * 103 C/5.00 C/s
= 1.88 * 103 s = 31.3 min
And finally… something funny
(what not to do)
I named my new dog Stay. Somehow he
got confused when I called him: “Come
here, Stay! Come here, Stay!”