Final review problems

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Final review problems

• Or breathe easy, youre almost done!

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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?

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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.

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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.

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More problem 1 solution

• Also note that on this exam, I didnt 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.

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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 Ill set up an ICE table on the next page.

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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

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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

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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

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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

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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.

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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 --gt mol of silver --gt mol of
  electrons
• --gt coulombs of charge required --gt time
  required
• Remember this process, and you can solve any
  related problem!

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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.

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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

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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!