5 Questions you want to review from Exam #2

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Taking a look at how Section 001 did on Exam #2, I find approximately 5 questions in which there was a slight statistical deviation from the overall class average. I list these here along with a discussion of each problem.

I strongly recommend that you review these before NOW as well as before the final exam. That way you'll reinforce the logic behind these problems.

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Here we go:

Question #4: Which of the following result in a buffered solution?
- In any titration type of question (yes, it's a titration...remember that buffers and titrations are intimately related), we let any stoichiometric reactions occur and see what we have left before proceeding.
  1. 100 mL of 0.10 M HCl + 100 mL of 0.1 M NH₃
    HCl is a strong acid, so this reaction proceeds completely to the right (it's not an equilibrium). The reactants are both present in a 1:1 ratio (0.1 L)(0.10 mol/L) = 0.01 mol, so there is no limiting reagent. At the end we'll simply have 0.01 mol of NH₄Cl. This is not a buffer.
  2. 100 mL of 0.10 M HCl + 200 mL of 0.1 M NH₃
    The reaction proceeds as in A, except that HCl is the limiting reagent. At the end of the reaction we'll have 0.01 mol of NH₄Cl and 0.01 mol of NH₃ will be left unreacted. NH₄Cl is the conjugate acid of a weak base, ammonia, so this is a buffer system.
  3. 200 mL of 0.10 M HCl + 100 mL of 0.1 M NH₃ This is the same reaction again, only now ammonia is the limiting reagent. At the end of the reaction we'll have 0.01 mol of NH₄Cl and 0.01 mol of HCl will be left unreacted. This is not a buffer, but merely an acidic solution!
  4. 100 mL of 0.1 M NH₃ You might try to argue that this can react with water to generate NH₄⁺, but because the K_b is so small, hardly any would exist in solution. Remember, for a buffer to be effective, it must contain relatively high concentrations of a weak acid/base and its conjugate.
Question #11: What is the pH of the solution that resulls when 0.20 moles of HCl are added to 1.0 L of a 0.40 M sodium acetate, CH₃COONa, solution? K_a for acetic acid = 1.8 x 10^-5 (assume constant volume).
- In any titration type of question we let any stoichiometric reactions occur and see what we have left before proceeding. (Gee, where have we heard that before?)
  The reaction of HCl with sodium acetate will be irreversible because HCl is a strong acid. I've omitted the spectator ions in this equation:
  
  H⁺ (aq) + CH₃COO^- (aq) ----> CH₃COOH (aq)
  
  We can set up a chart to determine what's present now:
  
  [H⁺] [CH₃COO^-] [CH₃COOH]
  
  Initial 0.20 0.40 0
  
  Change -0.20 -0.20 +0.20
  
  Final 0 0.20 0.20
  
  We can now look at the ionization reaction of either acetic acid or acetate ion and set up another chart. However, an easier method is to recognize that we have a buffer and can simply use the Henderson-Hasselbach equation:
  
  pH = pK_a + log [conj base]/[acid].
  
  The pK_a is simply - log[K_a] = -log[1.8 x 10^-5] = 4.74. The log term disappears because log(0.20/0.20) = log(1) = 0.
  
  Therefore, the pH = 4.74.
Question #13: Calculate the solubility in mol/L, of AgCl in 0.015 M MgCl₂. K_sp for AgCl = 1.6 x 10^-10.
- The key to a solubility problem is always writing up a good chart. We have MgCl₂ already in the solution, and we know that it dissociates in water according to:
  MgCl₂ (aq) ----> Mg²⁺ + 2 Cl^-
  
  Notice that each mole of magnesium chloride gives two moles of chloride ion. Therefore, the [Cl^-] is twice the [MgCl₂], or 0.030 M.
  
  We now add AgCl. The reaction we are concerned with is:
  
  AgCl (s) < ===> Ag⁺ (aq) + Cl^- (aq)
  
  We can now construct a table, using the unknown variable s to represent the amount of AgCl that can dissolve. Each AgCl dissociates to one silver and one chloride ion, so s also represents the concentration of these ions:
  
  [Ag⁺] [Cl^-]
  
  Initial 0 0.030
  
  Change s s
  
  Final s 0.030 + s
  
  We know that the K_sp expression is given by K_sp = [Ag⁺][Cl^-]. It's products over reactants as always, and the reactant in this case is a solid and is not included in the expression.
  Plugging our equilibrium values into the equation we get 1.6 x 10^-10 = (s)(0.030 + s). We can solve this using the quadratic formula, but it is much easier to say that s is a small number and the quantity 0.030 + s can be approximated as 0.030.
  
  Doing this gives us 1.6 x 10^-10 = 0.030s. Solving for s gives us 5.3 x 10^-9.
  
  Lastly, note that our approximation that s is small compared to 0.030 is valid.
Question #10 What is the pH of a solution after 10 mL of 0.10 M NaOH has been added to 20 mL of 0.25 M HBr. Assume volumes are additive.
- In any titration type of question we let any stoichiometric reactions occur and see what we have left before proceeding. (Gee, where have we heard that before?)
  This reaction of a strong acid and strong base will be irreversible:
  
  NaOH (aq) + HBr (aq) ---> NaBr (aq) + H₂O (l)
  
  We also know that NaBr does not affect the final pH because both Na⁺ and Br^-, being the conjugate of a strong base and acid respectively, do not react with water.
  
  So all we need to do is find the limiting reagent. The number of moles of NaOH is (0.01 L)(0.10 mol/L) = 0.001 moles. The number of moles of HBr is (0.02 L)(0.25 mol/L) = 0.0005 moles. If we wanted to we could write a chart:
  
  NaOH HBr NaBr
  
  Initial 0.001 mol 0.005 mol 0
  
  Change -0.005 mol -0.005 mol +0.005 mol
  
  Initial 0 mol 0.004 mol 0.005 mol
  
  Important: When we do titrations we must remember to account for the change in volume (hence the note in the question, "Assume volumes are additive"). The new volume is therefore 0.01 + 0.020 = 0.030 L.
  
  The concentration of HBr is therefore (0.004 mol)/(0.030 L) = 0.13 M. Since HBr is a strong acid (completely dissociates) and NaBr does not react with water, the [H⁺] = 0.114 M.
  
  Therefore the pH = -log[H⁺] = 0.88
Question 5: What is the pH of a buffer solution made of 0.15 M CH₃NH₃Cl and 0.25 M CH₃NH₂? K_b for CH₃NH₂ is 4.4 x 10^-4.
- We are given the concentrations of the two components of a buffer, so we can go ahead and use the Henderson-Hasselbalch equation:
  pH = pK_a + log [conj base]/[acid].
  
  The pK_b = -log(K_b) = 3.36. Since pK_a + pK_b = 14, we can therefore determine that pK_a = 10.64.
  
  Solving, this gives us pH = 10.64 + log(0.25/0.15) = 10.64 + 0.22 = 10.86.
  
  Be careful not to mix up the conjugate base and acid or you'll get 10.64 - 0.22 as your answer!

	[H⁺]	[CH₃COO^-]	[CH₃COOH]
Initial	0.20	0.40	0
Change	-0.20	-0.20	+0.20
Final	0	0.20	0.20

	[Ag⁺]	[Cl^-]
Initial	0	0.030
Change	s	s
Final	s	0.030 + s

	NaOH	HBr	NaBr
Initial	0.001 mol	0.005 mol	0
Change	-0.005 mol	-0.005 mol	+0.005 mol
Initial	0 mol	0.004 mol	0.005 mol

Learn from your mistakes so you don't repeat them! Watch how the logic flows so you can learn how to approach new or unfamiliar problems.

Be sure to try not only the TopClass exercises provided here, but the extra ones at a variety of at other universities.