5 Questions you want to review from Exam #2.2

These are the 5 most commonly missed questions from the Spring 1999 CHE 107 Exam #2. We 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.

Here we go:

1. Question #14: A 0.25 M aqueous solution of NaX has a pH of 9.24. What is the value of the acid ionization constant, K_a for HX?
   • Based on our knowledge of salt hydrolysis, we know that Na⁺ does not affect pH, so all we need to worry about here is CN^-. CN^- is a base; if we didn't know that we could write the reaction of it with water to see that it must accept H⁺ and form both HX and HO^-

     X^-(aq) + H₂O HX(aq) + HO^-

     Since this is a base dissociation (i.e it makes HO^- when we put it in water), we can find K_b. If we know K_b, then we can use the relationship K_aK_b = K_w = 1 x 10^-14 to find K_a!

     To find K_b, we need the equilibrium values of the species in the equation shown above. This calls for our ever-famous ICE chart. Let's start by putting in our initial values:

     	[CN^-]	[HX]	[HO^-]
     Initial	0.25	0	0
     Change	?	?	?
     Equilibrium	?	?	?

     Before we go crazy sticking in x's all over the place, let's ask if we know any other values? Hmmm... we are given pH. Can I get [OH^-] from this? Yes! pH + pOH = 14, so pOH = 14 - pH = 4.76.

     We know that pOH = -log[OH^-] = 4.76, so if we take the inverse log of -4.76 we get [OH^-] = 1.738 x 10^-5 M. So I can stick that into my chart for the equilibrium value. And if I started out with 0 [OH^-], then my change in concentration must also be +1.738 x 10^-5.

     	[CN^-]	[HX]	[HO^-]
     Initial	0.25	0	0
     Change	?	?	+1.738 x 10^-5
     Equilibrium	?	?	1.738 x 10^-5

     In fact, since I form one HX for every OH^-, I must also have a change of +1.738 x 10^-5 for [HX] change! And the amount of X^- consumed must also be 1.738 x 10^-5! Which means we can complete our chart!

     	[CN^-]	[HX]	[HO^-]
     Initial	0.25	0	0
     Change	-1.738 x 10^-5	+1.738 x 10^-5	+1.738 x 10^-5
     Equilibrium	0.25	1.738 x 10^-5	1.738 x 10^-5

     Notice that 0.25 - 1.738 x 10^-5 = 0.24998, which is 0.25 as far as we are concerned.

     The K_b expression for our equilibrium is simply [HX][OH^-]/[X^-]. Substituting in the numbers from the table above gives us K_b = 1.21 x 10^-9. Plugging that into our K_aK_b = K_w equation gives us K_a = 8.3 x 10^-6.

     A wise student would have realized that this question was identical to problem 8 from CAPA set #5!

2. Question 15. I which one of the following acid-base reactions does the Lewis acid appear first?
   1. 6 CN^- + Fe³⁺ Fe(CN)₆³⁺
   2. NH₃ + BF₃ H₃NBF₃
   3. B(OH)₃ + H₂O B(OH)₄^- + H⁺
   4. H₂O + HCl H₃O⁺ + Cl^-

      Let's take these one at a time. But first, remember our definitions. A Lewis acid is an electron pair acceptor and a Lewis base is an electron pair donor. Also notice that we are told all of these are acid-base reactions!

   1. CN^- is a Bronsted base (we know this from question 14), so therefore it must also be a Lewis base (all Bronsted bases are Lewis bases, but not vice-versa). So this can't be right.

   2. As in A., ammonia is a Bronsted base, so it is also a Lewis base. Alternatively, we could note that NH₃ has a lone pair of electrons on it (draw the Lewis structure and see!), and that BF₃ is 2 electrons short of an octet and has an empty orbital (again, draw the Lewis structure). In fact, this particular example is directly out of your textbook on page 473!

   3. H₂O donated HO^- which means it acted as a Lewis base....therefore the B(OH)₃ must have been the Lewis acid! Did we mention that this example is also directly out of your textbook? If you think about it, you'll realize that any sp²-hybridized B or Al compound will be a Lewis acid because it is electron-deficient.

   4. We hope you recognize HCl as a strong Bronsted acid....which means that it is therefore a Lewis acid as well. That means H₂O is a base in this example and this can't be our answer.

      Therefore, the correct answer to this question was answer C.

3. Question 20. What volume of 0.155 M HCl is required to neutralize 25.0 mL of 0.0100 M Ca(OH)₂?
   • To determine this, we need to know how many moles of Ca(OH)₂ are present. We know that Ca(OH)₂ is a strong base and so it must completely dissociate in water as follows:

     Ca(OH)₂(s) Ca(aq) + 2 HO^-(aq)

     Notice that each mole of Ca(OH)₂ gives TWO moles of hydroxide ion! So all we need to do is calculate the number of moles of Ca(OH)₂ and then double it to get our [HO^-].

     moles Ca(OH)₂ = (0.025 L)(0.0100 mol/L) = 0.00025 moles.

     moles HO^- = (2 mol HO^-/mol Ca(OH)₂)(mol Ca(OH)₂) = 2(0.00025 moles) = 0.00050 moles

     Neutralize means that we add an equal amount of acid. In this case, our acid is strong, so each mole of HCl yields one mole of H⁺. Therefore, we need 0.00050 moles of HCl to react with our 0.00050 moles of HO^-:

     volume HCl required = (0.00050 mol)/(0.155 mol/L) = 0.00323 L

     volume HCl required = (0.00323 L)(1000 mL/L) = 3.23 mL (answer B).

     A wise student would have realized that this problem is very closely related to Question 19 from the S 98 Exam (and was one of those top 5 missed questions reviewed in another TopClass module!). And it was identical to Question 4 from CAPA #5!

4. Question 23. What is the pH of a solution obtained by mixing 217 mL of a 0.118 M sodium hydroxide solution with 94.0 mL of a 0.452 formic acid solution? K_a for formic acid = 1.7 x 10^-4. You may assume that volumes are additive.
   • We have a weak acid plus a strong base, so the first thing we do is write a balanced equation. We don't need to know the formula of formic acid...we could simply call it HA if we wanted to.

     HCOOH(aq) + NaOH(aq) HCOONa(aq) + H₂O(aq)

     What we need to in any titration or acid-base reaction is determine the limiting reagent and let the reaction occur. Then we can see what we have left and decide what to do from there (i.e. see if any of the products affect pH).

     To do a limiting reagent question, we must first convert everything to moles:

     moles NaOH = (0.217 L)(0.118 mol/L) = 0.025606 mol

     moles formic acid = (0.094 L)(0.452 mol/L) = 0.042488 mol

     These react in a 1:1 ratio, so the limiting reagent is whichever one has the least number of moles. In this case, that's NaOH. If we wish, we could write an ICE chart (in moles) to handle this:

     	mol NaOH	mol HCOOH	mol HCOO^-
     Initial	0.025606	0.042488	0
     Change	-0.025606	-0.025606	+0.025606
     Equilibrium	0	0.016882	0.025606
     [Equilibrium]	0	0.05428 M	0.082334 M

     We can ignore Na⁺ since it is simply a spectator ion that does not affect pH.

     In the last line of this table, we converted to molarity by dividing the moles of each component by the total volume (0.217 + 0.094 = 0.311 L). Note: in the approach we use below you could have skipped converting to molarity because volumes cancel out.

     When we look at what we have, it is a mixture of formic acid (a weak acid) and its conjugate! In other words, this is a buffer. Therefore, we can simply use the Henderson-Hasselbalch equation to get our pH! For this we need the pK_a which we get from pK_a = -logK_a = -log(1.7 x 10^-4) = 3.76955:

     

     A wise student would have realized this was exactly like Question 11 from 1998 Exam 2, another one of those "Top 5" listed in our TopClass exercises. This problem is also quite similar to Question 6 from CAPA set #5.

5. Question 24. What is the pH at the equivalence point when 50.0 mL of 0.250 M benzoic acid is titrated with 0.100 M sodium hydroxide? Assume volumes are additive. K_a for benzoic acid, a monoprotic acid, is 6.5 x 10^-5.
   • As in the previous question, we need not know what the formula for benzoic acid is. And as in any titration, we do a limiting reagent approach first and then deal with what we have left. Fortunately, we are told this is this equivalence point, so neither reagent is limiting and both are consumed.

     NaOH, a strong base, will consume all of benzoic acid, a weak acid, to make sodium benzoate, NaOOCC₆H₅. We can then put that into water and ask what happens.

     The concentration of sodium benzoate can first be found by calculating the moles. We used (0.050 L)(0.250 mol/L) = 0.0125 moles of benzoic acid, so we made 0.0125 moles of sodium benzoate.

     We need to work in concentrations. To get the volume of NaOH added, we simply recognize that 0.0125 moles of NaOH were also required. Therefore, the volume of NaOH used was (0.0125 moles)/(0.100 mol/L) = 0.125 L and our total volume was therefore 0.050 + 0.125 = 0.175 L.

     Thus, we have a (0.0125 mol)/(0.175 L) = 0.07143 M solution of sodium benzoate. This is a weak base; if we put it in water it would remove H⁺ from water to form hydroxide ion. We can ignore Na⁺ since it is a spectator ion:

     H₅C₆COO^-(aq) + H₂O H₅C₆COOH(aq) + HO^-

     This is a base dissociation. The K_b expression for our equilibrium is simply [H₅C₆COOH][OH^-]/[H₅C₆COO^-]. From K_a = 6.5 x 10^-5 and K_aK_b = K_w = 1 x 10^-14 we get that K_b = 1.5385 x 10^-10.

     We don't know any of our equilibrium values or concentration changes, so we set up an ICE chart to handle this:

     	[H5C6COO^-]	[H5C6COOH]	[HO^-]
     Initial	0.07143	0	0
     Change	-x	+x	+x
     Equilibrium	0.07143-x	x	x

     We plug our equilibrium values into our K_b expression and solve for x. The approximation that x is small compared to 0.07143 turns out to be valid:

     

     Therefore, our [HO^-] is 3.315 x 10^-6. The negative log of this gives us pOH = 5.4795. Our pH = 14 - pOH = 8.52 (Answer A).

     There are other ways to get this answer. You could have worked it as an acid dissociation, saving yourself the bother of converting K's and pOH etc.

     You could have done this problem without any calculations whatsoever in about 5 seconds if you realized that we were forming a solution of sodium benzoate, a weak base. That rules out answers B and C which are acidic as well as answer D which is a strongly basic solution!