(This is a reprint from the March 2011 issue of Chem 13 News, pages 4-5.)
Question #20 from the 2010 CHEM 13 NEWS Exam was, statistically speaking, the question that best discriminated between the “better” students and the “weaker” students. The question is reproduced in the box below. About 88% of students selected a response, but only 23% selected the correct one (response D). Somewhat surprising, at least to me, is that 24% selected “E”. When setting the question, I felt that responses A-C are all obviously wrong (and thus E is also wrong) and, by the process of elimination, response D is the only reasonable choice.
Which of the following occurs if a 0.10 mol/L solution of a weak acid is diluted to 0.010 mol/L at constant temperature?
A The hydrogen ion concentration decreases to 0.010 mol/L.
B The pH decreases.
C The ionization constant, Ka, decreases.
D The percentage ionization increases.
E all of the above
Before examining why “D” is the correct answer, and discussing ways to demonstrate to students that the degree of ionization increases with dilution, let’s briefly discuss how responses “A”, “B” and “C” can be quickly eliminated as possible answers and also consider possible reasons why students selected these answers.
From the statement of the question, we note that we are dealing with a solution of a weak acid. At a concentration of 0.010 mol/L, a weak acid will not be fully ionized and therefore, [H+] must be less than 0.010 mol/L. Answer “A” cannot be correct. Had we been dealing with a solution of a strong acid, which ionizes completely in solution, dilution to 0.010 mol/L would indeed cause the hydrogen ion concentration to decrease to 0.010 mol/L. Students who selected “A” may not fully understand how the ionization of a weak acid is different from that of a strong acid.
To decide whether or not response B is correct, we consider the effect of dilution on pH. A very concentrated solution of acid will be “very acidic” and have a pH significantly less than that of water. A very dilute solution of acid will be, at best, only mildly acidic and have a pH close to that of pure water. Thus, dilution of a solution of acid causes the pH to increase.
Why would students select “C” as the answer? Could it be because they do not understand the difference between the “reaction quotient” (Q) and the equilibrium constant (K)? Perhaps we, their teachers, contribute to their misunderstanding. I am willing to bet that, when discussing the ionization of an acid (HA) in water, most of us have written the following, or something close to it.
HA(aq) ⇌ H+(aq) + A−(aq), Ka = [H+][A−]/[HA]
The expression given above for Ka is easily misinterpreted because, taken literally, it indicates that the value of Ka changes if [H+], [A−] or [HA] changes. Of course, the value of Ka does not change,1 even if [H+], [A−] or [HA] changes. How are students to interpret the expression? They must understand that the left-hand side of the expression represents a number and the right-hand side is the reaction quotient, Q = [H+][A−]/[HA]. By setting Ka equal to Q, we are indicating that, at equilibrium, the reaction quotient Q will have a certain value. Although dilution causes a momentary change in the value of Q, the value of Ka never changes. Explaining the difference between the reaction quotient and the equilibrium constant seems simple enough. However, as they say, “the proof of the pudding is in the eating” and my experience has been that even with careful explanation, students just don’t get it. On a recent first-year exam at our university, a similar question was asked and 51% of students answered “all of the above”, indicating that the majority believed that when concentrations are changed, the value of the equilibrium constant changes.
Finally, let’s turn our attention to explaining the effect of dilution on the degree of ionization. The effect can be stated quite simply: the degree of ionization increases with dilution. This effect can be rationalized in a number of ways. First, let’s apply Le Chatelier’s principle to the following equilibrium.
HA(aq) + H2O(l ) ⇌ H3O+(aq) + A−(aq)
According to Le Chatelier’s principle, the addition of water will cause the equilibrium to shift to the right. Because we are not changing the amount of acid in the solution, a shift to the right corresponds to an increase in the degree of ionization. Unfortunately, some students focus on changing the amount of acid instead of focusing on the addition of water. (To make a solution more concentrated, they say, we add more acid. To make a solution less concentrated, we “remove” acid.) If acid is “removed”, the equilibrium will shift towards the left in an attempt to replace the acid that was “removed”.2 However, it is a mistake to conclude that the degree of ionization has actually decreased. For a given total concentration, Co, of acid (HA), the fraction that exists in the ionized form is α = [A− ] / Co. In this instance, both [A−] and Co will decrease! (Co decreases because acid was “removed” and [A−] decreases because the “removal” of HA causes the equilibrium to shift to the left.) Because both [A−] and Co change, it is difficult to decide whether α increases, decreases, or stays the same.
The effect of dilution on α can be approached in a more quantitative way by deriving a relationship that shows how α and Co are related. An approximate relationship can be obtained by setting up the usual equilibrium summary (ICE table).
|Change||− x||-||+ x||+ x|
|Equilibrium||Co – x||-||x||x|
With the assumption that x is small compared to Co, we can write Ka ≈ x2 / Co and x ≈ [KaCo]1/2. For the degree of ionization we obtain the result α = [A−] / Co ≈ [Ka / Co ]1/2. Using this equation, we can make the following predictions:
(1) For a given acid (Ka), the degree of ionization increases as the initial (or total) concentration decreases; and
(2) For a given initial concentration (Co), the stronger the acid, the greater the degree of ionization.
We have justified the statement that the degree of ionization increases with dilution, and it is very tempting to stop the discussion here. However, the analysis above does not help us understand how the degree of ionization truly depends on Co and Ka. The graphs below provide a much more complete picture and can be used to help students understand several concepts. Figure 1 shows how the degree of ionization truly depends upon the initial concentration. It also shows that the approximate result predicts incorrectly that that the degree of ionization increases without limit with increasing dilution. Figure 2 shows the degree of ionization as a function of initial concentration for acids of varying strength. Notice that if Ka is greater than 101 and the initial concentration is less than 0.1 mol/L, the acid is essentially 100% ionized. Because it is relatively straightforward to calculate the data needed to make these graphs, teachers could have their students work in groups to carry out the necessary calculations and then have them pool their results to make the graphs. The graphs may be help students understand not only why the degree of ionization increases with dilution but also how the ionization of a strong acid differs from that of a weak acid.3
- The value of Ka is actually not constant. For example, its value will change if the temperature changes. We may treat Ka as constant provided (i) the temperature remains constant; and (ii) the concentrations of dissolved entities are low enough that concentrations can be used in place of chemical activities in the expression for Q.
- In this section, I have enclosed “removed” in quotes to emphasize that the removal of HA from solution is a hypothetical process. One might consider removing HA by adding strong base. Although the addition of strong base will certainly “remove” HA, it also adds A− because of the reaction HA + OH− → A− + H2O. The ionization of HA in the presence of excess A− is a much different problem than the one we are considering here.