applications anJ analogies
edited by
RONDELORENZO Middle Georgia College Cochran, GA31014
A Dating Analogy for Acid-Base Titration Problems Ronald DeLorenzo Middle Georgia College Cochran. GA31014 Although many students can easily and quickly calculate the answers to titration problems using the familiar V a a = VbNb formula,' they frequently have trouble understanding the conseauences of concentrating- or diluting unknowns before titrating them. I have used a combination of exoanded dimensional analvsis (1)and dating analogies to alieviate this confusion. e he use of other anaiogies help clarify other confusing concepts related to solution concentrations and stoichiometric ratios have been published recently by John Fortman (2). To illustrate my approach, consider the problem i n which 1L of a 0.1 N HCI solution is required to titrate 10 L of an unknown NaOH solution to the end point. What is the normality of the sodium hydroxide solution? To solve this problem, I divide the solution into three steps as shown below. Step 1
In the first step, use expanded dimensional analysis to calculate how many gram-equivalent weights of HC1 were added to the unknown base. (1L HCI soh added to NaOH soh1
x (0.1 GEW of HCII1 L HCI solnl = 0.1 GEW HC1 added to NaOH soln
Step 2
In the second step, calculate the number of gram equivalent weights of NaOH present in the unknown solution that was titrated. (0.1 GEW HCl added to NaOH soln) (1GEW NaOHpresent in NaOH soln) X (1GEWNaOH added NaOH soln) = 0.1 GEW NaOH present in NaOH soh
I
Figure 1. Male doml with one man per room. Female dorm with two women per room.
4W
M
Figure 2. Male dorm with one man per room. Female dorm with four women per room. trated i t would, therefore, require more HCI to titrate i t to the end point. Then I ask them what the effects would be if before performing the titration I first added water to the NaOH solution to increase its volume from 10 L to about 20 L. Most students respond that because the resulting NaOH solution is more dilute i t would require less HC1 to titrate i t to the end point. To help bring clarity to this discussion and to convince my students t h a t concentrating or diluting the original NaOH solution does not change t h e number of gram equivalent weights of HC1 needed to titrate to the end point, I offer the following dating analogy. Consider, a s shown in Figure 1,two dormitories, one for male students and one for female students. The concentration of men in the dorm for males i s one man per room, and there are four rooms of women in the dorm for females. The followingbalanced equation
Step 3
In the third step, calculate the concentration of the unknown NaOH using the definition that normality is the number of eram eauivalent weichts of solute ner liter of solution. (0.1 GEWNaOHpresent inNaOH soh$ (10 L Nal
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= 0.01 N NaOH
Even for s ~ u d e n t swho grasp this approach, they are confused when I then ask them what would hannen if before performing the titration I first boiled the OH solution until the amount of water present was reduced by evapostudents tell me ration from about 10 L to about 1 that because the resulting NaOH solution is more concen-
most
tells us that exactly one man (M) reacts with one woman (W) to become a dating pair (MW) i n much the same way that one gram equivalent of acid reacts with exactly one gram equivalent weight of base. I then ask the class to go through the same three-step approach given above to calculate the concentration of women per room in the dorm for females if i t takes eight rooms of men to date completely the women in four rooms. 'In this formula, V, and N, are the volume in liters and normality of the acid and V, and N, are the volume in liters and normality of the base. 1' am making the approximation that 1 L of a dilute solution of NaOH contains close to 1 L of water.
Volume 72 Number 11 November 1995
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Step 1
Fi
In the first step, they calculate how many men were added to the dorm for females. (8 rooms added)(lm a d l roam) = 8 men added Step 2
In the second step, they calculate the number of women that had to be present in the dorm for females. (8 men added)(l woman presenffl man added) = 8 women present Step 3
I n the third step, they calculate the concentration of women per room in the dorm for females. (8 women presentll(4rooms) = 2 womeniroom Figure 2 represents a situation similar to concentrating the original NaOH solution by evaporating water from it. In this analogy, however, two rooms are removed and the women become more concentrated, four to a room. Now most students can see easily that even though the women are more concentrated, I still need to add eight rooms of men to date all of them. As shown in Figure 3, more rooms have been added to the dorm for females to dilute the women so that there is one woman per room. Again, most
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Journal of Chemical Education
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Figure 3.Male dorm with one man per room. Female dorm with one woman per room. l y even though the women have studtmt~ain scc c , ~ s ~ that hccn dlluicd. 1 still must add e ~ g hrooms t ofmen to date a11 of the women in the dormitory for females. I have used this combination of dimensional analysis and dating analogy with my general chemistry classes for many years. The students have found that it gives them a better grasp of these titration problems. Literature Cited 1. OeLorenro, R. J. Chem Edur 1994, 71.789-791. 2. Fortman. J.J Chem Educ 1994, 71.27-28; 431432: 571-572.
