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Conductivity Lab Review. Configuration of the Lab Station.

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Presentation Transcript
configuration of the lab station
Configuration of the Lab Station

The left side image shows the entire station including the reagents, the burette for water and the computer required for the data acquisition and analysis. The right side image provides a closer look at the required reagents and hardware.

conductivity of nacl solution
Conductivity of NaCl Solution

After we delivered 25 ml of distilled water to the beaker we then tested for the conductivity of the control (only water). After the data was recorded by the computer we then delivered 5 drops of the NaCl solution to the test sample.

nacl test continued
NaCl Test Continued

The above images show from left to right the conductivity of the test solution after 5 drops, 10 drops and 30 drops of the NaCl reagent were added respectively. The linearity of the data confirms the increase in conductivity as the concentration of ions increases.

molecular potential
Molecular Potential

Before we tested NaCl we speculated on the expected performance. The molecular profile for this substance is one of an ionic compound. When NaCl (which is a salt) is dissolved in water we expect the “liberation” of the ions by what we like to think of as our water taxi.

NaCl(S) + H-O-H Na1+(aq) + Cl1-(aq)

Since conductivity requires the movement of charge and these liberated ions are now free to move the increase in conductivity confirmed our prediction.

conductivity of cacl 2
Conductivity of CaCl2

Before we test CaCl2 we will again speculate on the expected performance. The molecular profile for this substance is again ionic. When CaCl2 (which is again a salt) is dissolved in water we expect the “liberation” of the ions by the water taxi.

CaCl2 (S) + H-O-H Ca2+(aq) + 2 Cl1-(aq)

Since conductivity requires the movement of charge the reaction should liberate the captive ions an increase in conductivity is expected.

cacl 2 data
CaCl2 Data

From the image of the data it is clear that CaCl2 is conductive. The slope of the graph shows the solution to be more conductive than the similarly profiled NaCl.

cacl 2 data continued
CaCl2 Data Continued

After the final 5 drops were added the appearance of the graphs suggested further analysis. Both of the graphs were linear with fantastic C.O.R. values. The question that plagued us was why the two ionic compounds were not identical. They were both “binary” ionic compounds and both “liberated” metallic ions and chloride ions.

it s a question of number
It’s a Question of Number

When we considered the reactions we noted that the number of ions liberated by the salts was not the same. A ratio of 3:2 is observed for the CaCl2 salt compared to the NaCl salt. When we compared the slopes of the graphs we obtained a value of 1.5. Since the conductivity is generated by the movement of ions and the ions are “liberated” in a ratio of 3:2 the slope of 1.5 strongly supports the model.

molecular profile of hcl
Molecular Profile of HCl

To assist in generating a molecular profile for a substance we generally consult ASIMS. Hydrogen presents a half filled energy level so it is not expected to form ionic compounds. We expect this binary compound (HCl) to be molecular with bonds that are polar covalent. The reagent list noted the HCl system as hydrochloric acid. The behavior of this system may well help us to understand the important components of the acid profile.

hcl results
HCl Results

The 5 drops of HCl had a very steep slope. By the time 30 drops were added the relationship was clearly established.

naoh results
NaOH Results

The NaOH graph is the lower of the two presented in the image. Although both were made at the same time and at the same concentration the “years” of storage has caused the NaOH to become less basic. Over time bases absorb CO2 (which forms carbonic acid). What is important to note is that the ionic compound NaOH is conductive as expected.

the magic of titration
The Magic of Titration

When Mr. Zachmann asked the class to speculate on the conductivity of a system where a conductor like HCl is added to another conductor like NaOH. Many suggested an increase in conductivity would occur. The data however, does not confirm this hypothesis. During the titration of the basic solution we noted a decrease in the conductivity. This observation helps us to understand at the molecular level the act of “neutralization” when acids and bases interact.

neutralization
Neutralization

As we consider the act of neutralization we see the reformation of water by the liberated H+ ion and OH1- ions. The remaining Na1+ and Cl1- ions continue to support conductivity.

NaOH(S) + H-O-H Na1+(aq) + OH1-(aq)

HCl (aq) + H-O-H H1+(aq) + Cl1-(aq)

_________________________________________________

NaOH(S) + HCl (aq)Na1+(aq)+ Cl1-(aq)

titration review
Titration Review

The beaker shows the titration 1 drop before equivalence. Notice the slight pink color. The graph shows the same system one drop past equivalence (the pink color left the system). Notice what happens to the graph when more titrant is added (see next slide)

titration review continued
Titration Review Continued

The left image was taken at five drops past end point and the right image was for 10 drops past end point. Remember that indicators rely on color to convey data and this can be ambiguous. In the next slide we used an algebraic approach to determine the equivalence point for the titration.

algebraic approach to equivalence
Algebraic Approach to Equivalence

The ambiguity of the colorimetric determination of equivalence can be eliminated by finding the common (x, y) coordinate for the intersection of the two lines.

the algebraic solution
The Algebraic Solution

For the declining line y = -19.6 x + 742 and for the ascending line y = 55.4 x -749.5.

If we equate the two respective expressions for the common y value we get:

-19.6 x + 742 = 55.4 x – 749.5.

Collecting terms we get:

75 x = 1491.5

Solving for x we get:

X = 19.89 drops

Can we use this information to compute the concentration of the base? If you remember from our solutions unit VM = VM. Therefore the molarity of the base would be:

M(Base) =(VAcidMAcid)/ VBase = (19.89 drops)(.10M)/(30.0 drops) = .066 M

applied research
Applied Research

Using the same technique we measured the conductivity of solutions made from some common chemicals found in the home. Through the exercise we tried to use ASIMS or deductive process to make predictions. The data provided in the following slides will often retain the record of previously tested materials. This is done to remind you of the deductive process.

applications
Applications

The substances which appear in the image to the right were assessed for conductivity. Methanol solution was tested first. This test was followed by ethanol solution, ammonia solution, sugar in solution and finally baking soda (AKA sodium bicarbonate). Each substance was also tested with phenolphthalein. As the testing progressed deductive and inductive process was used to help with the predictive process.

methanol assessment
Methanol Assessment

The top image shows that the methanol was not conductive. The bottom image indicates that the phenolphthaleindid not have a positive reaction. We conclude from this result that the OH attached to the carbon of methanol was not liberated by the water solvent.

ethanol assessment
Ethanol Assessment

Because ethanol was structured very much like methanol we expected the same behavior. From the images to the right we can see that there was no difference in their respective behavior. We conclude that alcohols are not conductive.

ammonia assessment
Ammonia Assessment

The structure of ammonia is interesting as it presents a nonbonding electron pair. The conductivity test was positive as was the phenolphthalein test. We conclude ammonia was able to attack water or water attacked ammonia and as a result liberated the OH1- ion.

sugar assessment
Sugar Assessment

Like methanol and ethanol solution the sugar did not conduct nor did it present a positive phenolphthalein response. The structural formula for the glucose showed numerous OH groups attached to the carbon backbone. The tests indicate that these groups stay attached.

sodium bicarbonate assessment
Sodium Bicarbonate Assessment

The photo to the right shows the response to the addition of the sodium bicarbonate. The early spike in conductivity is the sodium bicarbonate. The later response was the ammonia response that was not removed so that we could compare the magnitude of the response. Again, we had a positive phenolphthalein response.

lab report expectations
Lab Report Expectations

In class we outlined the talking points that must appear in the report. I expect to see several paragraphs dedicated to these various talking points. I expect a drawing with suitable development to support your paragraph on the nature of conductivity. As you discuss the molecular potential of the various tests I expect you to write the reactions and make clear supporting comments that illuminate the components of the written reaction which support conductivity (or do not support conductivity). The titration deserves a separate paragraph in the introduction. Be sure to explain the titration process and our expectations with respect to the conductivity of the system as the titration progresses. Again you will need to write chemical equations to support your writing.

The discussion part of the report should have equal attention. Again, I expect well constructed paragraphs. I certainly expect to see downloaded images with supporting statements. This is particularly important for the titration part of the lab. The highly motivated student will devote a separate paragraph to each of the assessment substances. Again I expect to see downloaded results followed by a discussion of the formula that I provided on the board and how this formula interacted to provide or not provide conductivity.