Lab 6 Conductivity

EXPERIMENT A6: CONDUCTIVITY Purpose The purpose of this lab is to analyze different solutions and classify them as being a strong electrolyte, weak electrolyte, or a nonelectrolyte. They will be classified for purpose to learn what properties correlate with strong, weak or non-electrolytes. Also identify the conductivity of different solutions, and to extrapolate the rate rate of change of conductivity of solution based on concentration and type of electrolyte used and number of ions in solution in solution on the conductivity. Introduction Ultimately in this lab the main goal for completion is to learn if there is a relationship between electrical conductivity of a solution and the effects it has on molar concentration and number of ions. Solutions can be classified into three categories: strong, weak, and nonelectrolytes. These categories order solutions by their conductivity. Conductivity is the measure by how easily charge is transmitted or moved around. In other words these three categories are based on electrolytes ability to conduct electricity. Each solution will be measured on conductivity, then compared to the idea that conductivity in solutions increases in relation to the concentration of ions. Strong electrolytes such as HCl and NaCl have high conductivity as they dissolve in solutions almost completely. Comparatively, weak electrolytes, like acetic acid among others, undergo partial dissociation until they reach an equilibrium where the concentration of electrolytes at any time in the solution is approximately the same. Finally, nonelectrolytes do not dissociate in solution hence they do not affect the conductivity of a solution. Procedure To begin the lab initially preparation came by collecting a set of reagents including 0.05M of NaCl, CaCl2, AlCl3,HCl, H3PO4, CH3COOH, H3BO3, CH3OH, C2H6O2, tap water, and deionized water. On top of reagents also collected was a vernier kit which inside has a conductivity probe to identify the conductivity of different solutions with identical molarities. Finally to begin the lab the conductivity probe from the vernier kit was placed into a single solution to measure the conductivity. After a measurement was made and recorded the probe was cleaned with deionized water. After each cleansing of the probe the process was repeated until the conductivity of each reagent was recorded. Measuring the solely conductivity is part one of this lab. In the second part, those calculations will help identify how conductivity is affected when the molarity of a solution is increasing. Similar to part one, in part two 10 ml of three different solutions were collected 0.05 M of NaCl, CaCl2, and AlCl3. Again like part one the probe was used to measure the conductivity of each solution, however as stated above after each trial the molarity concentration was increased by 0.1 ml. The concentration started being measured at 0 then after 9 trial finishing at 0.8 ml. To see if there is a relationship between conductivity of the solution and change of molar concentration the measurement data is set up on a graph to clearly see a overview picture if there 2 is a noticed correlation which can be spotted by a linear regression line (“EXPERIMENT A6: CONDUCTIVITY.” ) . Data Tables Part 1: Classify The Given Solution As A Strong/Weak/Non-Electrolyte Table 1: Conductivity Of Given Solutions in (µS/cm) along with their level of Classification Solution Conductivity (µS/cm) Classification NaCl 2,145 Strong CaCl2 10,405 Strong AlCl3 12,996 Strong HCl 3,302 Strong H3PO4 1,128 Strong CH3COOH 355 Weak H3BO3 101 Non-Electrolyte CH3OH 101 Non-Electrolyte C2H6O2 101 Non-Electrolyte H2O (tap) 721 weak H2O (deionized) 103 Non-Electrolyte Part 2: Determine TheEffect Of Molar Concentration And Number Of Ions In A Solution On The Conductivity Of The Solution Table 2: Conductivity Of Solutions in (µS/cm) by Volume in (mL) Conductivity (µS/cm) Trial # Volume (mL) CaCl2 NaCl AlCl3 1 0 11 11 11 2 0.1 13 42 60 3 0.2 18 73 116 4 0.3 23 103 165 5 0.4 29 145 204 6 0.5 34 182 256 7 0.6 38 219 300 8 0.7 43 245 362 9 0.8 48 283 419 10 0.9 52 320 463 3 Data Analysis Part 1: Classify the given solution as a strong/weak/non electrolyte. 1. Based on the data list all the strong electrolytes in the given set of reagents. NaCl, CaCl2 ,AlCl3 , HCl, and H3PO4 2. For each strong electrolyte listed above, write the dissociation equation. NaCl (aq) → Na+ (aq) + Cl- (aq) CaCl2 (aq) → Ca+2  (aq) + Cl2 - (aq) AlCl3 (aq) → Al+3  (aq) + Cl3 - (aq) HCl (aq) → H+ (aq) + Cl3- (aq) H3 PO4 (aq) → H3 + (aq) + PO4-2 (aq) 3. Based on the data, list all the weak electrolytes in the given set of reagents. CH3COOH and H2 O 4. For each weak electrolyte listed above, write the dissociation equation. CH3 COOH ⇆ CH3 COO- (aq) + H+ (aq) H2O ⇆ H2+ (aq) + O-2 (aq) 5. Based on the data, list all the non-electrolytes in the given set of reagents. H3 BO3, CH3 OH, and C2 H6 O2 Part 2: Determine The Effect Of Molar Concentration And Number Of Ions In A Solution On The Conductivity Of The Solution 1. Calculate the concentration of the electrolyte in each trial. Trial 2: Initial molarity (M1) = 0.050 M Initial volume (V1) = 0.100 mL Final molarity (M2) = 0.0001425 M Final volume (V2) = 35.1 mL → 0.050 M × .100 mL = M2 × 35.1 mL → M2 = 0.000143 M M1V1 = M2V2 Trial 3: Initial molarity (M1) = 0.050 M Initial volume (V1) = 0.200 mL Final molarity (M2 ) = 0.0002841 M Final volume (V2) = 35.2 mL → 0.050 M × .200 mL = M2 × 35.2 mL → M2 = 0.000284 M M1V1 = M2V2 Trial 4: Initial molarity (M1) = 0.050 M Initial volume (V1) = 0.300 mL Final molarity (M2 ) = 0.0004249 M Final volume (V2) = 35.3 mL → 0.050 M × .030 mL = M2 × 35.3 mL → M2 = 0.000425 M M1V1 = M2V2 Trial 5: Initial molarity (M1 ) = 0.050 M Initial volume (V1) = 0.400 mL Final molarity (M2) = 0.0005650 M Final volume (V2) = 35.4 mL → 0.050 M × .400 mL = M2 × 35.4 mL → M2 = 0.000565 M M1V1 = M2V2 Trial 6: Initial molarity (M1 ) = 0.050 M Initial volume (V1) = 0.500 mL Final molarity (M2) = 0.0007042 M Final volume (V2) = 35.2 mL → 0.050 M × .500 mL = M2 × 35.5 mL → M2 = 0.000704 M M1V1 = M2V2 Trial 7: Initial molarity (M1) = 0.050 M Initial volume (V1) = 0.600 mL Final molarity (M2 ) =0.0008427 M Final volume (V2) = 35.6 mL → 0.050 M × .600 mL = M2 × 35.6 mL → M2 = 0.000843 M M1V1 = M2V2 4 Trial 8: Initial molarity (M1) = 0.050 M Initial volume (V1) = 0.700 mL Final molarity (M2 ) = 0.0009804 M Final volume (V2) = 35.7 mL → M2 = 0.000980 M M1V1 = M2V2 → 0.050 M × .700 mL = M2 × 35.7 mL Trial 9: Initial molarity (M1 ) = 0.050 M Initial volume (V1) = 0.800 mL Final molarity (M2) = 0.0011173 M Final volume (V2) = 35.8 mL → 0.050 M × .800 mL = M2 × 35.8 mL → M2 = 0.00112 M M1V1 = M2V2 2. Enter The Conductivity Values As A Function Of The Concentration Of The Electrolyte For Each Trial In The Table. Table 3: Conductivity of Solutions in (µS/cm) ordered by molar Concentration (M) Conductivity (µS/cm) Trial # Concentration, M NaCl CaCl2 AlCl3 1 0 11 11 2 0.000142 13 42 11 60 3 0.000284 18 73 116 4 0.000425 23 103 165 5 0.000565 29 145 204 6 0.000704 34 182 256 7 0.000843 38 219 300 8 0.000980 43 245 362 9 0.001117 48 283 419 Plot a graph of the concentration of the electrode vs. the conductivity of the electrode 5 4.What are the slopes of the lines corresponding to each electrolyte solution in the above graph? Electrolyte Slope NaCl 48.2 CaCl2 345 AlCl3 500 5. Write the dissociation equation for each of the electrolytes. NaCl (aq) ฀ Na+ (aq) + Cl- (aq) 2 moles of ions CaCl2 (aq) → Ca+2  (aq) + Cl2 - (aq) 3 moles of ions AlCl3 (aq) → Al+3  (aq) + Cl3- (aq) 4 moles of ions 6. Comment on the relationship between the slopes of the three lines and the number of ions in solution for each electrolyte. Based off the graph we can see that as the slope of electrolytes increase as the total number of ions in the solution increases. This can clearly be seen by looking at the graph and data table above. 7. Comment on the effect of a) molar concentration and b) number of ions in solution on the electrical conductivity of a solution. A) Molar Concentration As the molar concentration of NaCl, CaCl2 , and AlCl3 increases the three respective electrical conductivity increases, we can see this based off the Conductivity table above. B) Number of ions The number of ions in a solution affects the electrical conductivity of electrolytes by that as the number of ions increases their respective electrical conductivity value also increases. This can we seen as AlCl3 had the highest electrical conductivity and had the highest number of ions at four. Plus NaCl had the lowest number of ions at 2 and also had the lowest electrical conductivity. 6 Discussion Overall in part one the purpose was to discover what type of compounds create strong, weak, or nonelectrolyte. Theoretically, there should have been three groups for example A,B, and C, where group A solutions had the highest conductivity, followed next by group B, then group C . The data obtained in the lab supports the previously stated idea above. Group A’s . As group A is filled with NaCl, CaCl2 ,AlCl3, and HCl, these solutions clearly display the characteristics of ionic compounds. While solutions of group B and C display the properties of molecular acids and molecular compounds respectively. For part one of the lab as can be seen by Table 1, their was roughly five strong electrolytes solutions. The five strong electrolytes were were NaCl CaCl2 AlCl3 HCl, and H3 PO4 with values of conductivity of 2145, 1,0405, 12,996, 3,302, and 1,128 (µS/cm). Theoretically the strong electrolytes should all be ionic compounds, however H3 PO4 is one solution out of five labeled “strong electrolytes” that is not an ionic compounds. The outcome for this leads to two possible interpretations, one like any rule in STEM there is always an exception to the original rule, and two there was a possible error that occurred causing H3 PO4 to appear as a strong electrolyte when in fact it really is not. Reflecting back on the order of conductivity recorded, by looking at Table 1 was from top to bottom suggest it is likely an error occurred during transition from one solution to the next. The reason it is likely there was an error during transition between measuring solutions is because by looking at Table 1 as a whole it can easily be observed that all the strong electrolyte were measured before H3PO4. Knowing the remaining other strong electrolytes were measured before H3 PO4 creates the possible error one of the previous solutions was not completely cleaned off the probe and was incorrectly transferred into the H3PO4 solution. Having other solutions measure with H3PO4 causes an incorrect measurement of H3PO4  to be interpreted. Reflecting back on Table 1’s values for AlCl3, H3PO4, CH  3COOH, and H3BO3 their conductivity values can be observed to decreases in a linear fashion from 12,996 (µS/cm) to 3,302 (µS/cm) to 1,128 (µS/cm) to 355(µS/cm) to 101 (µS/cm) as the number of trials that previous measured an ionic compound increases. This second observation supports the idea that the probe was not completely cleaned of all previous solutions before entering a new solution. For part two of the lab, the overarching idea was how will different amounts of molar concentration and different number of ions in a solution affect the conductivity of the solution. Based off the data in Table 3, the effects on conductivity from different amounts of molar concentration has a possible positive tread of correlation. By examining all three different solutions on Graph “Conductivity of Solutions” a general upward trend is clearly there, as molar concentration increases , so does the conductivity of the solution. Not only does one increase with the other, but all three solutions have linear regression lines. Having three linear regression lines prove that there is a positive correlation between increase of Molar concentration and conductivity. The first idea proven in part two of the lab was how does concentration affect conductivity. The second idea tested in part two of the lab is there a relationship between number of ions and conductivity. By looking at table 3 again there are three different solutions, each with different number of ions. By analyzing the table by their individual rows a tread appears with no outliers. The tread that appears is that as the number of ions in solution increases, so does the conductivity. At any single concentrations, the conductivity is greatest for AlCl3 then followed by CaCl2 and lowest at NaCl. This thread correspond directly with the number of ions each compound releases while in solution. This thread clearly answers the question for part two, 7 which is how does the different amount of concentration and different number of ions affect conductivity. For both cases an increase in either concentration or number of ions will positively increase the value of conductivity for a solution. Possible sources that could have result during part two of the lab could have been incorrectly measuring the solution with a dirty probe. Also another possible source of error was that during the process of removing the probe and adding small amounts of concentration it was observed that when the probe was taken out of the solution the probe was also taking small amounts of solution outside with it. Clearly since there is no outliers in the data this states that the error occurred in this lab was a small amount that did not affect the overall result. Conclusion Lab six on conductivity was performed for the purpose to learn from part one that conductivity is highest for ionic solutions. This theory was proven by Table 1 where all the highest conductivity values for the given solutions were form ionic solutions. Next in part two to learn that a positive increase in concentration or number of ions will result in a increase of conductivity. This idea that an increase in concentration will result in a increase of conductivity was demonstrated to be correct from Table 3 and Graph “Conductivity of Solutions”, by the fact that when all three solutions were set up on the graph they each had a linear regression line between conductivity and concentration. Also by Table 3 the idea that the number of ions increase will cause a increase in conductivity was proven by isolation single rows of constant molar concentration. In each row the amount of conductivity increases by each increase in ions, not only in a couple rows, but in all of the rows. Having a clear trend of positive correlation proves that the number of ions increase will result in an increase in conductivity. One major source of error is random error as the data collect can not replicate the method of collecting data exactly the same each time. For example the first test will most likely have the cleanest probe, however by the last test there is a high possibility that small amounts of previous solution is still there. To prevent this possible error in the future it is best to thoroughly clean all equipment that will be used in a lab, before begging. Suggestion to improve this lab would be to instead of having three solutions in part two to have four, but one is unknown.Therefore based off our collected data we must give an educated guess to what the unknown might be.