The project will involve determining the roots of a virial equation of state with cubic order dependency on specific volume. This program can be used in the future for helping to….
Obtain the five Group C solutions or liquids.
Properties of Solutions: Electrolytes and Non-Electrolytes
The experiment herein sought to unearth the properties of strong electrolytes, non-electrolytes and weak electrolytes. Such an end was sought by assessing the behavior of varied substances when immersed in aqueous solutions. Further, the experiment sought to establish whether the substances explored here are molecular, molecular compound or iconic. The decision was pegged on how these substances conduct electricity. The hypothesis that had to be tested was that iconic compounds were strong electrolytes; it was also hypothesized that group B compounds would be molecular acids and they would embody both strong and weak electrolytes. Lastly, C would, unlike B, be compounds with non-electrolytes.
The Materials used included Goggles, Chrome book, Graphical Analysis 4 app, Vernier data-collection interface, Conductivity Probe, Electrode Support, 250 mL beaker, H2O (tap), H2O (deionized), Tissues; 0.05 M each of HCL, Fe2, NaCl, CaCl2, Fe2, (SO4)3, HC2H3O2, H3BO3, CH3OH, H3PO4, Fe(NO3)
1. Obtain and wear goggles! DANGER: Handle all the solutions in this experiment with care. They may be harmful if swallowed or in contact with the skin or eyes. Do not handle until all safety precautions have been understood Notify your teacher in the event of an accident.
2. Assemble the Conductivity Probe, utility clamp, and ring stand as shown in Figure 1. Be sure the probe is clean and dry before beginning the experiment.
3. Set the selector switch on the side of the Conductivity Probe to the 0–20000 µS/cm range. Connect the Conductivity Probe to Lab Quest and choose New from the File menu. If you have an older sensor that does not auto-ID, manually set up the sensor.
4. Obtain the Group A solution containers. The solutions are: 0.05 M CaCl2, 0.05 M NaCl, and 0.05 M Fe (NO3)3. Measure the conductivity of each of the solutions.
a. Carefully raise each vial and its contents up around the Conductivity Probe until the hole near the probe end is completely submerged in the solution being tested. Important: Since the two electrodes are positioned on either side of the hole, this part of the probe must be completely submerged.
b. Briefly swirl the vial contents. Monitor the conductivity reading displayed on the screen for 6–8 seconds, then record the value in your data table.
c. Before testing the next solution, clean the electrodes by surrounding them with a 250 mL beaker and rinse them with distilled water from a wash bottle. Blot the outside of the probe end dry using a tissue. It is not necessary to dry the inside of the hole near the probe end.
6. Obtain the four Group B solution containers. These include 0.05 M HC2H3O2, 0.05 M HCl, 0.05 M H3PO4, and 0.05 M H3BO3. Repeat the Step 5 procedure.
7. Obtain the five Group C solutions or liquids. These include distilled H2O, tap H2O, 0.05 M CH3OH, and 0.05 M C2H6O2. Repeat the Step 5 procedure.
1. Group A compounds qualified as ionic compounds because all of the tested substances in Group A had high conductivity values. They potentially completely dissociate when placed in aqueous solutions.
2. The Group A compounds all dissociate on varied levels and carry varied ions present in the aqueous solution; further, they all vary in conductivity values.
CaCl2➝Ca (aq) +Cl (aq) +Cl (aq)
Fe2 (SO4 )3➝ Fe (aq)+Fe(aq)+(SO4 )(aq)+(SO4 ) (aq)+(SO4 ) (aq)
NaCl➝Na (aq) +Cl (aq)
3 The substances in Group B qualify as molecular acids because some substances dissociate fully and some only dissociate partially.
When arranged from the strongest to weakest they appear as-H3PO4, HCL, HC2H3O2, and H3BO3
a. HC2H3O2 ↔ H+ (aq) +C2H3O2 (aq)
b. HCl →H (aq) +Cl (aq)
c. H3PO4→H3 (aq) +PO4 (aq)
d. H3BO3↔H3 (aq) +BO3 (aq)
5 The subscript 3 failed to contribute additional ions in solution. However, it resulted in molecules of H3 dissociate partially in the aqueous state.
6 In Group C all three compounds appear to be molecular compounds because they all have low levels of conductivity. Based on this answer it should be assumed that they will not dissociate at all.
7 Tap water embodies increased forms of salt in it than deionized water. The reason thereof explain why they conduct electricity more effectively than in deionized water devoid of salts.
Based on the data used, the hypothesis was proven to be correct; Group A was found to be ionic compounds; they dissociated easily. Group B failed to dissociate hence qualified as molecular acids. Lastly, Group C to be molecular compounds and dissociate some of the time. Substances that dissociated fully had high levels of ionic compounds; Group B exhibited a wide range of conductivity; Group C exhibited reduced conductivity.