Unformatted text preview: Experiment #2: Aqueous Solutions & Solution Stoichiometry A. SUMMARY In Part A of this lab we tested to see the differences in conductivity between different solutions. We found by this comparison that there is a great number of variations in conductivities of solutions because of the molecules of which they are composed. In Part B we conducted the titration of Ba(OH)2 with H2SO4. We measured the change in conductivity during this titration. We found that once the solution reaches equilibrium more acid can be added to raise the conductivity back to, and even beyond, the conductivity point at which the solution of Ba(OH)2 started. B. RESULTS Part A: SOULUTION HCl CH3OH Tap H2O CaCl2 CH3COOH Deionized H2O SOLUABLILITY (M) 0.1 0.1 1.0 0.1 0.1 1.0 CONDUCTIVITY (s) 4660 38 554 2473 401 38 C. GRAPHS AND CALCULATIONS Ba(OH)2 (aq) + H2SO4 (aq) 2 H2O + Ba SO4 (ppt) Mol = Liters * M X = .00972 * .07356 D. DISCUSSION Part A: Based on your conductivity values, which compounds appear to be strong electrolytes, weak electrolytes, or non electrolytes? Strong- HCl, CaCl2 Weak- Tap H20, CH3COOH Non- Deionized H2O, CH3OH :: X = 7.15x10^-4 mol What would you expect each of them to partially dissociate, wholly dissociate, or not dissociate at all? Strong- Wholly dissociate Weak- Partially dissociate Non- Not dissociate Why do the various solutions, each with the same concentration have such large differences in conductivity values? The weaker the conductivity the more near equilibrium the solution is because of the molecules that make up the solution. It is more likely to not dissociate, or dissociate completely, because the solution is more stable independent from the dissociation. How do you explain the relatively high conductivity of tap water compared to a low or zero conductivity to distilled water? The tap water does not have all of the elements removed from it, just the elements that would be harmful to drink. Distilled water has everything removed from it except the water itself. Part B: Determine the volume of H2SO4 added at the equivalence point. ~9.72 mL Calculate the number of moles of H2SO4 added at the equivalence point. Mol = Liters * M X = .00972 * .07356 :: X = 7.15x10^-4 mol Calculate the number of moles of Ba(OH)2 added at the equivalence point. Mol = Liters * M X = .07 * .01 :: X = 0.007 mol Compare your results to those of your classmates. Are the results reproducible? Yes, everyone's graphs are quite similar, so the results are reproducible. ...
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This note was uploaded on 04/21/2008 for the course CHEM 107 taught by Professor Generalchemforeng during the Fall '07 term at Texas A&M.
- Fall '07