Experiment On Aqueous Solutions Solution Stoichiometry Biology Essay

Experiment On Aqueous sources Solution Stoichiometry Biology EssayThe purpose of this experiment was to study the behavior of electrolytes and non-electrolytes through and through measuring the drawivity of different effects. The experiment was divided into two parts. In the first part, the conductivities of seven different solutions were examined. Those solutions were HCl, CH3COOH, NaCl, CaCl2, CH3OH, distilled water and tippytoe water. It was concluded that HCl, NaCl, and CaCl2 were self-colored electrolytes as they dissociate completely forming free ions. On the separate hand, CH3COOH, CH3OH, and the tap water were weak electrolytes. This is because they partly dissociated into ions. Finally, the distilled water was put in to be a non-electrolyte. In the second part of the experiment, the preoccupancy of the Ba(OH)2 was determined by a tit balancen routine, when reacting with H2SO4 solution. The titration process was held by monitoring the conduction of the base soluti on Ba(OH)2 passim the reaction. This monitoring happened by changing the volume of the well acid H2SO4 until the equation point was reached. Then calculations were done to find the concentration of Ba(OH)2 throughout the reaction, which was prime to be 0.1 moles/L. The results of this experiment helps in understanding what types of species are present in different solutions and the strength of conductivities for different types of solutions. Such an experiment is very portentous in the field of electricity, as the properties of strong and weak electrolytes play an big because they are considered to the basic components in many industrial products.Results incite (A)In this part of the experiment, the conduction of different solutions was examined. Those solutions were NaCl, CaCl2 , CH3COOH, HCl, CH3OH, distilled water system, and tap water. All of the solutions were of 0.1molarity. The following dining table fork overs the conductivity determine collected.Table 1 Conductiv ity Values of unlike SolutionsSolutions Conductivity (S/cm)NaCl 13373CaCl2 16923CH3COOH 775HCl 35765CH3OH 207Distilled H2O 198Tap H2O 205From the results plunge, it was concluded that the solutions well-tried varied in conductivity strength. For example, NaCl, CaCl2, and HCls conductivity values were comparatively high compared with the other solutions. This clearly implies that they are strong electrolytes. Those three solutions were found to be strong electrolytes because the atoms of each molecule completely dissociated into positive and negative ions inside the aqueous solution, leaving no remaining reactant molecules. Therefore, those ions became free to conduct electricity. The following chemical equations illustrate the dissociation of those three solutions.HCl ? H+ + Cl- Eq(1)NaCl ? Na+ + Cl- Eq(2)CaCl2 ? Ca+ + 2Cl- Eq(3)On the other hand, it was observed that CH3COOH, CH3OH, and the tap water H2O were weak electrolytes because their conductivity was increased, alone in a low rate. This happens to be a fact because those three solutions partially dissociated. This means that the atoms of those molecules formed ions, but to a limited extent, as some of those ions got attracted again to form the alike molecule. Therefore, only the remaining ions in the solution conducted electricity. The following chemical equations show the reactions of those solutions. The arrow directed to the reactant indicates the reaction of the ions after the partial dissociation.CH3COOH H+ + CH3COO- Eq(4)CH3OH CH3+ + OH Eq(5)H2O H+ + OH- Eq(6)Finally, the distilled Water was found to be a non-electrolyte as the conductivity value was 198, which means that the distilled water had a very poor conductivity power. Therefore, it did not dissociate. As a result, no equation is to be illustrated.Part (B)In this part of the experiment, the concentration of the solution Ba(OH)2 was to be found through titrating it by adding 0.08 M H2SO4 . Various conductivity values were taken by ad ding some milliliters of H2SO4 into the solution. The table below lists the values of the added H2SO4 and the corresponding conductivity values of Ba(OH)2 solution of every addition.Table 2 The Conductivity of the Solution Ba(OH)2 as H2SO4 is Added. mass (mL) Conductivity (S/cm)1.00 50723.00 37545.00 24527.00 12148.00 6589.00 27610.00 77210.20 86910.50 106710.60 10510.70 119110.80 149710.90 169811.00 189811.10 200811.40 200811.60 2435After conducting the titration process and doing the calculations, the concentration of of Ba(OH)2 was found to be 0.1 mol/L.CalculationsIn part (B), the concentration of Ba(OH)2 solution was found through serval calculation steps. First, the volume at the equation point of H2SO4 was found from the graph produced during the lab to be 9mL, which equaled to 0.009L. Using this value, the publication of H2SO4 moles was found from the following equation.Number of Moles = chiliad x Volume Eq(7)H2SO4 Moles = Molarity x Volume= 0.8 x 0.009= 0.0072 moles of H2SO4The equation of the reaction was,H2SO4 + Ba(OH)2 ? BaSO4 + 2H2O Eq(8)From equation 8, the ratio of H2SO4 to Ba(OH)2 is 11. This implies that the number of moles of both of them was equal. Therefore, as the volume of Ba(OH)2 added was 70mL, the concentration of Ba(OH)2 was found as follows.Molarity= Number of moles / Volume Eq(9)Concentration of Ba(OH)2= Number of moles / Volume= 0.0072moles/ 0.07L= 0.1 moles/LConclusionIt was found out that some solutions were considered to be strong electrolytes like HCl, CaCl2 , and NaCl. This was because of their complete dissociation into the solution, which resulted in forming free ions that conductied electricity. On the other hand, other solutions like Ch3COOH, CH3OH, and tap water were weak electrolyte as they partially dissociated in the solution. Some of the free ions reacted again leaving only few ions. Therefore, the solution conducted electricity, but weakly. The distilled water did not conduct electricity. Therefore, it was a none lectrolyte. Finally, the titration process was used to determine the concentration of Ba(OH)2 inside the solution, which was 0.1 mol/L.