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Experiment On Aqueous Solutions Solution Stoichiometry Biology Essay

Paper Type: Free Essay Subject: Biology
Wordcount: 1103 words Published: 1st Jan 2015

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The purpose of this experiment was to study the behavior of electrolytes and non-electrolytes through measuring the conductivity of different solutions. 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 tap water. It was concluded that HCl, NaCl, and CaCl2 were strong electrolytes as they dissociate completely forming free ions. On the other hand, CH3COOH, CH3OH, and the tap water were weak electrolytes. This is because they partially dissociated into ions. Finally, the distilled water was found to be a non-electrolyte. In the second part of the experiment, the concentration of the Ba(OH)2 was determined by a titration process, when reacting with H2SO4 solution. The titration process was held by monitoring the conductivity of the base solution Ba(OH)2 throughout the reaction. This monitoring happened by changing the volume of the strong acid H2SO4 until the equivalence point was reached. Then calculations were done to find the concentration of Ba(OH)2 throughout the reaction, which was found 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 significant in the field of electricity, as the properties of strong and weak electrolytes play an important because they are considered to the basic components in many industrial products.


Part (A)

In this part of the experiment, the conductivity of different solutions was examined. Those solutions were NaCl, CaCl2 , CH3COOH, HCl, CH3OH, distilled H2O, and tap H2O. All of the solutions were of 0.1molarity. The following table shows the conductivity values collected.

Table 1: Conductivity Values of Different Solutions

Solutions Conductivity (¿½S/cm)

NaCl 13373

CaCl2 16923


HCl 35765

CH3OH 207

Distilled H2O 198

Tap H2O 205

From the results found, it was concluded that the solutions tested varied in conductivity strength. For example, NaCl, CaCl2, and HCl¿½s conductivity values were relatively 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, but 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 same 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 ¿½partial¿½ 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 adding few 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.

Volume (mL) Conductivity (¿½S/cm)

1.00 5072

3.00 3754

5.00 2452

7.00 1214

8.00 658

9.00 276

10.00 772

10.20 869

10.50 1067

10.60 105

10.70 1191

10.80 1497

10.90 1698

11.00 1898

11.10 2008

11.40 2008

11.60 2435

After conducting the titration process and doing the calculations, the concentration of of Ba(OH)2 was found to be 0.1 mol/L.


In part (B), the concentration of Ba(OH)2 solution was found through serval calculation steps. First, the volume at the equivalence point of H2SO4 was found from the graph produced during the lab to be 9mL, which equaled to 0.009L. Using this value, the number of H2SO4 moles was found from the following equation.

Number of Moles = Molarity x Volume Eq(7)

H2SO4 Moles = Molarity x Volume

= 0.8 x 0.009

= 0.0072 moles of H2SO4

The 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 1:1. 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/L


It 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 nonelectrolyte. Finally, the titration process was used to determine the concentration of Ba(OH)2 inside the solution, which was 0.1 mol/L.


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