Titration of Sodium Hydroxide Using Hydrochloric and Sulphuric Acids - Lab Report Example

Titration of sodium hydroxide using hydrochloric and sulphuric acids Acid-base titration to find the molarity of two acids Aim The aim of this lab practical is to carry out the titration of a base, sodium hydroxide with two acids: firstly hydrochloric acid and then sulphuric acid so as to determine the molarities of the two acids. Background theory When a base is titrated with an acid, the pH gradually changes until it reaches the equivalence point, where there is a sharp change. At this point there is an equal amount of both the acid and the base. As the titration proceeds, the point at which the colour of the solution that contains the indicator changes is called the end point (Levine, 2001). In the case of the titration of NaOH with dilute HCl and dilute H2SO4 this point is marked by the sudden and stable change in the color of the base-indicator mixture following titration with the acid. This experiment will determine the volume of the acid used when the end point is reached after 0.1M sodium hydroxide is titrated with hydrochloric acid and sulphuric acid of unknown molarities using phenolphthalein indicator. NaOH and HCl react as follows: NaOH (aq) + HCl (aq) -----------> NaCl (aq) + H2O (l) H+ (aq) + OH- (aq) ------------> H2O (l) NaOH and H2SO4 react as follows: 2NaOH (aq) + H2SO4 (aq) -----------> Na2SO4 (aq) + 2H2O (l) 2H+ (aq) + 2OH- (aq) ------------> 2H2O (l) Methods and safety While wearing safety glasses and a lab coat 100 cm3 of NaOH (aq) was collected in a large beaker. A burette was clamped firmly and filled with dilute HCl ­(aq) until the bottom of the meniscus of the acid was on the zero line. 25.00 cm3 of the NaOH solution was pipetted (with caution) into a conical flask and then four drops of phenolphthalein indicator added. This flask was then placed on a white tile that was placed directly underneath the burette’s tap. Acid from the burette was added rapidly into the conical flask until the red colour disappeared. The volume of the acid added was noted down as the rough value. The burette was then filled again with dilute HCl (aq) to the zero mark and another 25.00 cm3 of NaOH (aq) cautiously pipetted into a second conical flask and four drops of phenolphthalein added. The acid was then added rapidly until within 2 cm3 of the reading noted earlier, at which point it was added slowly until the colour of the solution in the conical flask just disappeared. The volume of acid used was recorded. These last two steps of the experiment were repeated two more times, after which the entire experiment was repeated again, using dilute H2SO4 (aq) instead of dilute HCl (aq). Results Titration Volume of NaOH (cm3) Initial Reading (cm3) Final Reading of HCl (cm3) Volume of HCl (cm3) Final Reading of H2SO4 (cm3) Volume of H2SO4 (cm3) Rough 25.00 cm3 0.00 cm3 25.5 cm3 25.5 cm3 15.7 cm3 15.7 cm3 First 25.00 cm3 0.00 cm3 24.5 cm3 24.5 cm3 13.3 cm3 13.3 cm3 Second 25.00 cm3 0.00 cm3 24.5 cm3 24.5 cm3 13.0 cm3 13.0 cm3 Third 25.00 cm3 0.00 cm3 24.2 cm3 24.2 cm3 13.0 cm3 13.0 cm3 Average 24.4 cm3 24.4 cm3 13.1 cm3 13.1 cm3 At the end point the colour of the solution in the conical flask changed from deep red to colourless. To calculate the molarity of the dilute HCl: The moles of NaOH (aq) are given by: 0.025 dm3 x 0.1 mol dm-3 = 0.0025 moles of NaOH Since mole ratio of NaOH: HCl = 1:1the number of moles of HCl is also 0.0025 moles. Using the formula Molarity = moles (n) / Volume (V): Molarity of HCl = 0.0025 mol/0.0244 dm3 = 0.1024 mol dm-3 Molarity of H2SO4 = 0.0125 mol/0.0131 dm3 = 0.0954 mol dm-3 The number of moles of H2SO­4 is 0.0125 (half that of NaOH since the mole ratio of H2SO4: NaOH is 2:1). The percentage error for HCl calculation is ((24.5-24.4)x2)+(24.4-24.2)/24.4x100) /3=0.546% The percentage error for H2SO­4 calculation is ((13.1-13.0)x2)+(13.3-13.1)/13.1x100) /3=1.075% Discussion In this experiment, phenolphthalein was used as the indicator. During the titration, the acid was allowed to mix with the base. In doing so, the pH values changed from basic (>7) towards 7. At the end point, the pH was neutral as this is the point when there are equal amounts of base and acid in the mixture. The chemical equations are: NaOH (aq) + HCl (aq) -----------> NaCl (aq) + H2O (l) 2NaOH (aq) + H2SO4 (aq) -----------> Na2SO4 (aq) + 2H2O (l) Hydrochloric acid is a monoprotic acid and it yields one H+ ion per acid molecule, and sulphuric acid is a diprotic acid and it therefore yields two H+ ions for every acid molecule. This extra ion means that it takes a lesser number of moles of sulphuric acid to react and neutralize NaOH than hydrochloric acid (Khopkar, 1998). The percentage errors are greater for the titration involving sulphuric acid (1.0753%), while that of hydrochloric acid titration is 0.5462%. The greater difference in the errors probably resulted from the improper rinsing of the glassware prior to beginning the titration using the sulphuric acid, and also by not ensuring that the lower side of the meniscus of the acid in the burette is level with the zero mark. Overall the results obtained for the titration are accurate as the percentage errors are small (less than 2%). This experiment can be improved upon by ensuring that the lower side of the meniscus of the reagent in the burette is exactly level with the zero mark. Also, the glassware should be properly rinsed and dried before beginning the second titration. Conclusion The molarities of the two acids were successfully determined. The values with huge errors were the third and first titrations for hydrochloric acid and sulphuric acid respectively. Titration is used in the neutralization of acids and bases to form salts. References Khopkar, S. (1998). Basic Concepts Of Analytical Chemistry . New Age International. Levine, I. (2001). Physical Chemistry. McGraw-Hill. Read More