Titration Research Paper

Quantitative Chemistry –Titration Determination of the Molarity of an Unknown Solution through Acid-Base Titration Technique 1. Introduction 1. 1 Aim The aim of this investigation was to determine the precise molarity of two (NaOH(aq)) sodium hydroxide solutions produced at the beginning of the experiment through the acid-base titration technique. 1. 2 Theoretical Background Titration is a method commonly used in laboratory investigations to carry out chemical analysis. The most frequent chemical analysis performed through titration is when determining the exact concentration of a solution of unknown molarity.

This technique is usually used in redox and acid-base reactions. Redox reaction is when reduction – lost of oxygen – of one of the substances present in a reaction occurs and subsequently oxidation – gain of oxygen – of the second substance in the same reaction takes place. On the other hand, acid-base reaction is when a solution of known molarity2 and volume present in a conical flask is titrated against a solution of unknown molarity in a burette until neutralization is reached. As I have shown in eq. 1, in this investigation it was an acid reacting with a base, hence, an acid-base titration. q. 1 – Hydrochloric Acid + Sodium hydroxide Sodium Chloride + water HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) In this investigation the latter reaction was carried out, having hydrochloric acid (HCl(aq)) as the analyte in the conical flask and sodium hydroxide (NaOH(aq)) as the titrant in the burette. The analyte was also designated as the standard solution of the experiment, since it has known values of volume and concentration, the figures that allowed the molarity of the titrant to be calculated.

In an acid-base titration, the titrant in the burette is gradually added to the analyte in the conical flask until neutralisation happens, thus, the reaction reaches completion. When neutralisation happens the substances present at the end point are stoichiometrically equivalent, in other words, the value of moles of NaOH(aq) present at the end of the reaction is equivalent to the value of moles of HCl(aq) in the same solution as shown on eq. 2 below. eq. 2 – HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 1 : 1

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The end point of a titration reaction can be obtained through two major methods. Firstly is by using a pH meter which works by introducing electrodes to the flask containing the standard solution. Once in the conical flask, these electrodes would measure the H+ ions present in the conical flask since they change as the titrant in added, until neutralisation happens, as a result, determining the pH of the solution. Knowing that neutralization happens when the pH of the solution is equal to 7, consequently, at the end point the pH meter will read 7.

The second method would be using a colour indicator this could be paper or in liquid form. In an acid-base titration it would be convenient to use an indicator in liquid. For instance, phenolphthalein is a recurrent indicator in this type of reaction which is colourless in an acidic solution and turns pink when in a basic solution. This indicator works by adding a few drops into the conical flask containing the acidic analyte and titrate the basic titrant drop-wise until colour of the solution formed in the conical flask changes to pink.

All things considered, the colour indicator was used in this experiment since it is the most accessible method to measure the end point of an acid-base titration. The purpose of this investigation was to determine the unknown molarity of NaOH(aq) from acid-base titration. The preparation of NaOH(aq) was done by the students performing this investigation. The students were allocated mass of NaOH(s) that was diluted in water and hence obtained the solution NaOH(aq), in this case the titrant. However, the analyte was not produced by the student but provided.

Therefore, after the titration was performed as explained on the previous paragraphs, the data needed to calculate the molarity of NaOH(aq) was obtained. 1. 3 Preliminary calculations 1. 3. 1 The first important value to be obtained from the investigation was the volume of NaOH(aq) used. This was done by the following equation: eq. 3 – for 1st solution produced Average volume volume of 2nd trial - volume of 1st trial2= V1 eq. 4 – for 2nd solution produced Average volume volume of 2nd trial - volume of 1st trial2= V2 1. 3. The next step when determining the molarity of NaOH(aq) was to calculate the moles of HCl(aq) by using the volume HCl(aq) provided on the lab scripts and the molarity obtained from the bottle of HCl(aq) used during the investigation. The eq. 5 and eq. 6 below was used to calculate: eq. 5 – moles1 = V1 (dm3) ? molarity (M) eq. 6 – moles2 = V2 (dm3) ? molarity (M) 1. 3. 3 The third important equation, for both solutions, worth noting are the number of moles of NaOH(aq) present in the reaction. This was obtained by using ratio of the moles of NaOH(aq) : HCl(aq) used during the investigation.

This can be recalled by eq. 2 eq. 2 – HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 1 : 1 1. 3. 4 Hence, moles of both solutions of NaOH(aq) will be the same as the number of moles of HCl(aq) since the mole ratio is 1:1. That is for every one mole of HCl(aq) reacted, one mole of NaOH(aq) would have reacted with to neutralise the acid. Hence, the following equation will be used: eq. 7 – moles of HCl(aq)= NaOH(aq) 1. 3. 5 Lastly, the equation of the molarity of NaOH(aq): eq. 8 – Molarity for solution 1 =moles1volume1 q. 9 – Molarity for solution 1 =moles1volume1 2. Investigation 2. 1 Apparatus Due to the nature of the investigation very technical and precise laboratory apparatus were used to ensure best accuracy in results. For instance, in order to measure the acid, a pipette of exactly 20ml was used. This was very useful since it helped in diminishing the chance of measuring either more or less of acid needed for the investigation. Similarly, the burette used to titrate the NaOH(aq) had 50 ml of volume this allowed enough volume of NaOH(aq) to be titrated, since the exact volume to neutralise the acid was unknown.

Another precise apparatus was the magnetic stirrer. Being magnetic and electric it allowed the solution in the flask to be mixed continuously and vigorously and hence allow the exact volume of acid to be obtained. The other apparatus that were also used in this experiment were the solutions – titrant and analyte – themselves. The sodium hydroxide was given in pellets whereas the hydrochloric acid was provided in liquid form with the molarity of 1M. Moreover, volumetric flasks of volume of 100ml were also provided. This were used to produce the NaOH(aq) solutions, hence the reason for allowing 100ml of NaOH(aq) to be produced.

In its turn, conical flasks of 250 ml of volume were also provided. As mentioned on the introductory paragraphs, the analyte is deposited in the conical flask. In this case, a volume of 250 ml was allowed to host the acid and the titrated base giving enough space for the solution to be formed. Lastly, phenolphthalein indicator was provided together with it a pair of gloves to avoid accidental stain on students’ hands. The apparatus setting is shown below in fig. 1. fig. 1 – diagram of apparatus used in the investigation 2. 2 Safety In terms of safety, the investigation involved very strong solutions.

For instance, the sodium hydroxide pellets, although they were in solid form, after dissolving in water it could cause severe burns if put in direct contact with skin or eyes. Hence, as a pre-cautionary measure some gloves as well as goggles were provided to students. It was important to point out that if in case of accident in eyes, swallow or skin contact it should be rinsed vigorously in abundant water and seek medical attention. As for the hydrochloric acid, it was a very acidic solution that if swallowed it would be very harmful.

Similarly to sodium hydroxide it could cause severe burns if in contact with eyes or skin. For prevention of any accident, lab coats, goggles and gloves were provided. However, in case of accident, medical advice had to be immediately provided to student. 2. 3 Procedure This experiment, it involved two different solutions of NaOH(aq), for this reason, it was allowed to students to work in pairs in order to save time, since only 3 hours were allowed to perform investigation. The first part of the investigation was to prepare two NaOH(aq) solutions. Hence, each student was allocated a mass of NaOH(s) to measure.

In this investigation performed, 2g and 5g of NaOH(s) pellets were meant to be weight using a 2 decimal place weight balance. However, since relatively large pellets were provided and not powder, it made not possible to measure the exact mass intended, instead, 2. 07g and 5. 19g were weighed. After weighting the masses of NaOH(s), the pellets weighting 2. 07g and 5. 19g each mass was put in a separate 250ml volumetric flask, water was added to the flask and then shook in order to let the pellets dissolve to for a solution A and solution B of NaOH(aq) respectively. Secondly, the apparatus shown in fig. was as shown in the figure. Thirdly, 20ml of HCl(aq) was measured as accurate as possible by using a pipette if 20 ml of volume. This HCl(aq) measured was put in a 250ml conical flask. After preparing the acid, in this case, the analyte, 7 drops of phenolphthalein indicator was added to the conical flask where the analyte was added. The conical flask was places on the magnetic stirrer as shown in fig. 1. The forth part of the investigation was when a 50ml burette was filled with solution A. Following this, the magnetic stirrer was switched on, stirring the solution present in the conical flask moderately.

Hence, using the tap present on the burette, the solution A was added to the conical flasks in a drop-wise fashion until one drop was added to turn the solution pink permanently. When the solution in the conical flask turned permanently pink, the end point of the titration had been reached, thus, the volume of solution A used from the burette was recorded. Then, the conical flask was rinsed in abundant water. The sixth part of the experiment was to repeat third to fifth part of the experiment to obtain a second reading of the volume used to titrate solution A.

After the sixth part was finalised, second part to sixth part of the procedure was repeated, however, this time solution B was used in the place of solution A. By the end of the experiment, two values of volume of solution A and two values of volume of solution B titrated against the acid were obtained and recorded in table 1. 3. Treatment of Raw Data 3. 1 Results table | Solution A| Solution B| Molarity of HCl(aq)| 1 moldm-3| 1 moldm-3| Volume of HCl(aq)| 20 ml| 20 ml| | Indicator used| phenolphtlalein| | Volume of NaOH(aq) | Trial I| 41. 1 ml| 16. 8 ml| Trial II| 38. 4 ml| 17. 1 ml| Average volume of NaOH(aq) used| 39. 8 ml| 17. 0 ml| | Observations| *Calculations*Average mass of NaOH(aq) used was calculated using the following formula: Trial I + Trial II2| * Not the mass expected was weighted – mass for solution A -0. 1g difference; mass for solution B +0. 7g difference. * Bubbles given out when dissolving the NaOH(s) * Volumetric felt warm when mixing the NaOH(s) with water * Some residous seen in the acid * Conical flask was rinsed with tap water * Some acid was spilled on the table, i. e. not all 20 ml was put in the conical flask| . Treatment of Results 4. 1 Processing raw data 4. 1. 1 – Balanced equation The balanced equation of the reaction taken place in this investigation was reviwed in the introductory paragraphs, eq. 2 below: eq. 2 – HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) mole ratio: 1 : 1 eq. 2 show that one mole of hydrochloric acid reacts with exactly 1 mole of sodium hydroxide to form salt – sodium chloride and water, hence, the mole ratio between the substances is 1:1. 4. 1. – moles of HCl(aq) for solution A Again, the formula used to calculate the moles of hydrochloric acid for solution A has been reviewed in the beginning of the investigative report. Therefore, in order to find the moles of HCl(aq) eq. 5 was used: eq. 5 – moles1 = V1 (dm3) ? molarity (M) 0. 020 dm3 ? 1 moldm-3 = 0. 020 mol moles1 = ? V1 (dm3) = 20. 0 ml 20. 0ml 1000 = 0. 020 dm3 molarity (M) = 1 moldm-3 4. 1. 3 – moles of HCl(aq) for solution B the formula used to calculate the moles of hydrochloric acid for solution B was the same as the formula calculated for solution A.

Therefore, in order to find the moles of HCl(aq) eq. 6 was used: eq. 6 – moles2 = V2 (dm3) ? molarity (M) 0. 020 dm3 ? 1 moldm-3 = 0. 020 mol moles2 = ? V2 (dm3) = 0. 020 ml 20. 0ml 1000 = 0. 020 dm3 molarity (M) = 1 moldm-3 4. 1. 4 – moles of NaOH(aq) for solution A From the molar ratio between hydrochloric acid and sodium hydroxide it was seen that one mole of acid reacted completely with one mole of the base. This is shown on eq. 2 – HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) mole ratio: 1 : 1 nd hence, eq. 7 moles of HCl(aq)= NaOH(aq) it is just to say that the number of moles of NaOH(aq) produced in this reaction was 0. 020 mol the same as the number of moles of HCl(aq). 4. 1. 5 – moles of NaOH(aq) for solution B Again in the solution B, the molar ratio between hydrochloric acid and sodium hydroxide is the same as the above solution. Hence: eq. 2 – HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) mole ratio: 1 : 1 and thus and hence, eq. 7 moles of HCl(aq)= NaOH(aq) t is just to say that the number of moles of NaOH(aq) produced in this reaction was 0. 020 mol the same as the number of moles of HCl(aq). 4. 1. 6 – molarity of NaOH(aq) for solution A The molarity of NaOH(aq) from solution A was calculated using eq. 8 reviewed at the introductory paragraph of this investigative report. Hence: eq. 8 – Molarity for solution A = moles1volume1 0. 020 mol0. 0398 dm3 = 0. 5 moldm-3 moles1 = 0. 020 mol volume1 = 39. 8ml 39. 8ml 1000 = 0. 0398 dm3 4. 1. 7 – molarity of NaOH(aq) for solution B The molarity of NaOH(aq) from solution B was calculated using eq. also reviewed at the introductory paragraph of this investigative report. Hence: eq. 9 – Molarity for solution B = moles1volume1 0. 020 mol0. 0170 dm3 = 1. 17 moldm-3 moles1 = 0. 020 mol volume1 = 17. 0ml 17. 0ml 1000 = 0. 0170 dm3 5. Discussion of Results The results obtained from the calculations carried out in this investigative report were somehow near the actual value expected to get. For instance, the molarity of solution A was 0. 5 moldm-3 and the assigned molarity was also 0. 5 moldm-3. However, for the molarity of solution B, the value was a slightly offset, the actual value assigned was 1. moldm-3 and from the above calculation the molarity of the solution was 1. 17 moldm-3. Analysing the data from table. 1 it is possible that the difference on the molarity of solution B is due to many errors that occurred during the experiment. The main source of error in this experiment was human error. For instance, when the volume for solution B was being poured into the conical flask, some of the content was spilled on the desk. This means that not all volume of acid was reacted with the base, as intended to be at the beginning of the experiment.

This type of error mentioned on the above paragraph can be minimized have having more practice with using pipette. Another way is by utilizing sophisticated pipettes that will not let the content out unless the person utilizing chooses to do so by pressing a button. 6. Conclusion In this experiment, the unknown molarities of two solution of sodium hydroxide were to be investigating. The aim was to produce two solutions of NaOH(aq) and titrate them against an acid. The production of NaOH(aq) was successful which gave the opportunity to titrate against the acid.

However the molarities calculated were very similar to the molarities intended. This suggests that the investigation was successful, although it was not fully successful due to errors that occurred during the investigation. Nevertheless, the results can be used as the actual molarity of the solution taken into account that the initial masses were not as well as the masses intended to be weight. 7. References * Jones, L; Atkins, P. (2000). Chemistry’s accounting: Reaction Stoichiometry. Chemistry: Molecules, Matter and Change. 4th ed. New York: W. H. Freeman and Company. p160-162 * Clark, J. (2000).

Basic Calculations Involving solutions. Calculations in AS/A Level Chemistry. Pearson Education Limited. p61 – 66 * LoveToKnow. (1996-2011). Titration – Definiton of Titration. Available: http://www. yourdictionary. com/titration. Last accessed 14th Nov 2011. * ChemBuddy. (2005). Concentration lectures – definition. Available : http://www. chembuddy. com/? left=concentration&right=concentration. Last accessed 15th Nov 2011. * Harold, C. (2011-2012). Experiment 2 : Acid-Base Titration. CHE-00027/29 General and Organic Chemistry Laboratory Handbook. Keele University. p17-20 -------------------------------------------- 1 ]. Concentration – number of molecules present in a specific volume of a solution [ 2 ]. Molarity – concentration of a solute per mole; also known as molar concentration [ 3 ]. Analyte – the solution with known values of its volume and concentration [ 4 ]. Titrant – the solution with unknown values of volume and concentration [ 5 ]. End point – the stage at which enough titrant has been added to neutralise the analyte [ 6 ]. pH meter – a laboratory electronic equipment used to measure the pH of a solution [ 7 ]. pH – the negative logarithm value of H+ present in a solution, determining the acidity of the solution

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