PH Titrations Process - Lab Report Example

pH Titrations Background: In general, titration refers to an experimentally conducted process in chemistry lab, normally for the purpose of determining substance concentration of interest in a solution. Typically, acid-base titrations are applied in order to estimate the molarity either of an acid or a base used through a setup that consists of a burette that holds the ‘titrant’ and flask filled with ‘analyte’ or the solution with unknown concentration. During titration, it is essential to figure out the ‘equivalence point’ where moles of acid and moles of base added become equivalent. To aid an experimenter in this situation, a halochromic organic compound known as ‘indicator’ (s.a. phenolphthalein, methyl red, methyl orange, and bromothymol blue) is incorporated where appearance of color change from clear solution signifies the ‘end point’ or the point assumed to be nearest the point of equivalence. At this stage, the volume of reading from the burette is noted down the instant the indicator changes color. With experimental outcomes comprising known molar concentrations and volumes of acid and base used, pH is calculated. In place of an indicator, glass electrode is utilized so that multiple pHs may be read and graphed from which to determine the pH at the equivalence point. For the neutralization reaction occurring between HCl (aq) and NaOH (aq): For the neutralization reaction occurring between CH3COOH (aq) and NaOH (aq): Hypothesis: The equivalence point is achieved once the titration has gone to the extent when the acid (analyte) and the base (titrant) are in equimolar amounts which takes place on yielding a neutral pH along a steep region in the titration curve Aim: To locate the ‘equivalence point’ through the measure of pH based on the titrant volume vs. pH-titration curve obtained in the experiment and identify the remaining species (in excess) as well as find the pH of the resulting solution after each titration attempt Method: The burette was filled with 1.0 NaOH to the zero mark. A 250- conical flask was rinsed with distilled water, emptying the flask and refilling it with about 50 of deionised water, after which, a 25 of 0.1 HCl was poured into the flask via pipette and the flask was placed on the magnetic stirrer. With the stirrer bar dropped in the flask, the glass electrode was immersed in the flask then the burette was positioned over the flask’s mouth such that the NaOH would be added into it without obstruction from the glass electrode. The magnetic stirrer was turned on and the flask was adjusted in the manner that would enable the stirrer bar to move freely, having no obstructive contact with the glass electrode throughout the titration process. Measuring of the pH prior to titration was initiated then successive additions of the base titrant were carried out starting with 1 until the equivalent point was arrived at, followed by 0.2-additions that sum up to approximately 8 of NaOH. Final addition of two consecutive 1- titrant amounts was made, taking into account the pH reading for each addition from the onset of titration to this point. Another trial (using the same analyte -- HCl) was performed, then titration was repeated, this time using a different analyte (0.1 CH3COOH) and applying the same quantity (50 ) of deionised water and of 1.0 NaOH. As conducted for HCl, pHs of neutralization reaction between HOAc (analyte) and the titrant (NaOH) were correspondingly read after each base addition. Results and calculations: Note: ( 1 cm3 ? 1 mL ; 1 mol / dm3 ? 1 M ) pH @ equivalence point is about 6.8 based on the graph. This corresponds to NaOH volume of about 5.5 mL. mmoles NaOH = (1.0 M) (9.00 mL) = 9.0 mmoles NaOH mmoles HCl = (0.1 M) (25.00 mL) = 2.5 mmoles HCl Upon neutralization, 9.00 mmoles - 2.5 mmoles = 6.5 mmoles NaOH left total volume = 50 + 25 + 9 = 84 mL [OH-] = = 0.07738 M ---? pOH = -log (0.07738) pOH = 1.11 so that pH = 14 - 1.11 ---? pH = 12.89 Since NaOH and HCl are strong base and strong acid, respectively, pH at the equivalence point is ideally 7.0 (neutral) % error (pH @ equivalence pt.) = * 100% = 2.86% % error (pH of final solution) = * 100% = 6.13% pH @ equivalence point is about 6.7 based on the graph. This corresponds to NaOH volume of about 4.8 mL. mmoles NaOH = (1.0 M) (8.00 mL) = 8.0 mmoles NaOH mmoles HCl = (0.1 M) (25.00 mL) = 2.5 mmoles HCl Upon neutralization, 8.00 mmoles - 2.5 mmoles = 5.5 mmoles NaOH left total volume = 50 + 25 + 8 = 83 mL [OH-] = = 0.06626 M ---? pOH = -log (0.06626) pOH = 1.18 so that pH = 14 - 1.18 ---? pH = 12.82 % error (pH @ equivalence pt.) = * 100% = 4.28 % % error (pH of final solution) = * 100% = 5.93 % pH @ equivalence point is about 8.6 based on the graph. This corresponds to NaOH volume of about 4.6 mL. mmoles NaOH = (1.0 M) (7.00 mL) = 7.0 mmoles NaOH mmoles CH3COOH = (0.1 M) (25.00 mL) = 2.5 mmoles CH3COOH Upon neutralization, 7.00 mmoles - 2.5 mmoles = 4.5 mmoles NaOH left total volume = 50 + 25 + 7 = 82 mL [OH-] = = 0.05488 M ---? pOH = -log (0.05488) pOH = 1.26 so that pH = 14 - 1.26 ---? pH = 12.74 Theoretically, [ CH3COO- ] = = 0.05779 M CH3COO- (aq) + H2O (l) ?? OH- (aq) + CH3COOH (aq) Initial: 0.05779 0 0 Change: - x + x + x_______ Equilibrium: 0.05779 - x x x Kb = = = 5.56 * 10-10 = = ---? x = [OH-] = 5.67 * 10-6 M pOH = -log [OH-] = -log (5.67 * 10-6) = 5.25 so that pH = 14 - 5.25 = 8.75 However, pH (experimental) @ equivalence pt. = 8.6, % error (pH @ equivalence pt.) = * 100% = 1.71 % % error (pH of final solution) = * 100% = 5.10 % pH @ equivalence point is about 8.0 based on the graph. This corresponds to NaOH volume of about 4.3 mL. mmoles NaOH = (1.0 M) (4.80 mL) = 4.8 mmoles NaOH mmoles CH3COOH = (0.1 M) (25.00 mL) = 2.5 mmoles CH3COOH Upon neutralization, 4.8 mmoles - 2.5 mmoles = 2.3 mmoles NaOH left total volume = 50 + 25 + 4.80 = 79.8 mL [OH-] = = 0.02882 M ---? pOH = -log (0.02882) pOH = 1.54 so that pH = 14 - 1.54 ---? pH = 12.46 Theoretically, [ CH3COO- ] = = 0.05422 M CH3COO- (aq) + H2O (l) ?? OH- (aq) + CH3COOH (aq) Initial: 0.05422 0 0 Change: - x + x + x_______ Equilibrium: 0.05422 - x x x Kb = = = 5.56 * 10-10 = = ---? x = [OH-] = 5.49 * 10-6 M pOH = -log [OH-] = -log (5.49 * 10-6) = 5.26 so that pH = 14 - 5.26 = 8.74 However, pH (experimental) @ equivalence pt. = 8.0, % error (pH @ equivalence pt.) = * 100% = 8.47 % % error (pH of final solution) = * 100% = 9.07 % Discussion: According to the experimental results, titration curves exhibit that in the first and second trials of titrating HCl and NaOH, the pHs found, as obtained from the graphs, are correspondingly 6.8 and 6.7. These values are each slightly different from the expected (literature) neutral value of 7.0 which occurs on neutralization reaction between a strong base and a strong acid, particularly when mole amounts reach the extent of equivalence. Errors of 2.86% and 4.28% estimated out of the two respective attempts account for any unnoticed occurrence of inappropriate (meniscus) reading of the burette or lack of proper glass electrode calibration. Similarly, the pHs at the equivalence point in the titration of the weak () with the strong base, incurred errors of 1.71% and 8.47% both of which could have been made possible by overlooking procedure parts that require significant focus. Compared to organic indicators, nevertheless, the glass electrode appeared to play its role more effectively on detecting pH upon each addition of titrant unlike indicators that are merely observed to work at a single instant via color change (@ end point). Conclusion: Evidently, the generated curves each contained the point of equivalence somewhere along the mid-segment of the part with steep slope where the pH at such point approximates to or coincides with the theoretically solved pHs. In the future, the titrimetric method employed herein may quite be an indispensable basis on improving pH measurements of titration systems either with normal or peculiar behaviour considering diversity of acids and bases involved as well as any new schemes for neutralization or ‘redox’ dealing with component concentration and pH analysis. References Petrucci, R.H., Harwood, W.S., and Herring G. (2001). General Chemistry: Principles and Modern Applications. Maryland: Prentice Hall. Waser, J. (1967). Acid – Base Titration and Distribution Curves. Journal of Chemical Education. 44(5), p274. Read More