Obtaining pH curves for acid / alkali titrations Planning In order to obtain a pH curve, I must first devise a way in which to measure both the pH and the volume of alkali being added to the weak acid in order to draw a graph from my results and then from this graph find Ka. Method I will set up the equipment as shown in the diagram below: I will use a pipette and pipette filer to take 25 cm 3 exactly of the weak acid (0. 1 M ethanoic acid, CH 3 COOH) and place it in a beaker. I will then take a burette and place in it 50 cm 3 of 0. 1 M sodium hydroxide (NaOH).
I will then measure the exact amount of NaOH added to the acid whilst using an accurate pH meter rather than indicator to measure the change in pH. Before using the pH meter, I will first dip it into two buffer solutions of pH 4 and pH 7 and calibrate it so that it reads accurately. I will record the values for the pH and the volume of NaOH added. Theoretically, this should give me a pH curve from which to work.
In order to obtain a value for Ka, I shall find the equivalence point (at around 25 cm 3 where the graph goes up vertically) and find the pH at this point, I shall then divide this by two to give me the equivalence point. Having obtained this value, I can assume that: [CH 3 COOH] = [CH 3 COO-] = [H 3+O] And as: Ka = [CH 3 COO-] [H 3+O] [CH 3 COOH] Therefore: Ka = [H 3+O] Or: pKa = pH (at equivalence point) Or: pH = -Log 10 Ka Prediction I know that the curve for a weak acid / strong alkali should start higher up the axis than a strong acid / strong base curve. This results in the vertical section of the graph being shorter and due to salt hydrolysis, the curve will rise to 14 far quicker than it should. This is due to the following mechanism: CH 3 COOH + NaOH CH 3 COONa + H 2 O CH 3 COONa CH 3 COO- + Na H 2 O H+ + OH- The H+ ions will then combine with the CH 3 COO- ions to produce CH 3 COOH: H+ + CH 3 COO- CH 3 COOH The removal of H+ ions leaves OH- ions in excess. This excess of dissociated OH- ions leaves a strongly alkaline solution as the solution has the composition of NaOH with Na+ and OH- ions in solution. Although my equivalence point should be at exactly 25 cm 3, this may not work out experimentally due to variations in the concentrations of the acid and alkali and inaccuracies associated with the use of a pH meter which may well be wrong by the same amount every time, thus creating a set of results either higher or lower than they should be.
The Essay on Acid Base Titration Hcl Naoh Water
Titration of an Acid with a Base Jeff Barker Chem. A-1 Wednesday, May 17, 2000 Introduction: When you combine strong acids and bases, perfect ionization is achieved. This means that all that is left after the reactions are water and the spectator ion. The definition of an acid is a substance that produces hydronium ions when dissolved in water. Acids have a pH range from 1-6. Bases release ...
Results Volume (cm 3) +- 0. 1 pH (1. d. p) +- 0. 1 pH (1. d.
p) +- 0. 1 Average pH (2. d. p) +- 0. 10 0 2.
7 2. 7 2. 70 2 3. 3 3.
4 3. 35 4 3. 7 3. 7 3.
70 6 3. 9 4. 0 3. 95 8 4. 1 4.
1 4. 10 10 4. 3 4. 3 4. 30 12 4. 4 4.
5 4. 45 14 4. 6 4. 6 4. 60 16 4. 7 4.
8 4. 75 18 4. 9 4. 9 4.
90 20 5. 0 5. 1 5. 05 22 5. 2 5. 3 5.
25 23 – 5. 4 5. 40 24 5. 6 5. 5 5. 55 25 – 5.
7 5. 70 26 5. 8 6. 0 5.
90 27 – 6. 5 6. 50 28 7. 7 10. 0 8. 85 29 – 11.
3 11. 30 30 11. 7 11. 8 11. 75 31 – 12. 0 12.
00 32 12. 1 12. 1 12. 10 34 12. 2 12. 2 12.
20 36 12. 3 12. 3 12. 30 38 12.
4 12. 4 12. 40 40 12. 5 12. 5 12. 50 Calculation to find Ka Equivalence point = approximately 8.
8 equivalence point = approximately 4. 4 At equivalence point: [CH 3 COOH] = [CH 3 COO-] = [H 3+O] As: Ka = [CH 3 COO-] [H 3+O] [CH 3 COOH] Therefore: Ka = [H 3+O] Or: pKa = pH (at equivalence point) Or: pH = -Log 10 Ka 4. 4 = -Log 10 Ka Ka = 10-4. 4 Ka = 3. 9 x 10-5 Conclusion As I predicted the graph started higher up on the pH scale than that of a strong acid / strong base titration, and the vertical section was shorter and it rose to 14 more quickly than it should due to salt hydrolysis. I was also correct in assuming that the equivalence point may not be exactly 25 cm 3, as in actual fact it was around 28 cm 3.
The Essay on Determination Of Zinc And Nickel Concentration
I. Introduction In this experiment, the zinc and nickel contents of unknowns were tested using two methods. In the first method, nickel and zinc were separated through ion-exchange chromatography and analyzed through chelometric titration. In the second method, the unknown was analyzed through the atomic absorption spectroscopy (AAS) of the mixture. In and ion-exchange column, the ions are ...
This can be explained by errors in the pH meter, and slight variations in the concentration of the solutions of CH 3 COOH and NaOH. I calculated the acid constant (Ka) or the concentration of hydrogen ions [H 3+O] to be 3. 9 x 10-5 that seems reasonable having checked the calculation. Evaluation Although I have obtained reasonable results, I feel there are several ways in which my results could be improved. The pH meter I used, despite taking quick readings was not the most accurate possible. If I wanted to improve the overall accuracy of my results, I could use a more accurate pH meter that remained accurate after calibration, a problem I encountered with my pH meter.
Furthermore, although I can blame my pH meter for inaccuracy in my pH readings, it is more likely that the inaccuracy in the equivalence point is due to slight differences in the concentration of the acid and alkali. This is a problem that is difficult to remedy as I was provided with the solutions of 0. 1 M sodium hydroxide and 0. 1 M ethanoic acid.
However, if I made up these solutions accurately myself, I would have more control over the accuracy of the concentration. If I eliminated these errors, I think my experiment would be near perfect. I could use phenolphthalein to show the exact point at which the graph goes vertical using the colour change. This would be beneficial as it would back up my pH readings and enable me to calculate the equivalence point more easily. 363.
