However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic. This slight excess of NaOH is not much beyond the end point. The volume of the base is recorded and used to determine the molarity of the acetic acid solution. Experimental Data Standardization of NaOH solution | Trial 1| Trial 2| Trial 3| Mass of KHP| 0. 297 g| 0. 325 g| 0. 309 g| Initial buret reading, NaOH| 0. 00 mL| 0. 50 mL| 7. 70 mL| Final buret reading, NaOH| 32. 0 mL| 34. 0 mL| 38. 7 mL| Volume used, NaOH| 32. 0 mL| 33. mL| 31. 0 mL| Molarity of NaOH solution| 0. 0454 M| 0. 0475 M| 0. 0488 M| Average molarity of NaOH| 0. 0472 M| Titration of unknown | Trial 1| Trial 2| Trial 3| Initial buret reading, NaOH| 2. 70 mL| 19. 9 mL| 0. 00 mL| Final buret reading, NaOH| 19. 9 mL| 36. 2 mL| 19. 8 mL| Volume used, NaOH| 17. 2 mL| 16. 3 mL| 19. 8 mL| Molarity of acetic acid solution| 0. 0780 M| 0. 0769 M| 0. 0935 M| Average molarity of acetic acid solution| 0. 0828 M| Sample Calculations The following calculations were used for each Trial, but only inputs for Trial 1 will be shown below.

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Volume = Final buret reading – Initial buret reading i. Volume of NaOH = Final buret reading of NaOH – Initial buret reading of NaOH ii. Volume of NaOH = 32. 0 mL NaOH – 0. 00 mL NaOH iii. Volume of NaOH = 32. 0 mL Molarity = Moles/Liters i. Molarity of NaOH solution = (mass of KHP/molar mass of KHP) / Volume of NaOH ii. Molarity of NaOH solution = (0. 2966 g/204. 22 g)/0. 032 L iii. Molarity of NaOH solution = 0. 0454 M Molarity of acetic acid = (Molarity NaOH * Volume NaOH) / Volume Acetic Acid i. Molarity of acetic acid = (0. 0472 M * 0. 0172 L)/ 0. 1 L ii. Molarity of acetic acid = 0. 0780 M percent error = Experimenal value-Accepted valueAccepted value*100 i. Percent Error of Molarity of NaOH = 0. 0472 M-0. 05 M0. 05 M*100 ii. Percent Error of Molarity of NaOH = 5. 6% i. Percent Error of Molarity of acetic acid = 0. 078 M-0. 080 M0. 080 M*100 ii. Percent Error of Molaarity of acetic acid = 2. 5% Discussion The results obtained from the experiment proved to the principle that using the indictor we can find the end point, which is very close to the equivalence point of an acidic solution.

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Introduction: The way to get the full results of this lab was through the process of osmosis. Osmosis is the movement of water across a membrane into a more concentrated solution to reach an equilibrium. When regarding cells osmosis has three different terms that are used to describe their concentration. The first of these words is isotonic. Cells in an isotonic solution show that the water has no ...

Then using that point we were able to calculate the unknown molarity which was one of the goals of the experiment. The calculations also verify Boyle’s Theory. When we calculated the molarity of the acetic solution, an average value of 0. 078 M was obtained. The true value of the molarity of the acetic acid solution was 0. 08 M. Although it isn’t right on, it is very close to the true value which leads me into discussing the percent error. We found the percent error of the molarity of NaOH to be 5. 6%, and the percent error of the molarity of acetic acid to be 2. 5%, which are both pretty small.

The error may have occurred when adding NaOH solution. Occasionally slightly more pressure was put on tilts of the piece on the buret to allow the solution to flow through. This means that more of the solution may have been used than needed. Overall, experiment agrees with the formulated hypothesis. Pre-Lab and Post Lab Questions Pre-Lab 1. Molarity of NaOH solution = (mass of KHP/molar mass of KHP) / Volume of NaOH a. Molarity = (0. 2816 g/204. 22 g)/29. 68 mL Molarity = 4. 64*10-5 M 2. Molarity of acetic acid = (Molarity NaOH * Volume NaOH) / Volume Acetic Acid b.

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... If the Density of the Vinegar Solution is 1.00 g/mL, what is the mass percent of Acetic Acid Present? Na= (0.55 mmol/mL) * (7.2 ... (Nb) (Volume b) Volume a 1 mol of NaOH= 40 g NaOH 1 mmol of NaOH= 0.04 g NaOH Na= (0.5 mmol/mL) * (7.9 mL of ... Percent Composition of an Unknown First Solution Knowing the Molarity and Concentration of a Predetermined Second Solution, which I am absolutely Certain will ...

Molarity = ((4. 64*10-5 M)*20. 22 mL)/10. 06 mL Molarity = 9. 34*10-5 M Post Lab 1. A. TD B. TD 2. A graduated cylinder with calibration type TD could be used to deliver a certain amount of a liquid into another container. A graduated cylinder marked TC could be used to contain an accurate volume of a liquid that is to be mixed with another solution, where the experiment is to be done inside of that graduated cylinder. 3. 50g * 1mol /49. 997g = 1 mol 100g * 1mL / 1. 53g = 1L / 15. 3 1mol / (1L / 1. 53) = 1mol* 1. 53 / 1L = 15. 3 mol/L= 15. 3 M