1 / 1

Determination of an Unknown Substance by Titration Larry Sepulveda

Determination of an Unknown Substance by Titration Larry Sepulveda. Discussion. ABSTRACT. RESULTS. What did we know about the starting materials ? We know concentration (a form of density), which is weight versus volume.

dalila
Télécharger la présentation

Determination of an Unknown Substance by Titration Larry Sepulveda

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Determination of an Unknown Substance by Titration Larry Sepulveda Discussion ABSTRACT RESULTS • What did we know about the starting materials? • We know concentration (a form of density), which is weight versus volume. • Concentrations of different species: • [KOH] =mol L-1 = (mass)(molar mass)-1(Lsoln)-1=(4.105g KOH ) (39.10 g/mol)-1 (1 L) = 1.05 M • [Unknown] = 5.873g (60.01 g/mol)-1(1L)-1 = 0.09788 M • [HOAc] = 5.800g (60.01 g/mol)-1(1L)-1 = 0.09651M • What did we know about the finishing materials? • We know that the finishing solution has become more basic according to its change in pH. The higher the pH, the more basic a solution is. We also know the acid dissociation constant1 which helps us calculate the pH. • Acetic acid dissociation constant (Ka ) and pH Calculation: • Ka= [OAc-][H+]/[HOAc] = x2 / (0.09651-x) ≈ x2 / 0.09651 • So x = [H+] = (Kax 0.09651)1/2 = (1.78x10-5x0.09651)1/2= 0.001311 M • And therefore pH = -log 0.001311 = 2.88 • Using the same concentration of base and acid, reproduction of the experiment with acetic acid should yield similar results for Ka: • Ka= [OAc-][H+]/[HOAc] = [OAc-](10-pH)/[HOAc] = (10-2.88)2/[HOAc] =1.9x10-5 • HOAc pH ≈ unknown pH • But how did we account for error? • See table in “Results” section. • Objective: To determine the identity of an unknown solid and be able to reproduce the results. • Methods: gravimetric analysis and acid/base titration • Major Findings: • - The concentration of the dissolved substance in KOH and solutions was found to be 0.09877 with a standard deviation of 32 ppt. • - The substance was identified after analysis and reproduction to be acetic acid (H3CCOOH). INTRODUCTION • An unknown substance is suspected to be acetic acid, a chemical used in the manufacturing of household cleaning agents such as vinegar (also used in the food industry). Its chemical formula is H3C-COOH. • Gravimetric analysis is the study of the transfer of mass in a reaction. It is used in this experiment to compare the ratio of hydrogen ions to conjugate-base ions. This is done through converting mass to a unit that describes chemical ratios (1mole = molecular weight in grams). • Titration is a method of acid/base analysis, where a weak acid or base is treated with a strong base or acid respectively in order to determine the concentration of the weak species. • Acetic acid, suspected of being the unknown substance, reacts with a strong base in the following way: • AcOH+ OH-  AcO- + H2O • The acetate ion then undergoes the following equilibrium reaction: • AcO- + H2 O AcOH+ OH- • For every acid there is a constant of dissociation. If the constant for acetic acid is consistent with the experimental constant, then a match has been made. CONCLUSION EXPERIMENTAL • The unknown acid was determined to be Acetic Acid with an equivalence point pH of about 3.0 and a dissociation constant approximately equal to that of acetic acid. • Acetic acid is a chemical found in vinegar, used for pre-treating laundry , and cleaning floors and countertops. • Future work includes determining the concentration of Acetic acid in unlabeled bottles of vinegar by titration. • Sample Preparation: A KOH solution (~ 0.1 M) was prepared by adding • 4.105 g KOH to one L of water. 5.873 g of unknown solution was added • one liter of water under the assumption that an acetic identity would • yield a 0.01 molar solution. • Instrumentation: A titration funnel and stand were set up to titrate • the unknown. Phenylaline was used as a pH indicator. (see Figure 7-1)2 • Procedure: The unknown was titrated with KOH in order to find the • equivalence point volume. The same procedure was followed for acetic • acid.3 Works Cited 1. Sigmaaldrich.com. 2. Harris, D., Quantitative Chemical Analysis.2007, p. 122-127. 3. Titration Video. http://www.youtube.com/watch?v=g8jdCWC10vQ

More Related