Titration of Acids and Bases Essay Example for Free

Titration of Acids and Bases EssayIntroductionAccording to Arrhenius definition, acid is kernel that produces H3O+ ion mend base is meaning that produces OH- ions. The reaction between acid and base often yields the products of salt and water. The formation of water in this neutralization reaction is caused by the combination of H3O+ and OH- ions.In recite to determine the concentration of an unknown acid and base, a method called acid-base titration is used. The end of the titration has been reached when the moles of acid tints the moles of base. This is called the equivalent point. However, end point os the actual point that is reached in neutralization reaction. End point is signalled by the change in color of the solution because of the presence of pH indicator. The common used indicator is phenolphthalein it is colorless in acidic solution and pink in basic solution. In this lab, the concentration of sodium hydroxide (NaOH) was determined by using a known add together of Potassium Hydrogen phthalate (KHP). The balanced chemical equation for the reaction is KHC8H4O4(aq) + NaOH(aq) = H2O(l) + KNaC8H4O4(aq) (1)KHP was chosen as a good acid to standardize NaOH because it has high molecular charge and stable on drying. The standardized NaOH solution was then used to determined the share composition of KHP in an unknown substance 47.Experimental Information segmentation 1 Standardization of NaOHNaOH solution was prepared by taking 75mL. NaOH from stock solution and then diluted to 750 mL with DI water in Nalgene bottle. Three samples of KHP were weighed on analytical balance model BP2505, and placed in 250 mL Erlenmeyer flask. The samples of KHP were warmed up on hot plate to dissolve in Erlenmeyer flasks with 100 mL of DI water added. Two drops of indicator phenolphathalein were added to each flask after KHP completely dissolved.The 50 mL buret was rinsed with DI water and the prepared NaOH solution. The buret was then fill with NaOH solution with the initial good deal of each trial recorded. The NaOH solution was slowly dropped into the flask contained KHP solution for titration from the buret. The flask was swirled often for the reactants for mix thoroughly. The KHP solution in the flask turned light pink when the end point had been reached. The final pile of NaOH was recorded.The volume of titrant NaOH used was determined by subtracting the initial volume NaOH from the final volume NaOH marked on the buret. The mole of KHP was calculated by dividing the mass of KHP used in each trial to the molar mass of KHP. Moles KHP = mass, g x 1 mole204.23 g (2)The mole of NaOH was equal to the mole of KHP based on the balanced chemical equation (1), since the ratio is 11 Moles NaOH = moles KHP (3) The concentration of NaOH was calculated by dividing the mole of NaOH to the volume of NaOH used to titrate. NaOH = moles NaOHL, NaOH used to titrate (4)The volume of titrant NaOH used was determined by subtracting the initial volume NaOH from the final volume NaOH marked on the buret. The mole of NaOH was calculated by multiplying mean NaOH from part 1 by the volume NaOH used. Mole NaOH = mean NaOH x volume NaOH used (5)The mole of KHP was equal to the mole of NaOH based on the balanced chemical equation (1), since the ratio is 11. Moles KHP = moles NaOH (6)The mass of KHP in the sunstance 47 of each trial was calculated by multiplying the moles of KHP to the molar mass of KHP. Mass KHP = moles KHP x 204.23g1 mole (6)The per centum composition of KHP in the substance was calculated by dividing the mass of KHP found to the mass of substance of each trial. % KHP = grams KHPgrams of substance 47 x 100 (7)Results and DiscussionIn part 1, the concentration of NaOH in trial 1, 2, and 3 was determined to be 0.1054 M, 0.1052 M and 0.1048 M respectively. Therefore, the mean NaOH was 0.1051M +/- 0.0003. The amount of NaOH solution used to titrate was about 0.023 L per 0.49g KHP. The concentration of NaOH calculated from thr ee trials was pretty precise. However, trial 3 gave the conduce with the greatest in difference compared to trial 1 and 2. Thus, the possible source of error in trial 3 could be inaccurate reading og volume of NaOH solution used to titrate. Also, there could be loss in the amount of KHP when transferred to the flask after weighed. Those two factors could lead to inaccurate calculation of moles of NaOH as come up as concentration of NaOH.In part 2, the percent of KHP in substance 47 in trial 1, 2, and 3 was calculated to be 55.96%, 55.87%, and 55.37%. Therefore, the mean percent KHP was 55.73%. The amount of NaOH solution used to titrate was about 0.018L per 0.69g substance 47. The results of percent KHP of three trials compared to each other were precise. There was no literature value of percent of KHP in substance 347 provided to calculated percent error. However, the possible sources of error could be inaccurate reading of volume NaOH used, loss in amount of substance while tran sferred from weigh paper to Erlenmeyer flask or over titrating.The important steps to get the most accurate in calculation for NaOH and %KHP were to read and record info carefully. Also, avoiding over titration was extremely important. It could be done by carefully letting NaOH solution go down drop-by-drop and constantlt swirl the flask for the reactants to mix completely.ConclusionThe grand of NaOH solution based on three trials was 0.1051 +/-0.0003 M. The percent KHP in substance 47 was 55.7 +/-0.3%. The significant point of this lab was to determine the end point of a reaction between an acid and a base. The data collected from the end point gave the amount of base needed to react with an acid when once treasured to perform neutralization reaction. Another significant point was that by using NaOH solution with a known concentration, the percent KHP in a substance could be determined after perform the titration process.ReferencesAnliker, Keith et al. Experimental Chemistry II. Indianapolis Hayden McNeil , 2008, pp.47-52