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UIUC CHEM 203 - Characterization of a Weak Acid lab

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x̄ =x̄ = = 155.8gMean pKax̄ =x̄ = = 3.35Mean Concentration of NaOHx̄ =x̄ = = .1087 MTable 6 Standard Deviation of Molar Mass.Xi=152.14g/mol(Xi- x̄)(Xi- x̄)2∑(Xi- x̄)2Trial 1-2.968.7616Trial 2-4.4619.891641.32684.55Trial 3-3.5612.6736Sample calculationsTrial 1: (Xi- x̄) = (152.14-155.1) = -2.96(Xi- x̄)2 = (-2.96)2 = 8.7616∑(Xi- x̄)2 = (8.7616+19.8916+12.6736)2 = 41.326Standard Deviation = = = 4.55Xi=3.41(Xi- x̄)(Xi- x̄)2∑(Xi- x̄)2Trial 1.04.0016Trial 2.07.0049.0101.0711Trial 3.06.0036Sample calculationsTrial 1: (Xi- x̄) = (3.41-3.37) = .04(Xi- x̄)2 = (.04)2 = .0016∑(Xi- x̄)2 = (.0016+.0049+.0036)2 = .0101Standard Deviation = = = .0711Table 7 Standard Deviation for Concentration of NaOHXi=.1 M(Xi- x̄)(Xi- x̄)2∑(Xi- x̄)2Trial 3-.013.000169Trial 4-.01.0001.000278.0118Trial 5-.003.000009Sample calculationsTrial 1: (Xi- x̄) = (.1-.113) = .013(Xi- x̄)2 = (.013)2 = .000169∑(Xi- x̄)2 = (.000169+.0001+.0000009)2 = .0118Standard Deviation = = = .0118Table 8 Percent Error Molar MassActual Mandelic Acid Molar Mass = 152.14 g/molCalculated Molar Mass(g/Mole)Percent errorTrial 1155.11.95%Trial 2156.62.9%Trial 3155.72.33%Mean155.82.41%Sample Calculations for the mean= x 100 =2.41%Table 9 Percent Error Molar MassActual Mandelic Acid pKa3.41Calculated pKaPercent errorTrial 13.371.17%Trial 23.342.05%Trial 33.351.76%Mean3.351.67%Sample Calculations for the mean= x 100 =1.67%Characterization of a Weak AcidAbstractIn this lab, a manual titration and an electronic titration were performed in order to observe the characteristics of a weak acid. The manual titration used a known weak acid KHP, and the electronic used an unknown acid. For the unknown acid titration, the pH vs. Volume was graphed using lab quest. This information was used to determine the unknown acid. The acid was determined to be mandelic acid. IntroductionThe main goals of this experiment are to understand the titration curve of a weak acid, calculate the molar mass of the weak acid, calculate the Ka value of an unknown weak acid, and use the molar mass and Ka values to identify the acid. The determination of the dissociation constant Ka is an important concept in this lab. The dissociation constant is the extent that, for example, a weak acid dissociates in a solution into ions. Also the molar mass is the amount of grams in one mole of a solution. These will be determined by titrating a weak acid into a strong base, NaOH, first by using phenolphthalein, then a pH probe. The equivalence point is the point at which the concentration of the acid is equal to the concentration of the base. The amount of acid used will be massed, and then molar mass can be determined1. The general formula for the addition of a weak acid and a strong base is as follows for a monoprotic acid. A monoprotic acid is one that loses one proton.1 HA + OH-  H2O + A-(1)The Ka also can be expressed as follow. Also the H+ ions can be calculated by the observed pH at any moment with the following reaction1.Ka = [H+][A-]/[HA] (2)The concentration of HA can also be calculated at any time in the following equation.1 Original molesof acid−moles HA consumedvolume(3)In order to find the many things in a titration—when a measured amount of unknown substance is slowly added to another known solution until equivalence is reached—by finding the halfway point. The halfway point is the point at which the amount of solution added needed to reach equivalence point is half. At this point Ka=[H+] and pKa=pH. Also if the [H+] has been calculated the following can be used to determine pH1:2pH = -ln[H+] (4)For a diprotic acid—an acid that loses two protons per molecule. This follows the following equations for titration reaction for the first and second equivalence points respectively1:H2A + OH-  A2- + H2O (5)HA- + OH-  A2- + H2O (6)There are two dissociation constants Ka1, and Ka2 for a diprotic acid and can be defined bythe following1:Ka1 = [H+][HA-]/[H2A] (7)Ka2 = [H+][HA2-]/[HA-] (8)For a diprotic acid there are two equivalence points because it dissociates twice. In this case the second halfway point is important for the reason that moles of HA- is equal to moles of A2- and because of this concentrations are equal. Therefore Ka2=[H+] and pKa2 = pH. pKa measures the strength of an acid. Also a detailed introduction can be found in thelab manual1. [H+] is the acid [A-] is the base.Titrations and neutralization reactions can be seen in many areas in the science world to determine many different things. In the Czech Republic, some scientists used a neutralization micro-titration to determine the nitrate content in different drinking wells. This was done to determine a scheme to remove these ions from this water2. Also becauseof the acidification of soils and surface water in Australia, indicator bas-field titrations areused to monitor the pH of water in the area, and ensure the pH does not go uncontrolled, and keep the water neutralized3. Finally in an experiment, experiment 223, a bacteria was emitting H+ sulfuric acid. Alkaline cations were then dumped into the lake and it was found to neutralize the lake, therefore neutralizing the bacteria4. Materials and MethodsA detailed procedure can be seen in the lab manual. There were many slight deviations to this lab1. For the first titration, KHP was added to boiled DI water and each was titrated. The amount of KHP varied by a small amount each time but was weighed onan analytical scale by difference. The first trial of manual titration used .3004g KHP, the second .3004g, the third used .3002, the fourth .3020 the fifth .3000g. There were five trials run because the first two were overshot. The third fourth and fifth trial were the 3ones that mattered. The starting position of the stock solution varied from the manual only for the fourth and fifth trial. In the fourth trial the starting position of stock was 4.9mL and in the fifth, 3.1mL. For the electronic titrations, the first trial used .2994g was used, for the second, .3004g was used and for the third .2770g was used. For the calibration of the drop counter 9.3mL was run through the counter. The drop counter had to be recalibrated due to malfunction and the second time 9.1mL was run through. There was no other deviation from the lab manual. ResultsThe process of creating the stock solution was a simple addition of DI water to a concentrated NaOH to dilute it to a .1M solution. This was done by calculating how muchNaOH was needed and


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