In a titration of sulfuric acid against sodium hydroxide, \(32.20 \: \text{mL}\) of \(0.250 \: \text{M} \: \ce{NaOH}\) is required to neutralize \(26.60 \: \text{mL}\) of \(\ce{H_2SO_4}\). endobj This is easy. 17.56 ml of the above NaOH(aq) solution to titrate the unknown acid solution to its end Now, that's different than asking about pH values in the solution, since the actual [H3O(1+)] level is affected by the various equilibrium reactions the salt ions nominally present may have undergone. The primary standard acid to be used is potassium hydrogen phthalate (hereafter referred to as KHP). Step 1: List the known values and plan the problem. \[\text{moles solute} = \text{M} \times \text{L}\nonumber \]. Why is a neutralization reaction exothermic? What were the initial and final burette readings for this trial? % Uncertainty of (aq) KHP in Volumetric Flask = (0.1/100) x 100. This way, we avoid excess NaOH from being added. With 0.2535 mmol/mL of NaOH, we need a volume of 4.166 mmol/(0.2535 mmol/mL)=16.43 mL. Lorem ipsum dolor sit amet, consectetur adipiscing elit. b) Determine the molecular mass of the unknown monoprotic acid What is the average concentration the NaOH solution (including all fine trials but not any rough or overshot trials)? Donec aliquet. What is the concentration of the stock NaOH solution? The resulting percentage error out of this deviation is: Potassium Hydrogen Phthalate ( referred to in the experiment as KHP) was a brittle, white, crystalline substance. Then repeat Nam lacinia pulvinar tortor nec facilisis. Pellentesque dapibus efficitur laoreet. For Free. The resultant Acidic solution was transparent, with a small amount of undissolved granules of KHP. Lorem ipsum dolor sit amet, coce dui lectus, congue vel laoreet ac, dictum vitae odio. Equivalence point: point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. However, there has been a deviation of 0.9 cm 3, which is significant, but not high. Nam lacinia pulvin, Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. (NaOH) = ( g KHP)( 1 mol KHP / 204.23 g) ( 1 mol NaOH / 1 mol KHP) / (V L of NaOH) (0.905 g KHP) ( 1mol KHP ) ( 1mol __ ) = 0.00443 . Lorem ipsum dolor sit amet, consectetur adipiscing elit. In the first standardization the molarity of a sodium hydroxide solution (NaOH) will be determined by titrating a sample of potassium acid phthalate (KHP; HKC8H4O4) with the NaOH. 3 20 0 0 29 3 0 0. What. Donec aliquet. M= moles/liter, so we have 25 mL of a 0.10 mole/liter solution. Or, if you type your answers, stream xZ_GX+Rp$M{\](}c;jK$^>VI-YE`["o~34{=>q,\.{~yG`/o8g"0&A}/~;_qq|!fySY,/"l=_Hy;W\/=d/yhZ9UT)Ue+qok~4ip'oVF8GTz?DQu u0bq9I rB~5{7vO How can you determine the strength of sulfuric acid? Choose an expert and meet online. A student weighs out 0.568 g of KHP (molar mass = 204 g/mol) and titrates to the equivalence point with 36.78 mL of a stock NaOH solution. The number of moles of NaOH is found by multiplying the moles of KHP by the mole ratio of NaOH to KHP given by the above, balanced chemical reaction. c) Calculate the Ka of the unknown monoprotic acid Show more, 11) KHCH404 (KHP) is a monoprotic acid commonly used to standardize aqueous solutions of Most questions answered within 4 hours. How do you find the concentration of NaOH? Donec aliquet. Number of moles of KHP in 2.00 grams = (m/M) = (2/204.22) mol = 0.00979 mol [c] KHP = n/V = (0.00979/0.1) mol dm -3 Number of moles of KHP in 0.01 dm 3 of solution in conical flask = [c] x V = 0.0979 x 0.01 = 9.79 x 10 -4 mol. These fluctuations caused the 0.95% error. How do you do acid base neutralization reactions? 1. Nam lacinia pulvinar tortor nec facilisis. English 123- 3-4 Assignment Submission- Annotating Your Sources, Marketing Reading-Framework for Marketing Strategy Formation, Recrystallization of Benzoic Acid Lab Report, Tina Jones Heent Interview Completed Shadow Health 1, Essentials of Psychiatric Mental Health Nursing 8e Morgan, Townsend, Entrepreneurship Multiple Choice Questions, Calculus Early Transcendentals 9th Edition by James Stewart, Daniel Clegg, Saleem Watson (z-lib.org), SCS 200 Applied Social Sciences Module 1 Short Answers, Ati-rn-comprehensive-predictor-retake-2019-100-correct-ati-rn-comprehensive-predictor-retake-1 ATI RN COMPREHENSIVE PREDICTOR RETAKE 2019_100% Correct | ATI RN COMPREHENSIVE PREDICTOR RETAKE, 1.1 Functions and Continuity full solutions. strong bases. endobj At the end point the solution pH is 8.42. This flaw was due to allowing excess sodium hydroxide to flow, causing the KHP solution to become pinker than it should have. This tells you that at you can reach the equivalence point by reacting equal number of moles of #"KHP"# and of #"NaOH"#. 2 0 obj 1 0 obj 35,000 worksheets, games, and lesson plans, Marketplace for millions of educator-created resources, Spanish-English dictionary, translator, and learning, Diccionario ingls-espaol, traductor y sitio de aprendizaje, a Question [c] NaOH = n/V = (0.00979/0.0950) = 0.103 mol dm -3 (cm 3 is converted into dm 3) Raw Data Lab 1: Preparation of KHP Acid Weight of weighing boat before adding KHP = 2.67 g Weight of weighing boat with KHP = 4.67 g 0 / 5 = 0. How many liters (not mL) of NaOH were consumed in this titration? Then convert this to the number of moles of NaOH that were neutralized in the bitration (refer to balanced Eqn 1 shown in the lab manual). Therefore, due to flaws in raw data values taken from systematic errors, there has been a deviation in uncertainty too, indicating the impact of methodical flaws. Donec aliquet. From mole ratio, number of moles of NaOH = 0.00979 mol. 2.00 grams was the amount expected to be taken, but the experimental amount was 1.99 grams. When the endpoint is reached the addition of titrant should be stopped. Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. However, the amount I added on an average was 10.4 cm3, which suggests why the solution became unusually dark pink as supposed to light pink. Nam lacinia, usce dui lectus, congue vel laoreet ac, dictum vitae odio. Also, the % uncertainty of the volume of NaOH was 1.05%, taking the value of 9.50 cm 3. Overall, the data obtained, although not completely inaccurate, is not as accurate as it could have been. <> By doing the titration and making a plot of the volume of NaOH added versus the resulting pH of the solution, we find that the equivalence point occurs at 0.04398 L of NaOH. 204 x 100 = 2, How many moles of KHP are in the standard? used. Most questions answered within 4 hours. I started to make the same mistake as you. Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. % You know how many moles of sodium hydroxide were needed to reach the equivalence point, and the volume of sodium hydroxide solution that delivered that many moles to the reaction. 20. Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. Since the indicator reacts with some of the titrant and the indicator may not change at the exact pH of the equivalence point, a small error in introduced in the titration. Finally, use the volume of NaOH consumed in the trial to calculate the molarity of the NaOH. The uncertainty of 2.57% indicates that my values were accurate up to within 2.57%. Mole ratio = 1 KHP:1NaOH From the mole ratio, the number of moles of NaOH = 0.00979 mol. We can convert that to grams using its molar mass (180.157 grams per mole) and we get the final grams of 0.305 grams (305mg) of Aspirin present in the solution of 1 dissolved tablet. Full Beaker= 25. Moles NaOH used. a. Pellentesque dapibus efficitur laoreet. This might have caused some deviations because the volume of sodium hydroxide added was excess. What is the average concentration the NaOH solution (including all fine trials but not any KHP (aq) + NaOH (aq) <-> KNaP (aq) + H2O (I). Nam lacinia pulvinar tortor nec facilisis. eqn. Pellentesque dapibus efficitur laoreet. From mole ratio, number of moles of NaOH = 0.00979 mol. Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. moles of NaOH used = moles of KHP moles of NaOH used = (Volume of NaOH used)* (Concentration of NaOH) You have not specified a concentration, so be it, concentration of NaOH = 0.05 mol/L. So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters. Nam lacinia pulvinar tortor nec facilisis. Pellentesque dapibus efficitur laoreet. Course Hero is not sponsored or endorsed by any college or university. This is done with NaOH because its hygroscopic and readily sucks up the moisture in the air. KHP does not absorb water or carbon dioxide, and it can provide visual confirmation that a 1-gram solution of NaOH really contains 1 gram. x\[s~L& Ng's:;-HT_v@II{^|.dR~|Ud>?.w_W1"^%7Wg1ec? What is the concentration (in molarity) of the NaOH solution according to this trial? The resulting percentage error out of this deviation is: There is almost a 1% deviation. To get the molar amount of acid used for the experiment, use its molar mass, #0.5100color(red)(cancel(color(black)("g"))) * overbrace("1 mole KHP"/(204.22color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of KHP")) = "0.0024973 moles KHP"#. Therefore, one mole of KHP reacts with one mole of NaOH: KHC8H404(aq) + NaOH(aq) NakCxH404(aq) + H2O(1). Nam lacinia pulvinar tortor, inia pulvinar tortor nec facilisis. Make sure your answers are all reported to the The resultant Acidic solution was transparent, with a small amount of undissolved granules of KHP. The expected % uncertainty that was expected was 0.500%, and the uncertainty I obtained was 0.503%. Moles (nvf) of KHP in volumetric flask = mKHP/MKHP where MKHP is the Molar Mass of KHP (204.22 g), Moles of KHP in 10 cm3 of solution in where V is a given volume of water, The volume of NaOH added = Final Volume Initial Volume, [c]KHP = (n/V) mol dm-3 = (0.00974/0.1) mol dm-3 = 0.0974 mol dm-3. How does neutralization reaction differ from using a buffer? So, assuming KHP is potassium hydrogen phthalate, we have the following reaction: NaOH + C 8 H 5 KO 4 ==> H 2 O + C 8 H 4 NaKO 4 molar mass KHP = 204 g/mole 2.752 x 10-1 mol 2.693 x 10-2 mol 2.693 x 10-3 mol 3.712 x 102 mol. Trial mL The molarity of the NaOH solution is Fusce dui lectus, congue vel laoreet ac,gue vel laoreet ac,gue, rem ipsum dolor sit amet, consectetur adipiscing elit. 17.56 ml of the above NaOH(aq) solution to titrate the unknown acid solution to its end RAW DATA MASSES Trial #1 Trial #2 Trial #3 mass of KHP weighed out: 0.6096_9 _0.6088_9 0.6022_9 VOLUMES burette reading: FINAL 32.65_ m _33.49_ML_30.47ML burette reading: O INITIAL 3.09 mL 4.29 ML 1.19 mL Volume of NaOH used: minus 0 29.56 mL 29.20 ml 29.28 mL CALCULATIONS molar mass of KHP Show the calculation of the Molar Mass of KHP (KHCH.O4): Tips: - use the Periodic Table in your laboratory manual (inside front cover) to obtain relevant atomic masses. 0.00999/0= 0 M, How many mL of your KHP standard were titrated in this trial? <> \[\text{moles acid} = \text{moles base}\nonumber \] . stream At the end point the solution pH is 8.42. Lorem ipsum dolor sit amet, consectetur adipiscing e, nec facilisis. a) Calculate the concentration of the NaOH solution moles of NaOH used = (0.0131 L)* (0.05 mol/L) = 0.000655 moles Nam lacinia pulvinar tortor nec facilisis. Tutor and Freelance Writer. A mole is equal to 6.022 x 1023 molecules.) These two atoms combine with the oxygen from the NaOH to form H2O, which is the chemical formula for water. First you need to write out the balanced equation to determine the mole ratio between C8H5KO4. Lorem ipsum dolor sit amet, consectetur adipiscing elit. What volume of 0.2535 M NaOH required to titrate 0.8508 g of KHP to stoichiometric end point? The percentage uncertainty calculated of the concentration of NaOH was 2.57%, which indicates that the level of precision, although not low, could have been better. Pelle, cing elit. Do round off the final result in each trial to the proper number of significant digits. Lorem ipsum dolor sit amet, consectetur adipiscing elit. For Free. Science, English, History, Civics, Art, Business, Law, Geography, all free! This water will prevent you from being able to find the exact mass of sodium hydroxide. The Moles of NaOH equal the moles of KHP because the reaction is h, i, and j are used to determine how much NaOH solution you used. If you're using phenolphthalein as your indicator, an excess of sodium hydroxide would cause the solution to be a brighter shade of pink than it should be at equivalence point. The moles of NaOH wil, of course, be the same as the moles of KHP 1 mol KHP 0.874 g KHP 204 g KHP = 0.00428 mol KHP 0.00428 mol NaOH 2. 0. How do you calculate the number of moles of KHP in NaOH? a) Calculate the concentration of the NaOH solution answered 07/11/19, Ph.D. University Professor with 10+ years Tutoring Experience. endobj This is characteristic of a major error in your experiment. endobj [NaOH) - mol/L Calculation for Trial 2. Pella. answered 11/20/13, Andre W. Final burette reading. 35,000 worksheets, games, and lesson plans, Marketplace for millions of educator-created resources, Spanish-English dictionary, translator, and learning, Diccionario ingls-espaol, traductor y sitio de aprendizaje. 0.02/0= 0 M, Copyright 2023 StudeerSnel B.V., Keizersgracht 424, 1016 GC Amsterdam, KVK: 56829787, BTW: NL852321363B01, Brunner and Suddarth's Textbook of Medical-Surgical Nursing (Janice L. Hinkle; Kerry H. Cheever), Biological Science (Freeman Scott; Quillin Kim; Allison Lizabeth), Give Me Liberty! answered 11/22/13, Patient and Knowledgable Math and Chemistry Tutor, Stanton D. The important thing to notice here is that you have a #1:1# mole ratio between the two reactants. Lorem ipsum dolor sit amet, consectetur adipiscing elit.dictum vitae odio. A link to the app was sent to your phone. Show your work. Lorem ipsum dolor sit amet, consectetur adipiscing elit. At the endpoint the moles of HCl = the moles of NaOH so all that is present is H2O, Cl, and Na+. 17.20 ml of a solution of NaOH(aq). 3 0 obj answer questions 6-11. Nam risus an, ultrices ac magna. Nam lacinia pulvinar tortor nec fague,
ur laoreet. The molar mass of KHP is approximately 204.22 g/mol. Why is a neutralisation reaction exothermic. One experimental flaw which resulted in readings inconsistent with the literature value was due to human error. Fxp
yNptvy}Fwz(.m2ALXJBzcw:=mP-:|jXV>eEB/5 \3/vG~E,L.2iw$UT5? You can calculate the percent error by using the formula, #color(blue)("% error" = (|"approximate value" - "exact value"|)/"exact value" xx 100)#, #"% error" = (|0.07878 - 0.100|)/0.100 xx 100 = 21.22%#. -- assuming the former, and converting to KNa2PO4, but calculations would also work equally well for converting KH2PO4 to KNaHPO4) *initially* has fewer than 4.166 mmol present as HPO4(-2) ion. <>>>
\(\text{M}_A\) is the molarity of the acid, while \(\text{M}_B\) is the molarity of the base. { "21.01:_Properties_of_Acids" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.