From the following data, find the % (w/w) cream of tartar (KHC4H406. MM=188): Wt of sample = 1.4160 g N2OH titrant used = 20.87 mL HCl used for back titration = 1.27 mL 1.00 mL HCI = 1.12 mL N2OH 1.00 mL HCI = 0.01930 g CaCO3 O 95.59% (w/w) O 89.08% (w/w) O 47.75% (w/w) O 87.12% (w/w)
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- Answer if true or false , if false, put the correct answer 1. There is 11.34 mmol in 0.2011 of 0.5604 M HgCl2 2. The x axis of sigmoidal titration curve is p function of the analyte or titrant 3.The concentration of the secondary standard is relative to a primary standard 4. There is 110 mmol in 79.8 mL of 0.1379 M NH4VO3 (116.98 g/mol)Titration of a 0.824 g of 99.99% KHP (204.23 g/mol) with phenolphthalein required 18.3 mL of NaOH (40.00 g/mol) solution to reach the end point. The same titrant was used to analyze an impure acetic acid (CH3COOH, 60.06 g/mol) solution. A 10.0 mL aliquot of the sample required 12.9 mL of the titrant. What substance served as the primary standard?Which served as the indicator in the titration?What is the color at the end point?What is the concentration of the titrant?What is the molar concentration of the acetic acid solution?A dilute peroxide solution was prepared by quantitatively diluting 10 mL stock H2O2 (MW = 34.0147) to 250mL using a volumetric flask. 50 mL aliquot of the diluted peroxide solution was titrated using the previously standardized KMnO4 in problem 1. Titration of the sample required 29.00 mL titrant and the blank containing 50 mL 1:5 H2SO4 required 0.75 mL of the standard KMnO4. Calculate the concentration in %w/v of the stock H2O2. (Hint: H2O2 produces O2 under acidic condition). Follow sig. Fig
- Direct titration: 10.90 ± 0.02 mL of an unknown KHP solution required 11.72 ± 0.02 mL of 0.1090 ± 0.0006 M NaOH to reach the end point. What is the concentration of KHP in the unknown solution reported with absolute uncertainty? If the unknown solution was prepared by dissolving 5.128 g of the unknown KHP (204.22 g/mol) in 100.00 mL of water, what is the weight percent of KHP in the sample?A 1.067g sample of magnesium oxide of 84.736% were treated with 50mL of 1.017 N Sulfuric Acid, and a 5.195mL volume of sodium hydroxide is required in the back titration. 1. What is the equivalent weight consumed by the acidic titrant? A. 5.281 g-meq B. 55.150 g-meq C. 5.723 g-meq D. 50.850 g-meq 2. What is the difference of milliequivalent weight consumed in the reaction? A. 45.127g-meq B. 56.573g-meq C. 50.850g-meq D. 55.150g-meq 3. What is the amount (in mg) of the analyte that is equivalent to 1 milliliter of the tirant at its equivalence point? A. 22.060mg B. 40.680mg C. 44.12mg D. 20.340mgalculate the Ksp of a Ca(OH)2 solution using the data below: Titrated with 0.103MHCl Initial syringe value: 9.71 Final Syringe value: 1.53 Volume of Ca(OH)2 titrated 24.39 mL
- A 50.00 (±0.02) mL portion of an HCl solution required 29.71(±0.02) mL of 0.01963(±0.0032) M Ba(OH)2 to reach an end point with bromocresol green indicator. ? of HCL = 29.71?? ? 0.01963 ???? ??(??)2 ?? ? 2 ???? ??? ???? ??(??)2/ 50.00?? = 0.02333 ? Calculate the uncertainty of the result (absolute error).Calculate the coefficient of variation for the result.Molarity of titrant (NaOH): 0.4550 M HC2H3O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l) Trial # First Second Third Fourth Initial buret reading 0.15 mL 2.43 mL 1.32 mL 0.58 mL Final buret reading 18.62 mL 20.87 mL 20.03 mL 19.14 mL Volume of titrant used 18.47 mL 18.44 mL 18.71 mL 18.56 mL 4) Calculate the molarity of the acetic acid in the vinegar solution (Show your work). use FW for moles-->grams acetic acid. Molarity acetic acid = _____________ M 5) Calculate the weight % of acetic acid in the vinegar. How does this compare with the % listed on the label (5.00%)? (For this calculation assume that density of vinegar is 1.03 g/mL and of course, show your work). Weight % = ___________ 6) If you didn’t get the same weight % of acetic acid as listed on the vinegar label (5.00 %), what are two things (be specific) that could’ve happened during the experiment that could explain the variation from the expected weight %? To do…A 0.3012g sample of an unknown monoprotic acid requires 24.13mL of 0.0944MNaOH for neutralization to a phenolphthalein end point. There are 0.32mL of 0.0997MHCl used for back titration. a. How many moles of OH are used? How many moles of H+ from HCl? _______moles OH ________moles H+ b. How many moles of H+ are there in the solid acid? Use Eq.5. ____________ moles H+ in solid c. What is the molar mass of the unknown acid? Use Eq.4. ____________ g/mol
- Need solution to all parts if not answered I'll downvote solution Find the pH during the titration of 20.00 mL of 0.1910 M benzoic acid, C6H5COOH (Ka = 6.3 10-5), with 0.1910 M NaOH solution after the following additions of titrant. (a) 0 mL (b) 10.00 mL (c) 15.00 mL (d) 20.00 mL (e) 25.00 mLA 0.5027 g of KHP (KC8H5O4; MW = 204.22 g/mol) required 11.90 mL of NaOH titrant to reach the phenolphthalein endpoint. Calculate the standardized concentration of NaOH.A RbOH solution is titrated four (4) times against potassium hydrogen phthalate (KHP; FW=204.224) samples to the Phenolphthalein endpoint. Using the data below, determine the concentration of the RbOH solution? g of KHP Volume of Base Required 0.5373 g 42.49 mL 0.5856 g 43.88 mL 0.5790 g 48.56 mL 0.5856 g 44.60 mL (Report your answer as "mean +/- std dev") M What is the percent relative standard deviation? % What is the 99% Confidence Interval for the concentration of the solution (population mean)?