Standardization of NaOH against KHP Example 0.2500 II Wt. of KHP, g Final reading, mL NaOH Initial Reading, mL NAOH Volume, mL NaOH 12.15 0.00 12.15 Molar concentration of NaOH 0.1008 Normal Concentration of NaOH 0.1008 Determination of Concentration of HCI by Titration with Standardized NaOH| Example 25.00 Volume of HCI used, mL Final reading, mL NaOH Initial Reading, mL NaOH Volume of NaOH, mL NaOH 37.00 12.15 24.85 Molar concentration of HCI 0.1002 Normal Concentration of HCI 0.1002

Standardization of NaOH against KHP Example 0.2500 II Wt. of KHP, g Final reading, mL NaOH Initial Reading, mL NAOH Volume, mL NaOH 12.15 0.00 12.15 Molar concentration of NaOH 0.1008 Normal Concentration of NaOH 0.1008 Determination of Concentration of HCI by Titration with Standardized NaOH| Example 25.00 Volume of HCI used, mL Final reading, mL NaOH Initial Reading, mL NaOH Volume of NaOH, mL NaOH 37.00 12.15 24.85 Molar concentration of HCI 0.1002 Normal Concentration of HCI 0.1002

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

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Na2CO3 served as the primary standard in a titration experiment. Find the molarity of the titrant given the following data in 3 decimal places. Show solutions Primary Standard Used: Na2CO3Formula Mass of 1º standard: 105.99 g/mol% purity of 1º standard: 95% Trial 1 2 3 1º Standard weight, g 0.1005 0.1001 0.0997 Net volume of HCl, mL 9.30 9.00 8.90 Molarity of HCl X1 X2 X3
Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution.MnO4- + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2(g) + 8H2OCalculate the molar concentration of the KMnO4 solution.
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)?
Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution. 2MnO4- + 5C2O4-2 +16H+ -> 2Mn2+ + 10CO2(g) + 8H2O Calculate the molar concentration of the KMnO4 solution. Pls show step by step. What type of molecules will interfere in this analysis ?
11. A 0.485 g sample containing chloride was titrated with 36.8 mL of 0.1060 M AgNO3 stock solution. Express the results of this analysis in:a) Percent by weight of Clb) Weight percent Cl2c) Percent by weight of CaCl2d) Weight percent of AlC13
PART 2: A solution is prepared by mixing 150.0 mL of 0.0010 M magnesium nitrate and 250.0 mL of 0.100 M sodium fluoride (Ksp = 7.4x10-11). Is Ksp greater than , equal to, or lower than Qsp? Will a precipitate form, Y/N? What is the fluoride ion concentration (in solution) after mixing is complete? Give your answer in exponential form and to three sig. figs. (e. g. 1.23E-3). What is the magnesium ion concentration (in solution) after mixing is complete? Give your answer in exponential form and to three sig. figs. (e. g. 1.23E-3).
2) A solution of NaOH was standardized by titration of a known quantity of the primary standard potassium hydrogen phthalate (KHP, FM 204.221 g/mol). Titration of 0.824 g of potassium hydrogen phthalate required 38.314 mL of NaOH to reach the endpoint detected by phenolphthalein. Find the concentration of NaOH in molarity (M).
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 primary standard of (COOK)2.H2O was prepared up to 0.05M. This primary standard was then titred against an unknown concentration of 50cm3 of KMnO4. An average titre value of 13 was obtained. The balanced equation is as follows: 5(COOK)2.H2O + 2KMnO4 + 8H2SO4 = 6K2SO4 +2MnSO4 + 10CO2 +18H2O. Find the concentration of the KMnO4.
please stop rejecting. we pay for this service, so we deserve getting our answer! I do not want to have to put a claim in and report Use acid-base titration to determine theconcentration of:– A strong acid: HCl 0.100 M NaOH (standardized)•0.100 M HCl0.100 M CH3COOH Phenolphthalein is used for endpoint determination– Changes from colourless to pink as a solution becomes more basic• Only need 2-3 drops per titration• pH range: 8.3-10
Part A: Standardization of a Sodium Hydroxide Solution Titration 1 Titration 2 Titration 3 Mass of 125 mL flask 45.849g 46.715g 44.953g Mass of flask and KHP 46.849g 47.745g 46.003g Initial buret reading (mL) 0.5 ml 0.5 ml 0.5 ml Final buret reading (mL) 27.8 ml 26.5 ml 26.7 ml Volume of NaOH used (mL) 45.11 ml 45.06 ml 45.14 ml Calculations Titration 1 Titration 2 Titration 3 Moles of KHP Moles of NaOH Molarity of NaOH Average Molarity of NaOH: _______________
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. 20.340mg