ion of 0.1789 M phosphoric acid is used to standardize an unknown solution of sodium hydroxide according to the following equation: H3PO4 + 3 NAOH → NazPO4 + 3 H20 erforming a quick titration, the following data was collected: Trial 1 Trial 2 Initial volume H PO4 (mL) 0.13 8.75 Final volume H PO4 (mL) 15.73 24.08 Initial volume NaOH (mL) 11.40 0.08 Final volume NaOH (mL) 33.06 21.38 s the concentration of the sodium hydroxide in trial 1? Enter your numerical answer with a leading zero and appropriate sig figs. Do not enter a unit. s the concentration of the sodium hydroxide in trial 2? Enter your numerical answer with a leading zero and appropriate sig figs. Do not enter a unit. s the average concentration of the sodium hydroxide? Enter your numerical answer with a leading zero and appropriate sig figs. Do not enter a unit. s the unit for all of these concentrations?

ion of 0.1789 M phosphoric acid is used to standardize an unknown solution of sodium hydroxide according to the following equation: H3PO4 + 3 NAOH → NazPO4 + 3 H20 erforming a quick titration, the following data was collected: Trial 1 Trial 2 Initial volume H PO4 (mL) 0.13 8.75 Final volume H PO4 (mL) 15.73 24.08 Initial volume NaOH (mL) 11.40 0.08 Final volume NaOH (mL) 33.06 21.38 s the concentration of the sodium hydroxide in trial 1? Enter your numerical answer with a leading zero and appropriate sig figs. Do not enter a unit. s the concentration of the sodium hydroxide in trial 2? Enter your numerical answer with a leading zero and appropriate sig figs. Do not enter a unit. s the average concentration of the sodium hydroxide? Enter your numerical answer with a leading zero and appropriate sig figs. Do not enter a unit. s the unit for all of these concentrations?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Please help me fill out this table! Answering just one of the columns is good enough, thank you! Concentration of Standardized NaOH - 1.00 mol/L Mass concentration of acetic acid - 2.40 g/L Trial 1 Trial 2 Measured pH of the acetic acid solution 3.07 3.05 Mass of acetic acid solution taken for titration 20.0 g 20.0 g Initial buret reading of NaOH titrant 0.00 mL 0.00 mL Final buret reading of NaOH titrant 0.82 mL 0.79 mL Net volume of NaOH 0.82 mL 0.79 mL Millimoles of NaOH to end point of titration _?_mmol _____mmol Millimoles of acetic acid in sample _?_mmol ______mmol Molar concentration of acetic acid solution _?_mol/L ______mol/L
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…
An automatic titration of Cola product was analyzed and the following results were produced: Volume of Standardized NaOH Titrant used to achieve the first equivalence point in titration: Trial #1: 1.300 mL Trial #2: 1.137 mL Trial #3: 1.140 mL May you explain why trial #1 was skewed; for example, if pipetting of the 25.00 mL was done incorrectly, how would the molarity change if the volume examined was supposed to be 24.50 mL?
10.Calculate the pH of the solution prepared by dissolving 0.48 mol of NH3(Kb=1.77x10-5) and 0.750 mol of ammonium chloride in enough water to make 1.00L of buffer solution Group of answer choices A, 5.09 B, 9.05 C, 9.44 D, no correct answer E, 10.1 11,The equivalence point of titrating of the weak base with the strong acid will be observed at Group of answer choices A, pH>7 B, pH<7 C, pH=pKb D, pH>pKb E, pH=7 12,The titration curve of titrating a diprotic acid with a strong base will have two distinct equivalence points Group of answer choices A, if Ka1<<Ka2 B, if Ka1=Ka2 C, always D, in case of ta strong acid E, if Ka1>>Ka2
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 BS Medical Technology student was asked to determine the molecular weight of a diprotic acid using acid-base titration. This student performed fourtrials. Firstly, the student used potassium hydrogen phthalate (KHP, 204.22 g/mol) as standard to determine the exact concentration of the NaOHtitrant. The results of the student are summarized in the table below:Note: The stoichiometric relationship of KHP to NaOH is 1:1Standardization Trial 1 Trial 2 Trial 3 Trial 4KHP Mass, g 0.5033 0.5066 0.6989 0.6843Volume NaOH used, mL 24.32 25.61 24.67 24.56After the standardization, the student weighed different amounts of the unknown acid. In a similar fashion, the student performed acid-base titrationusing phenolphthalein as indicator. The results of the molecular weight determination are summarized below:Note: The stoichiometric relationship of Unknown acid to NaOH is 1:2Molecular WeightDeterminationTrial 1 Trial 2 Trial 3 Trial 4Unknown Mass, g 0.1234 0.1034 0.1178 0.1322Volume NaOH used, mL…
A BS Medical Technology student was asked to determine the molecular weight of a diprotic acid using acid-base titration. This student performed fourtrials. Firstly, the student used potassium hydrogen phthalate (KHP, 204.22 g/mol) as standard to determine the exact concentration of the NaOHtitrant. The results of the student are summarized in the table below:Note: The stoichiometric relationship of KHP to NaOH is 1:1Standardization Trial 1 Trial 2 Trial 3 Trial 4KHP Mass, g 0.5033 0.5066 0.6989 0.6843Volume NaOH used, mL 24.32 25.61 24.67 24.56After the standardization, the student weighed different amounts of the unknown acid. In a similar fashion, the student performed acid-base titrationusing phenolphthalein as indicator. The results of the molecular weight determination are summarized below:Note: The stoichiometric relationship of Unknown acid to NaOH is 1:2Molecular WeightDeterminationTrial 1 Trial 2 Trial 3 Trial 4Unknown Mass, g 0.1234 0.1034 0.1178 0.1322Volume NaOH used, mL…
A BS Medical Technology student was asked to determine the molecular weight of a diprotic acid using acid-base titration. This student performed fourtrials. Firstly, the student used potassium hydrogen phthalate (KHP, 204.22 g/mol) as standard to determine the exact concentration of the NaOHtitrant. The results of the student are summarized in the table below:Note: The stoichiometric relationship of KHP to NaOH is 1:1Standardization Trial 1 Trial 2 Trial 3 Trial 4KHP Mass, g 0.5033 0.5066 0.6989 0.6843Volume NaOH used, mL 24.32 25.61 24.67 24.56After the standardization, the student weighed different amounts of the unknown acid. In a similar fashion, the student performed acid-base titrationusing phenolphthalein as indicator. The results of the molecular weight determination are summarized below:Note: The stoichiometric relationship of Unknown acid to NaOH is 1:2Molecular WeightDeterminationTrial 1 Trial 2 Trial 3 Trial 4Unknown Mass, g 0.1234 0.1034 0.1178 0.1322Volume NaOH used, mL…
The experimental Ksp of calcium hydroxide is _________ x 10-6 if 23.09mL of the sat. calcium hydroxide solution was titrated with 12.53mL of 0.049M HCl(aq), considering 4 as an exact value for the calculation?
M of NaOH: 0.100 Volume of HCI 25mL = 0.025L Moles of HCI = 2.5 mol What is the Volume of NaOH equivalence point: What is Moles of NaOH equaivalence point: What is pH at equivalence point? What is calculated pH and experimental pH of NaOH at 0.00 mL , 12.50 mL, 25mL, and 30 mL. Also what is the Error percentage for each one?
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.
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