3. Titration: Strong Acid / Weak Base. (I warned you to be ready!) Here's the cool chart!Analyte: 20.00 mL 0.100 M Hydrazine (N₂H4)HdA: initialVertical (Value) Axis Title0B: half equivalenceC: equivalence5101520Titrant: mL of 0.100 M HCI added2520.00 mL of 0.100 M hydrazine (N₂H4, pKb = 5.77) is titrated with 0.100 M HCl.a) Find the initial pH (point A) and the pH at the equivalence point (point C).b) Find the pH at the half equivalence point (point B).

The objective of this question is to calculate the pH at different points during the titration of a…

Answered: 3. Titration: Strong Acid / Weak Base.…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.28QAP

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8-49 A solution of NaOH base was titrated with 0.150 M HCI, and 22.0 mL of acid was needed to reach the end point of the titration. How many moles of the unknown base were in the solution?
Quantity of Acetic Acid in Salad Dressing Titration Results: 1000 μL aliquots of salad dressing against a sodium hydroxide solution Please only fill in 3 concordant results! Titration number: 1 2 3 Error Final burette reading (mL) 7.9 15 22.2 ±0.05 mL Initial burette reading (mL) 0.9 7.9 17 ±0.05 mL Titre (mL) 7 7.1 7.2 ±0.10 mL Average (mean) volume of concordant titres: 7.1 Concentration of Acetic Acid (M) (to 3 sig figs): 6.88 mL IM Concentration of Acetic Acid (as %w/v) (to 3 sig figs): % Note: %w/v = (g/mL) x 100%
TITRATION First: 10 mL - unknown acid + 0.15 Phenolphthalein solution = 10.150 mL Next: 10.150 mL + 0.77 mL of 10.0 M NaOH = 10.92 mL unknown acid @ 25.0 degrees celsius What is the concentration (M) of the acid (Please use 3 significant figures) = ???
A 0.9600 M NaOH solution was diluted with a factor of 4.Then the diluted solution of NaOH was titrated against an unknown acid.The volumes of NaOH(aq) titrated are shown in the table below.Volumes of NaOH(aq):RoughtitrationFirsttitrationSecondtitrationThirdtitrationThirdtitrationFinal burette reading:(mL)19.00 34.05 20.86 36.01 23.57Initial burette reading:(mL)19.00 19.07 5.76 21.02 8.54The molarity of the unknown acid is 0.08108 M and the volume of the unknown acid used ineach titration is 14.80 mL.(i) Was the unknown acid monoprotic, diprotic or triprotic? Show your calculations andreasoning.(ii) What type of chemical reaction is this? _________________ or _______________
What is Analyte and Titrant? Explain each in 3-5 sentences.
2 Match the following acid-base titration plots with the correct titrant and sample. Veq = 20.00 mL mea a. b. Hd (1) Titrant: Strong Acid; Sample: Strong Base (2) Titrant: Strong Base; Sample: Strong Acid (3) Titrant: Strong Acid; Sample: Weak Base (4) Titrant: Strong Base; Sample: Weak Acid 3219 11 10 9 8 7 6 5 4 3 2 1 0 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 0 5 5 10 10 15 Volume 15 Volume 20 20 25 25 30 30
EXPERIMENT 5; Acid-Base Titration A. Trial 1 Trial 2 Trial 3 5.210 g Mass of KHP 5.210 g 5.210 g (C3H,O4) Mole of KHP Initial Burette 0.00 0.50 5.00 Volume NaOH (mL) Final Burette 31.00 26.25 29.95 Volume NaOH (mL) Total Volume of NaOH (mL) Moles of NaOH Molarity of NaOH (mol/L) Calculate the average Molarity of NaOH B. Titration of Vinegar (HC,H3O2) A 2.352 g sample of Vinegar is titrated with 0.08751M NaOH, and it requires 22.31 mL of NaOH to reach the endpoint. Calculate the mass percent of acetic acid (HC,H;O,) in the vinegar sample? Hint for Calculation of Mass percent of Vinegar (HC,H;O2) 1. Calculate mole of HC,H;O2 from concentration (M) and volume of NaOH (L) needed to titrate Vinegar 2. Calculate mass (g) of Vinegar from mole of Vinegar 3. Mass % Vinegar = (calculated mass Vinegar/ 2.352 g) X 100 C. A solution of malonic acid, H;C;H¿O4 ¸was standardized by titration with 0.100 M NaOH solution. If 20.76 mL of the NaOH solution were required to neutralize completely 13.15 mL of…
Vinegar? required to neutralize the acetic acid in each sample of vinegar, 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. Convert the average volume of NaOH required to reach endpoint from mL to L Average volume of NaOH required to reach endpoint in mL 15. Subtract the initial volume from the final volume to calculate the volume of NaOH I DATA 269 12.55 TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH MI TRIAL 1 TRIAL 2 Initial base buret reading Final base buret reading Volume of sodium hydroxide (NaOH) required to reach endpoint in mL Volume of vinegar used O15 12.55 TRIAL 3 12:55 24.11 EXTRA TRIAL 24.11 35.39 11.78 12.05 11.56 15.00 mL 15.00 mL 15.00 mL 15.00 mL (Show calculations here.) 12.05711-56+1178 -27.54 27.54ML (Show calculations here.) a7.54ML X IL 0-02754L :0.02754L nd bon 1600 Molarity of NAOH (recorded from the standard solution bottle) 0.1 I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of…
You collected the following data from a titration experiment using a 0.123M standardized NaOH solution to titrate a 27.36 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4). Initial Final Vol Acid Burette Burette Delivered Concentration Reading (mL) Reading (mL) (mL) (M) Trial 1 0.56 19.20 ?? Trial 2 0.57 19.7 Trial 3 0.92 19.56 For just Trial 1, determine the Molarity concentration of the H2SO4 to 3 significant digits. Do not include units. Answer:
Given for nos. 10-11: In an acid-base titration, you were given with a 0.1050 M sodium hydroxide solution placed in a buret and is needed to neutralize 20.0 mL of 0.1500 M sulfuric acid. 10. How much moles of sulfuric acid is found in the 20.0 mL solution of a 0.15M H2SO4? Answer: Computation: 11. What volume of 0.1050 M sodium hydroxide solution is needed to neutralize 20.0 mL of 0.1500 M sulfuric acid? Answer: Computation: 12. Draw and label a complete laboratory set-up of acid-base titration using phenolphthalein as the indicator, using HCl as the analyte/sample, and using NaOH as the standard solution. Cite complete source/(s) if picture will be copied from the internet.
Calculate the percentage CH3COOH in a sample of vinegar from the following data.Sample = 15.00 g, NaOH used = 43.00 ml ; 0.600 N H2SO4 used for back titration =0.250 ml ; 1.00ml NaOH is equivalent to 0.0315 g H2C2O4. 2H2O.
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: _______________

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