Experiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Trial 3 Rough trial Trial 2 Trial 1 Initial reading (mL) o.00mL 21.0SML O .0om o.00ML Final reading (mL) 43.15 mL 2l.6om L z1.45ML Z1.50ML Volume dispensed (mL) 21.55ML 21.Lo5mL 21.6OnL 21.SOML 21.OomL Average volume of NAOH used (Trials 1-3) O.2012 M Molarity of NaOH MOLES MOLARITY N A Unknown Vinegar # LITERS Calculation of moles HC2H302 in 5.00 mL vinegar: 21.6om L NaoH Moles of cetie acid

Experiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Trial 3 Rough trial Trial 2 Trial 1 Initial reading (mL) o.00mL 21.0SML O .0om o.00ML Final reading (mL) 43.15 mL 2l.6om L z1.45ML Z1.50ML Volume dispensed (mL) 21.55ML 21.Lo5mL 21.6OnL 21.SOML 21.OomL Average volume of NAOH used (Trials 1-3) O.2012 M Molarity of NaOH MOLES MOLARITY N A Unknown Vinegar # LITERS Calculation of moles HC2H302 in 5.00 mL vinegar: 21.6om L NaoH Moles of cetie acid

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter15: Complex Acid/base Systems

Section: Chapter Questions

Problem 15.29QAP

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In the experiment, we used titration to determine the total acid content ofsamples, which we reported in terms of molarity. Reporting acidity through pHmeasurements is quite different, in that we can only measure the amount of theacid in its ionized form. You were tasked to investigate a clear aqueous solutionof an unknown monoprotic acid. You decided to use two methods togather data.Method 1 – TITRATION: A 10. mL aliquot of the sample was diluted with 25 mLdistilled water. Two drops of phenolphthalein were added and then it was titrated3.54 mL of 0.048 M standardized NaOH to the endpoint.Method 2 – pH STRIP: You took 1 mL of the sample and used a pH strip toestimate the pH, which turned out to be around 3.3. Another 1 mL of the samplewas diluted with 9 mL of water. The pH was taken again and is now around 3.8.a) Calculate the molarity of the acid using the titration data.b) If we assume that the titrated unknown is a strong acid, predict thepH of the sample.c) Using pH strip results,…
Prepared buffer solution GivenAmmonia Volume = 68 mL | Ammonia Concentration = 0.17 MAmmonium Chloride Volume = 42 mL | Ammonium Chloride Concentration = 0.13 MDissociation Constant of Ammonia: 1.8x10-5What is the total vol. of buffer
20.0 mL of 1:1 buffer 5.00 mL of standard NaOH pKa=4.66 [NaOH]=0.1108M pH=5.19
Titration of unknown acid with NaOH. Determine the pKa of unknown acid: mass H2A=0.5 H2O=50 mL Volume @ eq point: 23.7 mL NaOH= 0.2084M pH @ Veq = 7.9 H2A+H2O <---->[HA-]+[H3O+]
Let's determine the concentration and the pH of a H2CO3 solution that was prepared by using standard NaOH solution with a concentration of 0.125 M using the following data. Trial 1 Trial 2 Trial 3 Volume of H2CO3 19.80 mL 20.05 mL 19.95 mL Initial volume of NaOH (mL) 1.90 16.88 31.98 Final volume of NaOH (mL) 16.88 31.98 47.00 What is the Molarity and the pH of the H2CO3 ?
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
Question: pH pf 50.0 mL H20 + 1.00 mL 0.6 M HCl (below is the data to answer the answer) Deionized water Acetate Buffer Ammonia Buffer pH of solution 6.62 4.59 9.49 Ph of solution after addition of 1 mL of 0.6 M NaOH 12.41 4.71 9.50 pH of solution after addition of 1 mL of 0.6 M HCl 1.38 3.87 9.24 Preparation of Buffer solutions Mass of sodium acetate in acetate buffer ~ 1.1981 Volume of 3.0 M acetic acid in acetate buffrer ~ 3.81 Mass of ammonium chloride ~ 1.2721 Volume of 5.0 M NH4OH ~ 6.00
Why is ethylene-di-amine tetra-acetic acid (EDTA) work as a base in complexometric titration? answer at your own words and answer should be to the point, not any irrelevant words
The data from my observations is given below: 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 0.82 mmol 0.79 mmol Millimoles of acetic acid in sample 0.82 mmol 0.79 mmol Molar concentration of acetic acid solution 0.043 mol/L 0.041 mol/L Calculated molar mass of acetic acid g/mol g/mol Average molar concentration of acetic acid for the samples titrated: 0.042 mol/L Average pH: 3.06 Question: Calculate the corresponding [H3O+], compute the corresponding concentration of A- and HA, and calculate the dissociation constant Ka Thank you!
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Consider how best to prepare one liter of a buffer solution with pH = 3.79 using one of the weak acid/conjugate base systems shown here. Weak Acid Conjugate Base Ka pKa HC2O4- C2O42- 6.4 x 10-5 4.19 H2PO4- HPO42- 6.2 x 10-8 7.21 HCO3- CO32- 4.8 x 10-11 10.32 How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?grams sodium salt of weak acid = grams sodium salt of conjugate base =
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?