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Experiment 9 Conductometric Titration and Gravimetric Determination of a Precipitate REPORT...
answer the question based on data available for each trial Table I. Gravimetric and Titration Data...
answer the question based on data available for each trial
Table I. Gravimetric and Titration Data for the Determination of Ba(OH). Concentration Trial 1 Trial 2 Trial 3 Volume at Equivalence Point (mL) 7.857 10.867 8.821 Mass of filter paper + precipitate (g) 0.4171 0.8905 0.8724 Mass of filter paper (g) 0.2577 0.6387 0.6294 Mass of precipitate (g) Molarity if HSO (M) 0.100 0.100 0.100 Moles of H.SO, used to reach equivalence point Average M Molarity of Ba(OH) (titration) d...
Name Section Date Lab Partner PRE-LAB ASSIGNMENT: CONDUCTIMETRIC TITRATION AND GRAVIMETRIC DETERM...
Name Section Date Lab Partner PRE-LAB ASSIGNMENT: CONDUCTIMETRIC TITRATION AND GRAVIMETRIC DETERMINATION OF A PRECIPITATE (Show work for All Problems) 1. Define the analyte and titrant in this experiment. How can you differentiate between them? 2. The following data below was gathered during a conductimetric titration involvinga standard solution of 0.100 M sulfuric acid, H.,SO, and 20.0 mL of barium hydroxide, Ba(OUsing LoggerPro, plot these data, perform a linear regression (refer to the Data Analysis section), and answer the questions...
What I have so far for data etc. My question I need help with though is:...
What I have so far for data etc.
My question I need help with though is:
This is my data and what I have so far but this is the
question I actually need help with:
-What ions are present in the titrated solution after the
beginning of the titration but before the equivalence point? How
about after the equivalence point when the conductivity is rising?
Lastly, compare your results from 2 and 4 above with the actual
molarity of...
What is the theoretical yield of H2SO4 + BaCl2 ---> BaSO4 + 2HCl Given data: Concentration...
What is the theoretical yield of H2SO4 + BaCl2 ---> BaSO4 +
2HCl
Given data:
Concentration of BaCl2 = 0.20 M
Concentration of H2SO4 = 6 M
volume of H2S04 = 5.00 mL
What is the theoretical expected yield of precipitate as
calculated using stoichiometry?
Compare actual yield and theoretical yield. Calculate percent
yield.
Data:
Trial 1: Mass of BaSO4 precipitate = 0.191 g
Trial 2: Mass of BaSO4 precipitate = 0.187 g
What is the Millimoles of BaSO4 for...
solve for Ka of acid please Data: Experiment 2; Determination of an Acid lonization Constant A. Determination of the...
solve for Ka of acid please
Data: Experiment 2; Determination of an Acid lonization Constant A. Determination of the Molar Mass of an Unknown Acid Trial 3 Trial 2 Trial 1 400 405 405 Sample Mass 0100m O.Sm) O.1D Initial reading NaOH buret 3H.3m 34ml 34 3ml OH buret Final reading o.0342 0-034 0-0343 Volume NaOH used O.0034 Moles NaOH used TLTOO O.0034 Moles H in sample O.0034 601 220.5glmal 23.lginmal 19.4glmol Unknown Acid Molar Mass (MM) B. Determination of...
Experiment 17A. A Solubility Product Constant Procedure Getting Started 1. Obtain a 10 ml pipet, ...
question#1
Experiment 17A. A Solubility Product Constant Procedure Getting Started 1. Obtain a 10 ml pipet, a 50 mL buret, and 2 pieces of 120 cm filter paper Preparing Saturated Solutions of M10J MIO), is an insoluble divalent iodate salt. The identity of the cation M-is unknown. 1 Prepare MIO,), by adding S0 ml of o.2 M KIO, to 20 mL of 1 Molar M(NO,J, in a 150 mt 2. Stir the mixture vigorously with a stirring rod. A white,...
Experiment 6: The Standardization of a Basic Solution and the Determination of the MM of an Acid
Experiment 6: The Standardization of a Basic Solution and the Determination of the MM of an Acid *Note: This experiment is relatively long unless you know precisely what to do. Read the experiment before coming to class and arrive on time. After reading through the experiment answer the following question: 1. 7.0 mL of 6.0 M NaOH are diluted with water to a volume of 400 mL. You are asked to find the molarity of the resulting solution. a. First find out how many...
Exp 9 titration of vinegar procedure b molarity of NaOH = .0859 ROCEDURE B. DETERMINATION OF...
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
1. What is a “back-titration”? 2. Why is a back titration necessary in this experiment? 3....
1. What is a “back-titration”? 2. Why is a back titration necessary in this experiment? 3. Calculate the number of moles of base equivalents in: a) 675 mg CaCO3 b) 135 mg Mg(OH)2 - Base equivalents = number of moles of acid (H+ ) consumed 4. Calculate the number of moles of acid (H+ ) in 33.6 mL of (a) 0.10 M HCl and (b) 0.10 M H2SO4. 5. Calculate the pH of each of the following solutions: a) 0.10...
EXPERIMENT 9 CHEMICAL REACTIONS: STOICHIOMETRY LABORATORY REPORT Your Name TA's Name You must rea...
EXPERIMENT 9 CHEMICAL REACTIONS: STOICHIOMETRY LABORATORY REPORT Your Name TA's Name You must read pages 69 and 70 to complete the following calculations. Lab Section 1411. Date Trial #1 Trial #2 Trial #3 vol . 0.100 M K2CO, solution -1200 mL 12.00 mL mmole K2CrO vol. 0.100 M Pb(NO,), solution 10.00ml mmole Pb(NO) 12.00m 14.00 mL 20 g mass product+paper 0.579 g0.654 g mass dry filter paper 0266 mass PbCrO, collected precip. 9 mmole Pbcro, CALCULATIONS AND QUESTIONS (USE THE...
answer the question based on data available for each trial
Table I. Gravimetric and Titration Data for the Determination of Ba(OH). Concentration Trial 1 Trial 2 Trial 3 Volume at Equivalence Point (mL) 7.857 10.867 8.821 Mass of filter paper + precipitate (g) 0.4171 0.8905 0.8724 Mass of filter paper (g) 0.2577 0.6387 0.6294 Mass of precipitate (g) Molarity if HSO (M) 0.100 0.100 0.100 Moles of H.SO, used to reach equivalence point Average M Molarity of Ba(OH) (titration) d...
Name Section Date Lab Partner PRE-LAB ASSIGNMENT: CONDUCTIMETRIC TITRATION AND GRAVIMETRIC DETERMINATION OF A PRECIPITATE (Show work for All Problems) 1. Define the analyte and titrant in this experiment. How can you differentiate between them? 2. The following data below was gathered during a conductimetric titration involvinga standard solution of 0.100 M sulfuric acid, H.,SO, and 20.0 mL of barium hydroxide, Ba(OUsing LoggerPro, plot these data, perform a linear regression (refer to the Data Analysis section), and answer the questions...
What I have so far for data etc.
My question I need help with though is:
This is my data and what I have so far but this is the
question I actually need help with:
-What ions are present in the titrated solution after the
beginning of the titration but before the equivalence point? How
about after the equivalence point when the conductivity is rising?
Lastly, compare your results from 2 and 4 above with the actual
molarity of...
What is the theoretical yield of H2SO4 + BaCl2 ---> BaSO4 +
2HCl
Given data:
Concentration of BaCl2 = 0.20 M
Concentration of H2SO4 = 6 M
volume of H2S04 = 5.00 mL
What is the theoretical expected yield of precipitate as
calculated using stoichiometry?
Compare actual yield and theoretical yield. Calculate percent
yield.
Data:
Trial 1: Mass of BaSO4 precipitate = 0.191 g
Trial 2: Mass of BaSO4 precipitate = 0.187 g
What is the Millimoles of BaSO4 for...
solve for Ka of acid please
Data: Experiment 2; Determination of an Acid lonization Constant A. Determination of the Molar Mass of an Unknown Acid Trial 3 Trial 2 Trial 1 400 405 405 Sample Mass 0100m O.Sm) O.1D Initial reading NaOH buret 3H.3m 34ml 34 3ml OH buret Final reading o.0342 0-034 0-0343 Volume NaOH used O.0034 Moles NaOH used TLTOO O.0034 Moles H in sample O.0034 601 220.5glmal 23.lginmal 19.4glmol Unknown Acid Molar Mass (MM) B. Determination of...
question#1
Experiment 17A. A Solubility Product Constant Procedure Getting Started 1. Obtain a 10 ml pipet, a 50 mL buret, and 2 pieces of 120 cm filter paper Preparing Saturated Solutions of M10J MIO), is an insoluble divalent iodate salt. The identity of the cation M-is unknown. 1 Prepare MIO,), by adding S0 ml of o.2 M KIO, to 20 mL of 1 Molar M(NO,J, in a 150 mt 2. Stir the mixture vigorously with a stirring rod. A white,...
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
EXPERIMENT 9 CHEMICAL REACTIONS: STOICHIOMETRY LABORATORY REPORT Your Name TA's Name You must read pages 69 and 70 to complete the following calculations. Lab Section 1411. Date Trial #1 Trial #2 Trial #3 vol . 0.100 M K2CO, solution -1200 mL 12.00 mL mmole K2CrO vol. 0.100 M Pb(NO,), solution 10.00ml mmole Pb(NO) 12.00m 14.00 mL 20 g mass product+paper 0.579 g0.654 g mass dry filter paper 0266 mass PbCrO, collected precip. 9 mmole Pbcro, CALCULATIONS AND QUESTIONS (USE THE...
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