3) Consider the dibasic compound "B". This compound has a pKo 4.00 and a pK,2 8.00...
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH after 3.7 mL of acid are added.
2. The dibasic compound B (pKb 4.00, pKb 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va -0,1, 10, 11,15, 19, 20, and 22 mL.
3. The dibasic compound B (pKbi = 4.00, pK2 = 8.00) was titrated with 1.00 M HCl. The initial solution ofB was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added: V,- 0, 1, and 10 mL. (40 points)
The
dibasic compound B (pKb1 5 4.00, pKb2 5 8.00) was titrated with
1.00 M HCl. The initial solution of B was 0.100 M and had a volume
of 100.0 mL. Find the pH at the following volumes of acid added and
make a graph of pH versus Va: Va 5 0, 1, 5, 9, 10, 11, 15, 19, 20,
and 22 mL.
pKb1= 4.00
pkb2= 8.00
11-23. The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated...
1. What is the definition of an 'equivalence point' in an acid/base titration? (1 point) 2. In part one of the experiment, you will prepare the acid solutions being titrated from a stock solution. Describe how you will accurately prepare 10.00 mL of 0.100 M HCl solution using a 1.00 M HCl stock solution. In your response to this question, be very specific about the quantities of stock solution and deionized water to be used in the dilution and the...
3. If 15.0 mL of 0.125 M phosphoric acid is titrated with 0.100 M NaOH, what volume of the titrant (in mL) must be added to completely neutralize the acid? Show all of your work (including the chemical equation). (1 point) Post-lab Questions: Experiment #9: Acid-Base Titrations Student Learning Objectives : Students will gain practice with the accurate preparation of solutions. Students will perform acid-base titrations and prepare titration curves. Students will identify strong and weak acids by the shapes...
#2 please. I have attached info for part 1 goal
#3
demonstrate the calculations
w OIVO SU UGLUUL iz. Demonstrate through mole calculations that iodate was indeed the limiting reactant in the production of triiodide in Part 1 - Goal 3 (titration) of this study. You have three reactants to consider KI, 103, and H. Refer to the titration for Molarities and quantities used for each reactant. Assume that 10 mL of Klau) contains 1 gram of Klin) Goal #3:...
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 1.40/1.30 28.20 29.10.26.1026.50 Volume of NaOH added Initial buret reading Final buret reading Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision PROCEDURE 1. Obtain approximately 40 mL of an unknown acid in an Erlenmeyer flask from your teaching assistant. 2. Clean your 250-ml volumetric flask (contains exactly 250.00...
Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 11.40/1.30 28.20 29.10.26.1026.50 Initial buret reading Final buret reading Volume of NaOH added Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision PROCEDURE 1. Obtain approximately 40 mL of an unknown acid in an Erlenmeyer flask from your teaching assistant. 2. Clean your 250-ml volumetric flask (contains exactly 250.00...
lab report titration help please!
additional images on lab report
3. Federal standards require that any commercial material called "vinegar" must contain as cast acetic acid. According to your results in question 4, does the vinegar sample you titrated meet the Federal standards? Explain why or why not. PART B: IDENTIFICATION OF AN UNKNOWN ACID DATA Molarity of NaOH used Mass of unknown acid used 0.050M 0.5989 14.43mc Average volume of NaOH used in the titration CALCULATIONS 1. Using the...