From the stock solutions of 1.00-mol L acetic acid and 1.00-molL of sodium acetate, six solutions were prepared with the acetic acid and acetate concentrations shown in the table below. The pk of acetic acid is 4.756. Solutio [Acetic acid). [Acetate). Final volume. mol L mol L ml 0.10 100 0.0000351 100 0.10 0.10 100 0.010 0.010 100 0.10 0.15 100 0.15 0.10 100 a. Determine the volume of the stock solutions of 1.00-mol L'acetic acid and 1.00-mol L of sodium acetate needed to prepare the solutions. Write your answer in the table below. Solution Vol. acetic acid, ml Vol. of acetate, mL b. Calculate for the initial pH of the solutions and write the pH in the table below. Solution Initial pH A. c. Which solution(s) have the same pH?

From the stock solutions of 1.00-mol L acetic acid and 1.00-molL of sodium acetate, six solutions were prepared with the acetic acid and acetate concentrations shown in the table below. The pk of acetic acid is 4.756. Solutio [Acetic acid). [Acetate). Final volume. mol L mol L ml 0.10 100 0.0000351 100 0.10 0.10 100 0.010 0.010 100 0.10 0.15 100 0.15 0.10 100 a. Determine the volume of the stock solutions of 1.00-mol L'acetic acid and 1.00-mol L of sodium acetate needed to prepare the solutions. Write your answer in the table below. Solution Vol. acetic acid, ml Vol. of acetate, mL b. Calculate for the initial pH of the solutions and write the pH in the table below. Solution Initial pH A. c. Which solution(s) have the same pH?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.26QAP

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If 500mL of 0.10M Ca2+ is mixed with 500mL of 0.10M SO42-, what mass of calcium sulfate will precipitate? Ksp for CaSO4 is 2.40 • 10-5. express answer to 3 significant figures. And include units.
By the use of Henderson Hasselbalch equation; pH = pKa + log{[acetate ion]/[acetic acid]} 4.5 = 4.75 + log{[0.10 M]/[acetic acid]} -0.25 = log{[0.10 M]/[acetic acid]} [Acetic acid] = 0.10 M/ 10-0.25 [Acetic acid] = 0.10 M/0.56 [Acetic acid] = 0.1786 M Moles of sodium acetate dissolved in 250 mL buffer solution = 0.10 M× (250mL/1000mL) × 1L = 0.025 mol Weight (w) of sodium acetate (purity 100%) dissolved to prepare 250 mL of solution with buffer concentration of 0.10 M is calculate as follow; w100% = 0.025 mol × 82.0343 g/mol = 2.051 g Weight (w) of sodium acetate (purity 99%) is calculate as follow; w99% = 2.051 g× (100/99) = 2.072 g What was the volume of 6.12 M acetic acid HC2H3O2 needed to prepare the 250 mL acetic acid/acetate ion buffer solution required in this part? Show your calculations.
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