A solution containing 40.00 mL of 0.0500 M metal ion buffered to pH = 12.00 was...

A solution containing 40.00 mL of 0.0500 M metal ion buffered to pH = 12.00 was titrated with 0.0400 M EDTA. Calculate the fraction (αY4-) of free EDTA in the form Y4−. Keep 2 significant figures.

Expert Solution

queen_honey_blossom answered 2 years ago

A solution containing 45.00 ml of 0.0500 M metal ion buffered to pH = 10.00 was...

A solution containing 45.00 ml of 0.0500 M metal ion buffered to pH = 10.00 was titrated with 0.0400 M EDTA. Answer the following questions and enter your results with numerical value only. Calculate the equivalence volume, Ve, in milliliters. Calculate the concentration (M) of free metal ion at V = 1/2 Ve. Calculate the fraction (αY4-) of free EDTA in the form Y4-. Keep 2 significant figures. If the formation constant (Kf) is 1012.00. Calculate the value of the...

A solution containing 45.00 mL of 0.0500 M metal ion buffered to pH = 11.00 was...

A solution containing 45.00 mL of 0.0500 M metal ion buffered to pH = 11.00 was titrated with 0.0400 M EDTA. Answer the following questions and enter your results with numerical value only. Calculate the equivalence volume, Ve, in milliliters. 56.2 You are correct. Your receipt no. is 157-1182 Help: Receipt Previous Tries Calculate the concentration (M) of free metal ion at V = 1/2 Ve. 0.0154 You are correct. Your receipt no. is 157-4788 Help: Receipt Previous Tries Calculate...

A solution containing 10.00 mL of 0.0500 M metal ion buffered to pH = 10.00 was...

A solution containing 10.00 mL of 0.0500 M metal ion buffered to pH = 10.00 was titrated with 0.0400 M EDTA. Calculate the fraction (αY4-) of free EDTA in the form Y4−. Keep 2 significant figures.

What is the pH of your solution after titrating with 12.00mL; 0.0500 M NaOH? 6.00 mL...

What is the pH of your solution after titrating with 12.00mL; 0.0500 M NaOH? 6.00 mL of 0.100 M butanoic acid; Ka = 1.52 x 10-5

A 50.0 mL sample containing Cd2 and Mn2 was treated with 40.4 mL of 0.0500 M...

A 50.0 mL sample containing Cd2 and Mn2 was treated with 40.4 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 19.2 mL of 0.0300 M Ca2 . The Cd2 was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 18.2 mL of 0.0300 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?

A biochemist has 100 mL of a buffered solution at pH 7.3. The concentration of the...

A biochemist has 100 mL of a buffered solution at pH 7.3. The concentration of the buffer is 0.1 M, and the pKa of the buffer is 7.8. If the biochemist adds 80 mL of a 0.1 N KOH solution to the above buffered solution, what will be the final pH to the nearest tenth of a unit?

A 10.00 mL solution of 0.0500 M AgNO3 was titrated with 0.0250 M NaBr in the...

A 10.00 mL solution of 0.0500 M AgNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E.(saturated calomel electrode) titration solution Ag (s). (Ksp AgBr(s) = 5.0e-13) Find the cell voltage (v) when the volume of titrant added is 5.00 ml. Find the equivalence volume (ml) of titrant added. Calculate the cell voltage (v) when the volume of titrant added is 24.58 ml Calculate the cell voltage (v) at the equivalence point. .

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium...

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium chloride (NaCl) and 0.05000 M potassium iodide (KI), what is the pAg of the solution after 15.00 mL of AgNO3 is added to the solution? Ksp, AgCl (s) = 1.82 x 10-10; Ksp, AgI(s) = 8.3*10-17. b. Same titration as in (a), what is the pAg of the solution after 25.00 mL of AgNO3 is added to the above solution? c. Same titration as...

Calculate the hydroxide ion concentration in a 0.0500 M NaOCl solution. The acid dissociation constant for...

Calculate the hydroxide ion concentration in a 0.0500 M NaOCl solution. The acid dissociation constant for HOCl is 3.0 x 10-8.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca2 (aq) and 0.0350 M...

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca2 (aq) and 0.0350 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate? Solubility-product constants, Ksp, can be found here. 1. Calculate [Ca^2+] = _____M 2. What percentage of the Ca2+ (aq) can be precipitated from the Ag+ (aq) by selective precipitation? ______%

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