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.

Expert Solution

queen_honey_blossom answered 1 year 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 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.

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 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. .

Consider titration of 10.00 mL of 0.120 M Cu2+ solution with 0.100 M EDTA at pH...

Consider titration of 10.00 mL of 0.120 M Cu2+ solution with 0.100 M EDTA at pH 10.0. Find the concentration of free Cu2+ at the equivalence point of this titration.

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 10.00 mL sample of a solution containing Fe3+ and Co2+ are added to 100 mL...

A 10.00 mL sample of a solution containing Fe3+ and Co2+ are added to 100 mL of pH 4.5 buffer and then titrated to a Cu-PAN endpoint with 17.62 mL of 0.05106 M EDTA. A 25.00 mL sample of the same solution are added to 100 mL of pH 4.5 buffer and then 10 mL of 1 M potassium fluoride are added to mask the Fe3+. This solution was titrated to a Cu-PAN endpoint with 12.44 mL of 0.05106 M...

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?

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...

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