Question

9 Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. ?Ba2+ = 1.48 × 10^-4. ?Y4- = 0.30 for pH 10. Kf = 7.59 × 10^7 for BaY2-.

Name: 9 Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. ?Ba2+ = 1.48 × 10^-4. ?Y4- = 0.30 for pH 10. Kf = 7.59 × 10^7 for BaY2-.
Uploaded: 2023-11-13T11:59:09-08:00
Duration: 2 min 56 s
Channel: Anand Jangid
Description: 9 Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. ?Ba2+ = 1.48 × 10^-4. ?Y4- = 0.30 for pH 10. Kf = 7.59 × 10^7 for BaY2-.

          9 Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. ?Ba2+ = 1.48 × 10^-4. ?Y4- = 0.30 for pH 10. Kf = 7.59 × 10^7 for BaY2-.

9 Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. ?Ba2+ = 1.48 × 10^-4. ?Y4- = 0.30 for pH 10. Kf = 7.59 × 10^7 for BaY2-.

Added by Christian I.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Anand Jangid

11/13/2023

Step 1

First, we need to find the moles of Ba²⁺ and EDTA in the solution. Moles of Ba²⁺ = 0.1 M * 0.050 L = 0.005 mol Moles of EDTA = 0.1 M * 0.035 L = 0.0035 mol Show more…

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Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. αBa2+ = 1.48 × 10^-4. αY4- = 0.30 for pH 10. Kf = 7.59 × 10^7 for BaY2-.