Question

Calculate [Ba2+] when 35.00 mL of 0.2000 M EDTA is added to 50.00 mL of 0.1000 M Ba2+ in the presence of 0.1 M nitrilotriacetate. = 1.48 x 10−4. = 0.30 for pH 10. Kf = 7.59 × 107 for BaY2−.

Answer 1

Kf = 7.59 × 10^7

= 0.30

Kf¹ = Kf x = 7.59 × 10^7 x 0.30

      = 2.277 x 10^7

millimoles of Ba+2 = 50 x 0.1 = 5

millimoles of EDTA = 35 x 0.2 = 7

molarity of [BaY^2-] = 5 / 50 + 35 = 0.0588 M

So, 5 mmols of MnCl2 will reacts with EDTA

remaining moles of EDTA = 2 mmol

Total volume of solution = 50 + 35 = 85

molarity of [EDTA] = 2 /85 = 0.0235 M

Ba+2    +   EDTA   --------------> BaY2-

                0.0235                       0.0588

Kf^1 = [BaY^2-]/[Ba2+][EDTA]

2.277 x 10^7 = 0.0588 / [Ba+2][0.0235]

[Ba+2] = 1.099 x 10^-7 M