Question

Calculate [Mn²⁺] when 50.00 mL of 0.1000 M Mn²⁺ is titrated with 17.00 mL of 0.2000 M EDTA. The titration is buffered to pH 11 for which = 0.81. K_f = 7.76 × 10¹³

Answer 1

At above pH = 10 the EDTA exists as mainly Y^4-

K' = Kf X α

K' = 0.81 X 7.76 X 10^13 = 6.29 X 10^13

Moles of Mn+2 = Molarity X volume X 10^-3 =0.1 x 50 X 10^-3 = 5 millimoles

Moles of EDTA = molarity X volume X 10^-3 = 0.2 X 17 X 10^-3 = 3.4 millimoles

The reaction will be

                Mn²⁺+ Y^4- --> MnY^2-

the concentration of Mn+2 left = Moles of Mn+2 - Moles of EDT / Total volume = 5-3.4 / 50+ 17 = 0.024 M