Question

A back-titration was peformed as follows: In a solution with an unknown concentration of nickel, 10 ml of 1.29 x 10^-2 M of EDTA were added. This analyte was titrated with 9.79 x 10^3 M of Zn. 7.2mL of Zn was used. Determine the concentration of Nickel in the analyte.

Answer 1

Zn^+2(aq) + EDTA^-4(aq) ---> Zn(EDTA)^-2

1 mol Zn+2 = 1 mol EDTA^-2

no of mol of Zn+2 REACTED = M*V

                          = (9.79*10^-3)*7.2

                          = 0.0705 mmol

no of mol of EDTA-4 reacted with Zn+2 = 0.0705 mmol

TOTAL NO OF of EDTA-4 TAKEN = M*V

                            = (1.29*10^-2)*10

                           = 0.129 mmol

Ni^+2(aq) + EDTA^-4(aq) ---> Ni(EDTA)^-2

no of mol of EDTA-4 reacted with Ni+2 = 0.129-0.0705 = 0.0585 mmol

no of mol of Ni+2 present in sample = 0.0585 mmol