Question

In this example, we are going to focus on the chelation of free Mn²⁺ ions in a solution at pH = 7.

(a) If there are 20.0 L of 0.00025 M Mn²⁺ in the solution before the addition of EDTA, how much EDTA is needed to complex all of the Mn²⁺ in moles?

(b) You have a 0.050 M solution of EDTA, how many Liters of the solution is needed to reach the equivalence point for the solution of Mn²⁺ described in part a?

(c) What is the concentration (M) of Mn²⁺ ions at the equivalence point? (Give answer in scientific notation using the format #*10^#)  

(d) If 0.20 L of EDTA is added (in total), what will the concentration of free Mn²⁺ be? (Give answer in scientific notation using the format #*10^#)

(e) How much total EDTA would be needed to reduce the concentration of Mn²⁺ to less than 5*10^‑12 M? Give this answer in Liters of 0.050 M EDTA to 1 significant figure.

Answer 1

(a) The moles of EDTA needed = 20 L * 0.00025 mol/L = 0.005 mol

(b) The Liters of the EDTA solution is needed to reach the equivalence = 0.005 mol/(0.05 mol/L) = 0.1 L

(c) Concentration (M) of Mn²⁺ ions at the equivalence point = 2.5*10^-4 M

(d) The concentration of free Mn²⁺ = 2*2.5*10^-4 M = 5*10^-4 M

(e) Total EDTA would be needed to reduce the concentration of Mn²⁺ = 0.005 mol/(5*10^-12 mol/L) = 1*10⁹ L