Question

A solution containing 10.00 mL of 0.0500 M metal ion buffered to pH = 10.00 was titrated with 0.0400 M EDTA. Answer the following questions and enter your results with numerical value only.

Calculate the equivalence volume, Ve, in milliliters.

Calculate the concentration (M) of free metal ion at V = 1/2 V_e.

If the formation constant (K_f) is 10^12.00. Calculate the value of the conditional formation constant K_f’ (=α_Y^4- * K_f) and enter your result as scientific notation form.

Calculate the concentration (M) of free metal ion M⁺ at V = V_e. (Use the K_f' calculated above)

Answer 1

At equivalence point number of mmols of EDTA and metal ions are equal and it forms metal-EDTA complex

so mmol = molarity * volume in mL

M1V1 = M2V2

M1 = molarity of metal ion

M2 = molarity of EDTA

V1 = volume of metal ion

V2= volume of EDTA

10*0.05 = V2 * 0.04

hence volume of EDTA at equivalence is 12.5

(ii)

At V = 1/2 V_e.

it is before equivalence point hence metal will be in excess

at 1/2 V_e. V = 6.25 mL

so mmols of EDTA delivered is 6.25 * 0.04 = 0.25

initial moles of metal = 10*0.05 =0.5 mmol

moles of metal left = 0.5 - 0.25 = 0.25

total volume at this point 10+6.25 = 16.25 mL

so concentration or molarity = moles / total volume = 0.25 / 16.25 = 0.0154 M

(iii) the value of α_Y^4- at pH 10 is 0.3 (taken from literature)

so Kf' = 0.3 * 10¹²

Kf' = 3 * 10¹¹

(iv) at equivalence point neither free EDTA nor Metal is left so we are having only M-EDTA complex

so moles of M-EDTA complex formed = 0.5 mmol

so [M-EDTA] = mmols / total volume = 0.5 / (10+12.5) = 0.022 M

so metal ion concentration is 2.69 *10^-7 M