Question 14.E.B: The apparatus in the figure can follow the course of an EDTA...
The apparatus in the figure can follow the course of an EDTA titration and was used to generate the curves in Figure 11-10. The heart of the cell is a pool of liquid Hg in contact with the solution and with a Pt wire. A small amount of HgY^{2−} added to the analyte equilibrates with a very tiny amount of Hg^{2+} :
Hg^{2+} + Y^{4−} \rightleftharpoons HgY^{2−}
K_{f} = \frac{[HgY^{2−}]}{[Hg^{2+}][Y^{4−}]} = 10^{21.5} (A)
The redox equilibrium Hg^{2+} + 2e^{−} \rightleftharpoons Hg(l) is established rapidly at the surface of the Hg electrode, so the Nernst equation for the cell can be written in the form
E = E_{+} − E_{−} = (0.852 − \frac{0.059 16}{2}\log(\frac{1}{[Hg^{2+}]})) − E_{−} (B)
where E_{−} is the constant potential of the reference electrode. From Equation A, [Hg^{2+}] = [HgY^{2−}] / K_{f}[Y^{4−}], and this can be substituted into Equation B to give
E = 0.852 − \frac{0.059 16}{2}\log(\frac{[Y^{4−}]K_{f}}{[HgY^{2−}]}) − E_{−}
= 0.852 − E_{−} − \frac{0.059 16}{2}\log(\frac{K_{f}}{[HgY^{2−}]}) − \frac{0.059 16}{2}\log[Y^{4−}] (C)
where K_{f} is the formation constant for HgY^{2−} . This apparatus thus responds to the changing EDTA concentration during an EDTA titration.
Suppose that you titrate 50.0 mL of 0.010 0 M MgSO_{4} with 0.020 0 M EDTA at pH 10.0, using the apparatus in the figure with an S.C.E. reference electrode. Analyte contains 1.0 × 10^{−4} M Hg(EDTA)^{2−} added at the beginning of the titration. Calculate the cell voltage at the following volumes of added EDTA, and draw a graph of millivolts versus milliliters: 0, 10.0, 20.0, 24.9, 25.0, and 26.0 mL.
The cell voltage is given by Equation C, in which K_{f} is the formation constant for Hg(EDTA)^{−2} (= 10^{21.5}). To find the voltage, we must calculate [HgY^{2−}] and [Y^{4−}] at each point. The concentration of HgY^{2−} is 1.0 × 10^{−4} M when V = 0 and is thereafter affected only by dilution because K_{f}(HgY^{2−}) \gg K_{f}(MgY^{2−}). The concentration of Y^{4−} is found from the Mg-EDTA equilibrium at all but the first point. At V = 0 mL, the Hg-EDTA equilibrium determines [Y^{4−}].
0 mL: \frac{HgY^{2−}}{[Hg^{2+}][EDTA]} = α_{Y^{4−}}K_{f}(for HgY^{4−}) = (0.30)(10^{21.5})
\frac{1.0 × 10^{−4} − x}{(x)(x)} = 9.5 × 10^{20} ⇒ x = [EDTA] = 3.2 × 10^{−13} M
[Y^{4−}] = α_{Y^{4−}}[EDTA] = 9.7 × 10^{−14} M
Using Equation C, we write
E = 0.852 − 0.241 − \frac{0.059 16}{2}\log\frac{10^{21.5}}{1.0 × 10^{−4}} − \frac{0.059 16}{2}\log(9.7 × 10^{−14}) = 0.242 V
10.0 mL: V_{e} = 25.0 mL, so \frac{10}{25} of Mg^{2+} is in the form MgY^{2−} , and \frac{15}{25} is in the form Mg^{2+}.
[Y^{4−}] = \frac{[MgY^{2−}]}{[Mg^{2+}]}/K_{f} (for MgY^{2−}) = (\frac{10}{15})/6.2 × 10^{8} = 1.08 × 10^{−9} M
[HgY^{2−}] = \underset{Dilution factor}{(\frac{50.0}{60.0})}(1.0 × 10^{−4} M) = 8.33 × 10^{−5} M
E = 0.852 − 0.241 − \frac{0.059 16}{2}\log\frac{10^{21.5}}{8.33 × 10^{−5}} − \frac{0.059 16}{2}\log(1.08 × 10^{−9}) = 0.120 V
20.0 mL: [Y^{4−}] = (\frac{20}{5})/6.2 × 10^{8} = 6.45 × 10^{−9} M
[HgY^{2−}] = (\frac{50.0}{70.0})(1.0 × 10^{−4} M) = 7.14 × 10^{−5} M
⇒ E = 0.095 V
24.9 mL: [Y^{4−}] = (\frac{24.9}{0.1})/6.2 × 10^{8} = 4.02 × 10^{−7} M
[HgY^{2−}] = (\frac{50.0}{74.9})(1.0 × 10^{−4} M) = 6.68 × 10^{−5} M
⇒ E = 0.041 V
25.0 mL: This is the equivalence point, at which [Mg^{2+}] = [EDTA].
\frac{[MgY^{2−}]}{[Mg^{2+}][EDTA]} = α_{Y^{4−}}K_{f}(for MgY^{2−})\frac{(\frac{50.0}{75.0})(0.010 0) − x}{x²} = 1.85 × 10^{8} ⇒ x = 6.0 × 10^{−6} M
[Y^{4−}] = α_{Y^{4−}}(6.0 × 10^{−6} M) = 1.80 × 10^{−6} M
[HgY^{2−}] = (\frac{50.0}{75.0})(1.0 × 10^{−4} M) = 6.67 × 10^{−5} M
⇒ E = 0.021 V
26.0 mL: Now there is excess EDTA in the solution.
[Y^{4−}] = α_{Y^{4−}}[EDTA] = (0.30)[(\frac{1.0}{76.0})(0.020 0 M)] = 7.89 × 10^{−5} M
[HgY^{2−}] = (\frac{50.0}{76.0})(1.0 × 10^{−4} M) = 6.58 × 10^{−5} M
⇒ E = −0.027 V