Alloys can be found in many objects we come across in our daily life. Due to their particular charac — Physical Chemistry — Thermodynamics Chemistry Question
Determination of copper and zinc by complexometric titration
Alloys can be found in many objects we come across in our daily life. Due to their particular characteristics (i.e., conductivity, mechanical or corrosion resistance), alloys are successfully applied in many advanced fields such as aeronautics, construction, electronics devices, and jewelry. That is why developing reliable methods of alloys analysis is of extreme importance.
Brass is an alloy of copper and zinc which is familiar to most students. In this experiment, a brass alloy containing Cu2+ and Zn2+ ions will be analyzed by complexometric titration with Na2H2EDTA. Since the stability constants of the complexes of these metals with EDTA are close, masking of the Cu2+ ions by a complexing agent (thiosulfate) is used. In the first titration, copper and zinc are titrated together with Na2H2EDTA. In the second titration, sodium thiosulfate is added to bind the Cu2+ ions, thus allowing titration of solely zinc ions with Na2H2EDTA.
Chemicals and reagents
* Brass sample, ~250 mg per student, or
Test solution (a standard solution containing ions Cu2+ (about 1.5 g dm-3) and Zn2+ (about 1 g dm-3) simulating a digested sample of brass)
* Nitric acid, HNO3, concentrated (~70% w/v)
* Na2H2EDTA standard solution, c = 0.0500 mol dm-3
* Acetate buffer solution, pH = 5.5 – 6.0, 0.1 mol dm-3 in acetate
* Sodium thiosulfate solution, Na2S2O3, ~10% (w/v)
* Metallochromic indicator 4-(2-pyridylazo)resorcinol (PAR), 0.1% aqueous solution (w/v)
(0.1% solution of Xylenol orange indicator may be used instead of PAR).
Apparatus and glassware
* Analytical balance (± 0.0001 g)
* Beaker, 10 cm3
* Hotplate
* Volumetric flask, 100 cm3
* Burette, 25 or 50 cm3
* Volumetric pipettes, 2, 5 and 10 cm3
* Erlenmeyer flask, 100 cm3 (3 ea.)
* Graduated cylinders, 10 and 25 cm3
Procedure
A. Brass digestion
a) Take a precise weight of the brass sample (~250 mg) and place it in a beaker.
Note. If no certified brass samples are available, you can use a test solution simulating the digested alloy.
b) Carefully add 5 cm3 of concentrated nitric acid (the experiment should be done under a fume hood, as NO2 gas evolves).
c) Heat the beaker slightly on a hotplate to provide for an effective dissolution.
d) When the digestion of the sample is complete, evaporate the solution to near dryness to remove the most part of the acid (avoid evaporating to dry salts, as hydrolysis may occur. If still so, add a minimal amount of HCl to dissolve the residue). Allow the beaker cooling down to room temperature.
e) Dissolve the contents of the beaker in distilled water, transfer it to a 100.00 cm3 volumetric flask and make it up to the mark.
B. Determination of the total amount of Cu 2+ and Zn 2+
f) Transfer 10.00 cm3 of the test solution into a 100 cm3 Erlenmeyer flask, add 20 cm3 of water, 5 cm3 of acetate buffer solution and 3 drops of PAR solution, mix thoroughly.
g) Titrate the content of the flask with a standard Na2H2EDTA solution (c = 0.0500 mol dm-3) until the color of PAR indicator changes from bluish-violet to blue or greenish-yellow (for Xylenol orange indicator, the color changes from red to green). Repeat the titration when necessary.
C. Determination of Zn 2+
h) Transfer 10.00 cm3 of the test solution into a 100 cm3 Erlenmeyer flask, add 10 cm3 of water, 5 cm3 of acetate buffer solution, 2 cm3 of Na2S2O3 solution and 3 drops of PAR solution, mix thoroughly.
i) Titrate the content of the flask with 0.0500 mol dm-3 standard Na2H2EDTA solution until the color changes from red to yellow (for Xylenol orange, the colors are the same).
D. Calculation of Cu 2+ concentration
j) The volume of Na2H2EDTA which is necessary for Cu2+ titration is calculated as the difference of the titrant volumes in titrations B and C.
Give balanced chemical equations for the reactions that take place when:
- brass dissolves in nitric acid;
- copper and zinc ions are titrated by Na2H2EDTA;
Model Answer
Cu + 4 HNO3 (conc.) → Cu(NO3)2 + 2 NO2 + 2 H2O
Zn + 4 HNO3 (conc.) → Zn(NO3)2 + 2 NO2 + 2 H2O
Cu2+ + Na2H2EDTA → CuH2EDTA + 2 Na+
Zn2+ + Na2H2EDTA → ZnH2EDTA + 2 Na+
Explain how Na2S2O3 masks the Cu2+ ion, giving the appropriate chemical equation.
Model Answer
Cu2+ ions present in the aqueous solution are reduced to Cu+ by thiosulfate. Moreover, the latter forms with Cu+ a soluble complex [Cu(S2O3)3]5–, which is more stable than Cu2H2EDTA:
2 Cu2+ + 8 S2O32– → 2 [Cu(S2O3)3]5– + S4O62–
Why should the pH value of the titrated solution be kept within 5 – 6?
Model Answer
Metal ions can be titrated with EDTA if the conditional stability constants β’ of the metal – EDTA complexes are not less than 108 – 109. The β’ values are connected with the real constants β as
β’ = αEDTA αM β,
where αEDTA and αM are molar fractions of H2EDTA2– and free metal ion, respectively. As the values of αEDTA and αM significantly depend on pH of the solution, there is an optimal pH range for the titration of metals. In the case of Cu2+ and Zn2+, the pH value within 5 to 6 is optimal. In such slightly acidic medium both metals do not form hydroxyl complexes (αM is high), whilst H2EDTA2– is not further protonated (αEDTA is high).
Calculate the molar fraction of H2EDTA2– at pH 6. EDTA is a weak acid with the following acidity constants:
K1 = 1.0 · 10–2, K2 = 2.1 · 10–3, K3 = 6.9 · 10–7, K4 = 5.5 · 10–11.
Model Answer
\alpha(H_2EDTA^{2-}) = \frac{K_1 K_2 [H^+]^2}{[H^+]^4 + K_1 [H^+]^3 + K_1 K_2 [H^+]^2 + K_1 K_2 K_3 [H^+] + K_1 K_2 K_3 K_4}
Using [H+] = 1.0 * 10^-6 M:
K1 = 1.0 * 10^-2, K2 = 2.1 * 10^-3, K3 = 6.9 * 10^-7, K4 = 5.5 * 10^-11, \alpha(H_2EDTA^{2-}) = 0.59
Derive the formulae for calculation of Cu2+ and Zn2+ concentrations in the test solution. Calculate the mass ratio of Cu and Zn in the alloy.
Model Answer
The first titration (B) gives the volume of titrant VCu+Zn, whilst the second one (C) gives VZn. Zn2+ concentration is calculated as follows:
c(Zn2+) (g dm–3 ) = VZn (cm3) × cEDTA (mol dm–3) × 65.39 g mol–1 × 0.1 cm–3
c(Cu2+) (g dm–3 ) = (VCu+Zn – VZn) cm3 × cEDTA (mol dm–3) × 63.55 g mol–1 × 0.1 cm–3
The mass ratio of the metals in alloy is calculated from c(Cu2+) and c(Zn2+) values in g dm–3:
m(Cu) / m(Zn) = c(Cu2+) / c(Zn2+)