A metal alloy which contains mainly Cu and Zn was analyzed for its metal content. An alloy sample of — Physical Chemistry — Kinetics Chemistry Question
Titration of Cu and Zn in Metal Alloy
A metal alloy which contains mainly Cu and Zn was analyzed for its metal content. An alloy sample of 2.300 g was placed in a 250 cm3 Erlenmeyer flask. To this flask, 5.00 cm3 of mixed acid (concentrated nitric acid and concentrated hydrochloric acid) was added in a fume hood for dissolution of the alloy. The resulting solution was transferred quantitatively into 250 cm3 volumetric flask and adjusted to volume with deionized water. A 25.00 cm3 aliquot of sample solution was adjusted to pH 5.5 and titrated with 0.100 mol dm-3 EDTA solution using 1-(2-pyridinezao-2-naphthol or PAN as indicator. The indicator changed color when 33.40 cm3 0.100 mol dm-3 EDTA was added to the sample solution. Another 25.00 cm3 aliquot of sample solution was adjusted to neutral pH and mixed with excess KI. The mixture was then filtered and the resulting solution was titrated against 0.100 mol dm-3 sodium thiosulfate using starch solution as indicator. The titration required 29.35 cm3 sodium thiosulfate to reach the endpoint. Note: Ksp(CuI) = 1.1×10-12
Write the oxidative dissolution reaction(s) of an alloy with nitric acid and hydrochloric acid.
Model Answer
Cu(s) + 4 HNO3(aq) → Cu2+(aq) + 2 NO3-(aq) + 2 NO2(g) + 2 H2O(l)
Zn + 2 HCl → Zn2+ + H2(g) + 2 Cl-
Determine the %w/w of Cu and Zn in alloy.
Model Answer
Total metal by EDTA titration: n(EDTA) used = 33.4 × 0.1 / 1000 = 3.34×10-3 mol
n(Cu) by redox titration:
2 Cu2+ + 4 I- → CuI(s) + I2
I2 + 2 S2O32- → 2 I- + S4O62-
n(Cu2+) = n(thiosulfate used) = 29.35 × 0.1/1000 = 2.935×10-3 mol
m(Cu) = 2.935×10-3 mol × 63.5 g mol-1 = 0.1864 g in 25.00 cm3 aliquot
Therefore, a 250 cm3 sample solution will contain 1.864 g
Thus w(Cu) = 1.864 g / 2.300 g = 0.81, i.e. 81.0 %
n(Zn) = n(total metal) – n(Cu) = 3.34×10-3 mol – 2.935×10-3 mol = 4.05×10-4 mol
m(Zn) = 4.05×10-4 mol × 65.4 g mol-1 = 2.649×10-2 g in 25.00 aliquot
Therefore, 250 cm3 sample solution will contain 0.2649 g of zinc.
Thus w(Zn) = 0.2649 / 2.300 = 0.115, i.e. 11.5 %