Low explosive pyrotechnics used in fireworks contain inorganic elements in the fuel or as oxidizers — Organic Chemistry Chemistry Question
Chemical elements in fireworks
Low explosive pyrotechnics used in fireworks contain inorganic elements in the fuel or as oxidizers or additives. Typical fuels are based on metal or metalloid powders and typical oxidizers are based on perchlorates, chlorates and nitrates with added alkali, alkaline earth and some transition metals. All these substances can easily be determined in an analytical laboratory.
Explain the principle of qualitative flame tests used for the detection of sodium, barium and lithium ions dissolved in aqueous solution. Which flame colours are associated with these elements?
Model Answer
An aqueous sample is introduced to a hot, non-luminous flame, where the tested compound is partially evaporated, atomized and free atoms are excited. During de-excitation, the energy difference between the atomic energy levels is emitted as a photon of an appropriate wavelength, characteristic of the particular chemical element. In this case, all three wavelengths are in the visible region of the spectrum and the corresponding colours for sodium, barium and lithium are yellow, lime green and red, respectively.
Ions of alkaline earth metals and transition metals can by determined by complexometric titrations with EDTA, which is a weak acid with pKa1 = 2.00, pKa2 = 2.67, pKa3 = 6.16 and pKa4 = 10.26.
Sketch the structure of a metal–EDTA complex. Which forms of EDTA will be present in a solution with a pH = 10 at molar concentration higher than 0.5% of the total?
Model Answer
The structure of a metal–EDTA complex is
[VISUAL]
δ(HY3−) = [HY3−] / c(EDTA) = β1 [H +] / (1 + β1 [H +] + β2 [H +]2 + β3 [H +]3 + β4 [H +]4)
δ(Y4−) = [Y4−] / c(EDTA) = 1 / (1 + β1 [H +] + β2 [H +]2 + β3 [H +]3 + β4 [H +]4),
where
β1 = 1 / Ka4, β2 = 1 / (Ka4 Ka3), β3 = 1 / (Ka4 Ka3 Ka2), β4 = 1 / (Ka4 Ka3 Ka2 Ka1)
δ(HY3−) = 1.82×1010 × 1×10−10 / (1 + 1.82×1010 × 1×10−10 + 2.63×1016 × 1×10−20 + 1.23×1019 × 1×10−30 + 1.23×1021 × 1×110−40) = 0.6453, i.e. 64.53 %
δ(Y4−) = 1 / (1 + 1.82×1010 × 1×10−10 + 2.63×1016 × 1×10−20 + 1.23 × 1019 × 1×10−30 + 1.23 × 1021 × 10−40) = 0.3546, i.e. 35.46 %
δ(HY3−) + δ(Y4−) = 99.99 % , hence other forms are present at molar ratios lower than 0.5 %.
The determination of calcium, strontium and barium ions by the reaction with EDTA is often performed in the presence of an ammonium buffer, which keeps the pH of the solution around 10.
What is the chemical composition of an ammonium buffer? What is the role of an alkaline pH in these reactions?
Model Answer
Ammonium buffer is a mixture of ammonia and ammonium chloride. Ions of alkaline earth metals form weak complexes with EDTA (log KMY between 7.7 and 10.7) and are present only in alkaline media (pH >= 9).
A combustible mixture used in fireworks (containing zinc, magnesium, lead, and no other multivalent ions) in a paper cartridge was analyzed in the following three steps:
i. The sample (0.8472 g) was dissolved and an excess of cyanide was added to mask the zinc in solution. This mixture was titrated with 0.01983 mol dm−3 EDTA and V1 = 35.90 cm3 was required to reach the equivalence point.
ii. Next, 2,3-disulfanylpropan-1-ol (DMP) was added and the released EDTA was titrated with 12.80 cm3 of 0.01087 mol dm−3 of Mg2+ standard solution to reach the equivalence point.
iii. Finally, formaldehyde was introduced to release the zinc ions, which were subsequently titrated with V2 = 24.10 cm3 of 0.01983 mol dm−3 EDTA to reach the equivalence point.
Write both ionic equations for the masking and the releasing of the zinc ions.
Model Answer
Zn2+ + 4 CN− → [Zn(CN)4]2−
[Zn(CN)4]2− + 4 HCHO + 4 H2O → Zn2+ + 4 HOCH2CN + 4 OH−
Explain the role of the DMP addition.
Model Answer
2,3-Disulfanylpropan-1-ol is used for masking lead ions.
Calculate the mass (in mg) of all three elements in 1 g of the original sample.
Model Answer
Step i: Zinc is masked in the cyanide complex, lead and magnesium react with EDTA
n(Pb) + n(Mg) = n1(EDTA)
Step ii: EDTA released from its complex with lead ions reacts with magnesium standard solution
n(Pb) = nstd(Mg)
Step iii: Zinc released from cyanide complex reacts with EDTA
n(Zn) = n2(EDTA)
Masses of the elements in the sample (m1 = 1 g):
n(Mg) = n1(EDTA) – nstd(Mg)
m(Pb) = [VISUAL] = 34.03 mg
m(Mg) = [VISUAL] = 16.43 mg
m(Zn) = [VISUAL] = 36.88 mg
10.00 cm3 of 0.0500 mol dm-3 Ca2+ solution was mixed with 50.00 cm3 of 0.0400 mol dm-3 EDTA in a 100 cm3 volumetric flask and after the pH was set to 6, the flask was filled up to the mark with distilled water. Calculate the concentration of free Ca2+ ions in the solution. Decimal logarithm of the stability constant for the complex of Ca2+ ions with EDTA is 10.61. Consider only the equilibria that have been mentioned so far.
Model Answer
Complexation equation: Ca2+ + Y4- → CaY2-
Final analytical concentrations after dilution are:
c(Ca2+) = 10/100 × 0.05 = 0.005 mol dm-3
c(Y4-) = 50/100 × 0.04 = 0.02 mol dm-3
The coefficient for EDTA side reactions
α(EDTA) = (1 + β1 [H +] + β2 [H +]2 + β3 [H +]3 + β4 [H +]4)
(definitions of βi are in 14.2); for pH = 6
α(EDTA) = (1 + 1.82×1010 × 1×10-6 + 2.63×1016 × 1×10-12 + 1.23 × 1019 × 1×10-18 + 1.23×1021 × 1×10-24) = 4.45×104
The conditional stability constant β' of Ca-EDTA complex is:
β' = [VISUAL] = 10^5.96
The final concentration of free Ca2+ ions in the solution is given by:
[Ca2+] = [VISUAL]
Based on the mass balances
[Y4–] = c(Y4–) − [CaY2–] = c(Y4–) − (c(Ca2+) − [Ca2+])
we get a quadratic equation with the following solution
βI[Ca2+]2 + (1 + βI c(Y4–) − βI c(Ca2+))[Ca2+] – c (Ca2+) = 0
[Ca2+] = 3.642 × 10–7 mol dm–3