— Physical Chemistry Chemistry Question
Lead compounds
I. Consider the following nuclide: 209Bi(I), 208Pb(II), 207Pb(III), 206Pb(IV).
9.1 Which nuclide is the last member of the decay series for 238U? Check in the appropriate box.
I II III IV
Model Answer
9.1 The correct answer is IV.
II. There are three natural decay series. They begin with Th-232(I), U-238(II), U-235(III) and end with Pb-208, Pb-206, Pb-207.
9.2 In which decay chain are there 6 α-decays and 4 β-decays? Choose the correct answer by checking in the appropriate box.
I II III none
Model Answer
9.2 The correct answer is I.
III. Pb(NO3)2 solution is slowly added into 20.00 cm3 of a mixture consisting of Na2SO4 (c = 0.020 mol dm-3); Na2C2O4 (c = 5.0 ⋅ 10−3 mol dm-3); KI (c = 9.7 ⋅ 10−3 mol dm-3); KCl (c = 0.05 mol dm-3) and KIO3 (c = 0.0010 mol dm-3). When the bright yellow precipitate of PbI2 begins to form, 21.60 cm3 of Pb(NO3)2 solution is consumed.
Use the following data: pKs(PbSO4) = 7.66; pKs(Pb(IO3)2) = 12.61; pKs(PbI2) = 7.86; pKs(PbC2O4) = 10.05; pKs(PbCl2) = 4.77. (Other processes of the ions are ignored).
9.3 Determine the order of precipitation?
Model Answer
Condition for precipitation of:
PbSO4 : c(Pb2+) ≥ 10^–7.66 / 0.02 = 1.09 ⋅ 10^–6 (mol dm^-3) (1)
PbC2O4 : c(Pb2+) ≥ 10^–10.05 / 5.0 ⋅ 10^–3 = 1.78 ⋅ 10^–8 mol dm^-3 (2)
PbI2 : c(Pb2+) ≥ 10^-7.86 / (9.7 ⋅ 10^-3)^2 = 1.47 ⋅ 10^-4 mol dm^-3 (3)
Pb(IO3)2: c(Pb2+) ≥ 10^-12.61 / (0.001)^2 = 2.45 ⋅ 10^-7 mol dm^-3 (4)
PbCl2: c(Pb2+) ≥ 10^-4.8 / (0.05)^2 = 6.34 ⋅ 10^-3 mol dm^-3 (5)
c(Pb2+)(2) < c(Pb2+)(4) < c(Pb2+)(1) < c(Pb2+)(3) < c(Pb2+)(5)
The order of precipitation: PbC2O4, Pb(IO3)2, PbSO4, PbI2 and PbCl2.
9.4 Calculate the concentration of Pb(NO3)2 solution?
Model Answer
When PbI2 begins to precipitate (assume I- has not reacted)
[SO4^2-] = K_s(PbSO4) / c(Pb2+)(3) = 10^–7.66 / 1.47 ⋅ 10^–4 = 1.49 ⋅ 10^–4
= 1.48 ⋅ 10^–4 = s(PbSO4)
(s is the solubility of PbSO4 in saturated solution). Hence PbC2O4, Pb(IO3)2 and PbSO4 have precipitated completely.
→ 21.60 × c(Pb(NO3)2) = 20.00 × ( c(C2O4^2-) + c(IO3^-) / 2 + c(SO4^2-) )
= 20.00 (5.0 ⋅ 10−3 + 2 × 0.0010 + 0.020)
→ c(Pb(NO3)2) = 0.025 mol dm-3
IV. One of the common reagents to detect Pb2+ species is K2CrO4, giving yellow precipitate PbCrO4, which is soluble in excess of NaOH. The solubility of PbCrO4 depends not only on pH but also on the presence of coordinating species.
9.5 Calculate the solubility product Ksp of PbCrO4 if the solubility of PbCrO4 in acetic acid solution (c = 1 mol dm-3) is s = 2.9 ⋅ 10−5 mol dm-3.
Use the following data:
pKa(CH3COOH) = 4.78; log β(Pb(CH3COO+ = 2.68;
log β(Pb(CH3COO)2 = 4.08; pKa(HCrO4–) = 6.50
Pb2+ + H2O ⇌ PbOH+ + H+ β = 10-7.8
Cr2O7 2- + H2O ⇌ 2 CrO4 2- + 2 H+ K = 10-14.64
Model Answer
PbCrO4 ⇌ Pb2+ + CrO4^2- Ks
CH3COOH ⇌ CH3COO− + H+ Ka = 10−4.76
Pb2+ + CH3COO− ⇌ Pb(CH3COO+ β1 = 10^2.68
Pb2+ + 2 CH3COO− ⇌ Pb(CH3COO)2 β2 = 10^4.08
Pb2+ + H2O ⇌ PbOH+ + H+ β = 10−7.8
CrO4^2- + H+ ⇌ HCrO4^- Ka^-1 = 10^6.5
2 CrO4^2- + 2 H+ ⇌ Cr2O7^2- + H2O K^-1 = 10^14.64
Let h be [H+]. A conservation of mass requires that:
s = [CrO4^2-] + [HCrO4^-] + 2 [Cr2O7^2-] = [CrO4^2-](1 + Ka^-1 h) + 2 K^-1 h^2 [CrO4^2-]^2 (1)
s = [Pb2+] + [PbOH+] + [Pb(CH3COO+] + [Pb(CH3COO)2] = [Pb2+](1 + β h^-1 + β1[CH3COO-] + β2[CH3COO-]^2) (2)
Because s = 2.9 ⋅ 10−5 mol dm-3 << c(CH3COOH) = 1 mol dm-3 → pH of the solution is largely dependent on the dissociation of CH3COOH:
CH3COOH ⇌ H+ + CH3COO− Ka = 10−4.76
1 – h h h
→ [CH3COO-] = [H+] = h = 1 ⋅ 10−2.38
Substitute [CH3COO−] = [H+] = h = 10−2.38 and s = 2.9 ⋅ 10−5 into (1) and (2), we have:
[CrO4^2-] = 2.194 ⋅ 10−9 and [Pb2+] = 9.051 ⋅ 10−6
→ Ksp = [Pb2+][CrO4^2-] = 1.99 ⋅ 10−14.
V. Lead-acid battery, commonly known as lead battery consists of two lead plates a positive electrode covered with a paste of lead dioxide and a negative electrode made of sponge lead. The electrodes are submersed in an electrolyte consisting of water and sulfuric acid H2SO4.
9.6 Write the chemical equations for processes on each electrode, overall reaction as the battery discharges and the cell diagram.
Model Answer
Cathode: PbO2 + 4 H+ + 2 e ⇌ Pb2+ + 2 H2O
HSO4- ⇌ SO4^2− + H+
Pb2+ + SO4^2- ⇌ PbSO4
Cathode reaction: PbO2 + HSO4- + 3 H+ + 2 e ⇌ PbSO4 + 2 H2O K1 (*)
Anode: Pb ⇌ Pb2+ + 2e
HSO4- ⇌ SO4^2- + H+
Pb2+ + SO4^2- ⇌ PbSO4
Anode reaction: Pb + HSO4- ⇌ PbSO4 + H+ + 2 e K2 ()
Overall reaction as the battery discharges: PbO2 + Pb + 2 HSO4- + 2 H+ ⇌ 2 PbSO4 + 2 H2O (*)
Cell diagram: (a) Pb│PbSO4, H+, HSO4-│PbO2 (Pb) (c)
VI.
9.7 Calculate the values of E0(PbSO4/Pb) and E0(PbO2/PbSO4)
Model Answer
According to (*):
10^(2 E0(PbO2/PbSO4) / 0.0592) = K1 = 10^(2(1.455)/0.0592) ⋅ 10^-2 ⋅ 10^7.66
→ E0(PbO2/PbSO4) = 1.62 V
According to ():
10^(-2 E0(PbSO4/Pb) / 0.0592) = K2 = 10^(-2(-0.126)/0.0592) ⋅ 10^-2 ⋅ 10^7.66
→ E0(PbSO4/Pb) = – 0.29 V
9.8 The potential V of the lead battery if c(H2SO4) ≈ 1.8 mol dm-3.
The following data are given: E0(Pb2+/Pb) = – 0.126 V; E0(PbO2/Pb2+) = 1.455 V;
pKa(HSO4-) = 2.00; pKs(PbSO4) = 7.66; at 25 oC: 2.303 RT/F = 0.0592 V
Model Answer
According to (*):
V = E(c) – E(a) = E0(PbO2/PbSO4) – E0(PbSO4/Pb) + (0.0592/2) log([HSO4-]^2 [H+]^2)
In which [HSO4-], [H+] are calculated as follows:
HSO4- ⇌ H+ + SO4^2- Ka = 10^−2
1.8 – x 1.8 + x x
[SO4^2-] = x = 9.89 ⋅ 10^-3 → [H+] = 1.81 ; [HSO4-] = 1.79
V = 1.62 + 0.29 + (0.0592/2) log(1.79^2 ⋅ 1.81^2) = 1.94 V