On January 30 in 2000, a dam failure in a gold mine spilled about 100 000 m3 of cyanide-containing w — Physical Chemistry — Kinetics Chemistry Question
Cyanide Pollution in the River Szamos
On January 30 in 2000, a dam failure in a gold mine spilled about 100 000 m3 of cyanide-containing waste water into the river Szamos. The pollution wave, which later reached the Central European rivers Tisza and Danube, killed massive amounts of fish. On February 15, a popular Hungarian TV news show presented a simple experiment: first a NaCN solution was prepared, the concentration of which was similar to those measured in the pollution wave. Fish were killed in this solution but survived when ferrous sulfate was also added. The TV show suggested that ferrous sulfate should have been used to lower the environmental impact of the cyanide solution. However, when the same experiment was repeated with an actual sample from the pollution wave, fish were killed even after ferrous sulfate was added. Unfortunately, this second experiment was not covered in any evening news.
To clarify the underlying chemistry, an expert designed a detailed series of experiments in which the use of a cyanide selective combination electrode was an important element. He first calibrated the electrode using 3 different concentrations at three different pH values. The temperature was 25 °C in all experiments. The instrumental readings were as follows:
Calculate the acid dissociation constant of HCN based on these measurements.
Model Answer
The electrode gives a Nernstian response with a slope of 59.1 mV/decade at all three pH values used for the calibration. The electrode potential in millivolts is:
E = E° − 59.1 x lg [CN−]
The equilibrium concentration of CN– can be given as a function of the analytical concentration and pH:
[CN–] = c / (1 + [H+]/Ka)
Applying these two equations for a pair of points with identical analytical concentration of cyanide gives:
E_pH1 – E_pH2 = 59.1 mV x lg(([H+]_2 - [H+]_1)/Ka + 1)
Applying this equation for pH1 = 12 and pH2 = 7.5 gives (at all three NaCN concentrations E_pH1 – E_pH2 = –101.6 mV):
Ka = 6.15 x 10^–10
To 100 cm3 of a test solution, which contained 49.0 mg dm–3 NaCN and was buffered to pH = 7.5 and 40.0 mg of solid FeSO4·7 H2O has been added. At this pH, the reaction between aqueous iron(II) and dissolved oxygen is quantitative under all conditions and gives an iron(III) hydroxide precipitate. Ignore possible complexation reactions between the precipitate and cyanide ions.
11.2 Write the balanced equation for this redox reaction.
Model Answer
4 Fe2+ + O2 + 10 H2O → 4 Fe(OH)3 + 8 H+
All the solutions used in the experiments initially contained 8.00 mg dm–3 dissolved oxygen. The electrode reading in this solution was 585.9 mV. Iron(II) only forms one complex with cyanide ion, which has a coordination number of 6.
11.3 Write the ionic equation describing the formation of this complex. Estimate the stability constant of the complex.
Model Answer
Fe2+ + 6 CN– → [Fe(CN)6]4–
The concentration of the NaCN solution is 0.0010 mol dm–3. From the calibration of the electrode it follows that Eº = 220.1 mV and:
[CN–] = 10^((Eº - E) / 59.1mV)
The electrode reading is 585.9 mV at pH = 7.5. From this:
[CN–] = 6.46 x 10^–7 ⇒ [HCN] = [H+][CN–] / Ka = 3.32 x 10^–5
The concentration of complexed cyanide is then:
[CN–]compl = [CN–]total – [CN–] – [HCN] = 9.66 x 10^–4
All complexed cyanide is in the hexacyano iron(II) complex, therefore its concentration is:
[Fe(CN)6 4–]eq = [CN–]compl / 6 = 1.61 x 10^–4
The total amount of iron(II) added is 1.44 x 10^–4 mol.
The amount of iron(II) that reacts with O2 is 1.00 x 10^–4 mol
The amount of iron(II) present as the hexacyano complex is 1.61 x 10^–5 mol. The concentration of free iron(II) is 2.78 x 10^–4 mol dm–3
The stability constant is:
β6 = [Fe(CN)6 4–]eq / ([Fe2+]eq [CN–]eq^6) = 7.99 x 10^36
The following toxicity data (LC50: median lethal concentration for 24-hour exposure) for fish can be found in tables:
* total non-complexed cyanide = [HCN] + [CN–]
The loss of dissolved oxygen is not a major problem for fish in the very small volume of the experiment, but it would probably be under natural conditions.
11.4 Are the experimental results and the toxicity data in agreement with the result of the experiment shown on the TV news show?
Model Answer
The concentrations of free CN– and HCN are 6.46 x 10^–7 mol dm–3 and 3.32 x 10^–5 mol dm–3, respectively. This corresponds to a total non-complexed cyanide concentration of 0.88 mg dm–3, which is less than half of the LC50 value. The concentration of practically non-toxic [Fe(CN)6]4– is high. These data are in agreement with the presented experiment, although prolonged exposure to these conditions would probably cause adverse health effects in fish.
A little known fact about the pollution wave was that it also contained metals, primarily copper (which is hardly surprising for a gold mine). Copper is often present in our environment as copper(II), but it was present as copper(I) in the pollution wave because of the presence of cyanide ions.
11.5 Write the chemical equation for the reaction between copper(II) and cyanide ion.
Model Answer
2 Cu2+ + 2 CN– → 2 Cu+ + (CN)2
Cu+ + n CN– = [Cu(CN)n](n–1)– (n = 2 – 3)
An actual sample from the pollution wave had a pH of 7.5, its total cyanide content (including complexed, non-complexed and protonated cyanide ions) was determined to be 26 ppm, its total copper content 21 ppm. The cyanide selective electrode gave a reading of 534.6 mV in this solution, and an electrochemical method showed that the concentration of free copper(I) is about 2 x 10^–15 mol dm–3. Copper(I) forms complexes with cyanide ion in a stepwise manner up to a coordination number of 3. The formation constant of [CuCN] is negligible compared to that of the other two complex ions. Dissolved oxygen, the concentration of which was 8.00 mg dm–3, coexists with cyanocopper(I) complexes.
11.6 Is there any copper(I)-cyanide precipitate in the solution? Ksp(CuCN) = 3.5 x 10^–19
Model Answer
From the electrode reading and the pH value: [CN–]eq = 4.77 x 10^–6.
Since [Cu+] = 2 x 10^–15, the product of the two concentrations is less than Ksp(CuCN) and, therefore, CuCN does not precipitate.
11.7 Determine the coordination number(s) of copper(I) complex(es) dominating in the sample studied. Estimate the stability constant(s) of the cyanocopper(I) complex(es).
Model Answer
[Cu+]tot = 0.021 g dm–3 / 63.55 g mol–1 = 3.30 x 10^–4 mol dm–3, practically all complexed.
[CN–]total = 0.026 g dm–3 / 26.02 g mol–1 = 1.00 x 10^–3 mol dm–3
The concentration of cyanide ions:
[CN–]eq = 4.77 x 10^–6 ⇒ [HCN]eq = 2.45 x 10^–4
[CN–]compl = [CN–]total – [CN–]eq – [HCN]eq = 7.50 x 10^–4
The average number of ligands in cyanocopper complexes:
[CN–]compl / [Cu+]tot = 2.27
This means that the mixture contains [Cu(CN)2]– and [Cu(CN)3]2– complexes.
[Cu(CN)3 2–]eq = [CN–]compl – 2 [Cu+]tot = 8.9 x 10^–5
[Cu(CN)2 –]eq = ([CN–]compl – 3 [Cu(CN)3 2–]eq) / 2 = 2.4 x 10^–4
Therefore:
β2 = [Cu(CN)2 –] / ([Cu+][CN–]^2) = 5.3 x 10^21
β3 = [Cu(CN)3 2–] / ([Cu+][CN–]^3) = 4.1 x 10^26
The toxicity of copper(I) cyano complexes is very similar to that of NaCN; [Cu(CN)2]– has an LC50 value of 4.5 mg dm-3. To 100 cm3 of the sample from the pollution wave, 40.0 mg of solid FeSO4· 7 H2O was added. The cyanide selective electrode gave a reading of 592.3 mV in this solution.
11.8 Estimate the concentrations of various complexes in this sample. Is this solution expected to be toxic? Does this expectation agree with the experiment not shown on TV?
Model Answer
From the electrode reading and the pH:
[CN–]eq = 5.04 x 10^–7 ⇒ [HCN]eq = 2.59 x 10^–5
[CN–]compl = [CN–]total – [CN–]eq – [HCN]eq = 9.74 x 10^–4
The total amount of iron(II) added is 1.44 x 10^–4 mol, of this 1.00 x 10^–4 mol reacts with O2. The total iron(II) concentration remaining in the solution is therefore 4.4 x 10^–4 mol dm–3.
From the definition of β6 and conservation of mass:
β6[CN–]eq^6 = [Fe(CN)6 4–]eq / ([Fe2+]tot – [Fe(CN)6 4–]eq)
Solving this equation gives:
[Fe(CN)6 4–]eq = 5.04 x 10^–5 and [Fe2+]eq = [Fe2+]tot – [Fe(CN)6 4–]eq = 3.9 x 10^–4
Combining the two complex formation reactions gives the following:
3 [Cu(CN)2]– + Fe2+ = 3 Cu+ + [Fe(CN)6]4–
KR1 = β6 / β2^3 = 5.34 x 10^–29
From this process the following ratio can be calculated:
[Cu+]^3 [Fe(CN)6 4–]eq / ([Cu(CN)2 –]^3 [Fe2+]eq) = KR1 / [CN–]eq^6 = 7.43 x 10^10
This means that practically all the copper is complexed in the solution. The concentration of cyanide complexed in copper complexes:
[CN–]Cu,compl = [CN–]compl – 6 [Fe(CN)6 4–]eq = 6.73 x 10^–4
Therefore the relative equilibrium concentrations of cyanocopper complexes are:
[Cu(CN)3 2–]eq = [CN–]Cu,compl – 2 [Cu+]tot = 1 x 10^–5
[Cu(CN)2 –]eq = ([CN–]Cu,compl – 3 [Cu(CN)3 2–]eq) / 2 = 3.2 x 10^–4
Precipitation of CuCN is possible, and this should be checked for:
[Cu+]eq = 7.45 x 10^–10 · [Cu(CN)2 –]eq = 2.4 x 10^–13
[Cu+]eq [CN–]eq = 1.2 x 10^–19 < Ksp(CuCN)
This solution contains toxic [Cu(CN)2]– in high concentration relative to the lethal concentration (37 mg dm–3 > LC50), so it is toxic in agreement with the experiment not shown on TV.