Chemistry Labs

International Chemistry Olympiad · 1998

Problems

  1. Problem 1A 0.4062 g alloy sample containing tin and lead is dissolved in hot HCl\ce{HCl}/HNOX3\ce{HNO3} (Pb → Pb(II), Sn → Sn(IV)); on cooling a precipitate of tin compounds and a lead compound appears. 25.00 cm325.00\ \mathrm{cm^3} of 0.2000 M NaX2HX2EDTA\ce{Na2H2EDTA} is added (the precipitate dissolves) and the solution is diluted to 250.0 cm3250.0\ \mathrm{cm^3}. A 25.00 cm325.00\ \mathrm{cm^3} aliquot, buffered at pH 6 with hexamine and Xylenol Orange indicator, is titrated with standard 0.009970 M Pb(NOX3)X2\ce{Pb(NO3)2}: the colour changes yellow→red at 24.05 cm324.05\ \mathrm{cm^3}. Then 2.0 g of solid NaF\ce{NaF} is added — the solution returns to yellow — and titration resumes to a second endpoint at 15.00 cm315.00\ \mathrm{cm^3}. (FX−\ce{F^-} binds SnXIV\ce{Sn^{IV}} strongly but not Pb(II) at pH 6.) (1.1) What lead compound precipitates in step 2? (1.3–1.8) Give the roles of hexamine, Xylenol Orange and NaF\ce{NaF}, and the key ionic equations of the two titrations. (1.10) Calculate the weight percentages of Sn and Pb in the alloy.Solutions: 1
  2. Problem 2Dating historical events using 210Pb^{210}\ce{Pb}. Nathan Thompson, an early settler of Lord Howe Island, planted European deciduous trees; the year is unknown. Pollen of European oak and elm (which pollinate in their first year) accumulated in lake sediment together with radioactive 210Pb^{210}\ce{Pb} (half-life = 22.0 years). In 1995 a sediment core was taken: oak/elm pollen first occurs at 50 cm depth, where the 210Pb^{210}\ce{Pb} activity is 1.40 Bq kg⁻¹, versus 356 Bq kg⁻¹ at the top of the core. (2.1) In what year were the seeds planted? (2.2) 210Pb^{210}\ce{Pb} is a daughter of X238X22238U\ce{^{238}U}, which stays in the earth's crust: X238X22238U→X234X22234U→X230X22230Th→X226X22226Ra→X222X22222Rn→(X218X22218Po,X214X22214Bi)→X210X22210Pb→X206X22206Pb\ce{^{238}U} \rightarrow \ce{^{234}U} \rightarrow \ce{^{230}Th} \rightarrow \ce{^{226}Ra} \rightarrow \ce{^{222}Rn} \rightarrow (\ce{^{218}Po}, \ce{^{214}Bi}) \rightarrow \ce{^{210}Pb} \rightarrow \ce{^{206}Pb} (stable). Which step explains how 210Pb^{210}\ce{Pb} reaches rainwater while its parent X238X22238U\ce{^{238}U} remains in the crust?Solutions: 1