Problem 1

A 0.4062 g alloy sample containing tin and lead is dissolved in hot

\ce{HCl}

\ce{HNO3}

(Pb → Pb(II), Sn → Sn(IV)); on cooling a precipitate of tin compounds and a lead compound appears.

25.00\ \mathrm{cm^3}

of 0.2000 M

\ce{Na2H2EDTA}

is added (the precipitate dissolves) and the solution is diluted to

250.0\ \mathrm{cm^3}

. A

25.00\ \mathrm{cm^3}

aliquot, buffered at pH 6 with hexamine and Xylenol Orange indicator, is titrated with standard 0.009970 M

\ce{Pb(NO3)2}

: the colour changes yellow→red at

24.05\ \mathrm{cm^3}

. Then 2.0 g of solid

\ce{NaF}

is added — the solution returns to yellow — and titration resumes to a second endpoint at

15.00\ \mathrm{cm^3}

. (

\ce{F^-}

binds

\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

\ce{NaF}

, and the key ionic equations of the two titrations. (1.10) Calculate the weight percentages of Sn and Pb in the alloy.

Step 4 of 4: Weight percentages

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w(\mathrm{Sn}) = \frac{10\times1.496\times10^{-4}\times118.69}{0.4062} = 43.7\%;\qquad w(\mathrm{Pb}) = \frac{10\times1.106\times10^{-4}\times207.19}{0.4062} = 56.4\%

Analysis

The aliquot is one tenth of the flask: $m(\ce{Sn}) = 10\times1.496\times10^{-4}\times118.69 = 0.1776$ g and $m(\ce{Pb}) = 10\times1.106\times10^{-4}\times207.19 = 0.2292$ g, giving Sn 43.7 % and Pb 56.4 % (total ≈ 100 %).