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 2 of 4: Chemistry of the two titrations

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\ce{Pb^{2+} + H2Y^{2-} -> PbY^{2-} + 2H+};\quad \ce{Pb^{2+} + XO_{(yellow)} -> PbXO^{2+}_{(red)}};\quad \ce{SnY + nF^- + 2H+ -> SnF_n^{(n-4)-} + H2Y^{2-}}

Analysis

In the first titration standard $\ce{Pb^{2+}}$ consumes the excess EDTA; at the endpoint the next drop of $\ce{Pb^{2+}}$ forms the red Pb–XO complex. $\ce{F^-}$ then strips EDTA from tin ( $\ce{SnY -> SnF_n + H2Y^{2-}}$ ), releasing $\ce{H2Y^{2-}}$ which destroys the red Pb–XO — hence the return to yellow — and the released EDTA is titrated in turn.