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 3 of 4: Amounts in the 25 mL aliquot

← Previous Next →

n(\mathrm{EDTA})_{excess} = 0.02405\times0.009970 = 2.398\times10^{-4}\ \mathrm{mol};\quad n(\mathrm{Sn}^{IV}) = 0.01500\times0.009970 = 1.496\times10^{-4}\ \mathrm{mol};\quad n(\mathrm{Pb}) = (5.000 - 2.398 - 1.496)\times10^{-4} = 1.106\times10^{-4}\ \mathrm{mol}

Analysis

The aliquot originally held $0.02500\times0.2000 = 5.000\times10^{-4}$ mol EDTA. The first titre gives the unreacted EDTA ( $2.398\times10^{-4}$ mol); the second gives EDTA released from Sn, i.e. $n(\ce{Sn^{IV}}) = 1.496\times10^{-4}$ mol; lead follows by difference.