Question

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A 0.4721 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO 4 H,O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 250.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.45 mL of 0.001518 M EDTA. Thiosulfate was used to mask the copper in a second 20.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.78 mL of the 0.001518 M EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 25.00 mL aliquot. Titration of the lead in this aliquot required 26.51 mL of the 0.001518 M EDTA solution. Determine the percent composition by mass of cach metal in the pewter sample. Zn: % Cu: % % Sn: Pb:

Answer 1

2 we are concentration using of each different volure, so measure on rathen than moles, Clast the non Pb + EDSA mmoles Pb 2 Pb EDTA Complex mmole ForA added [pb] x 28-00mL 2 0.001518 mx 26.51 ml Cpb) 2 o-001610m 20-0016096872 m. Ć last but one) Pb + Zn + GOTA - PbEOTA + 2nfora mmoles [Pb + 2n) immoles fora. ( [pb] + [an] x 20.00m 2 0.001518mx 34-78 m. 20-0026398027 Co-oolsomt (zn)) = 0.002640 m. Cand 2 0-002640 - 0-001610 < 0.001030 ist hinahon & LEOTA & znGOTA + Cutora pbt cut an t Gora mmoles [Pb tan tu) GOTR. immoles (CAD) + (zn) + (cu)] x 15.00m 2 0.001518 mx 34 45 ml z Total concentration & 0-003486 m. 2) [code 0-003486m - ((26)+ (2n)) 2 0-60348y- .00169 +0-00130) 2 0.0008465 6972

now calculate moles and then mass of each weral r. 6822 moles of Pb 2 (Pb) x von Lines) z o-ool6g9m x 200.0mL X 1631 2 0-40284163 smoles mass Pb 2 0-40 29x10 moles x 207-25 mol 2 0-0833869 moles za 2 0.001030m x 250-ome toll 2 0-2575 moles mass an 2 0-2575 x 103 moles *65-38 9/mo) on 20.016835389 moles @ wz 0-00 08466 x 250-ome x toy 2 0. 211835 103 mors mass w 2 0.2116 8 10 moles x 63.546 g/mc) LO 2 0-01344, SSS so mass sn & z Total mass 0-4729 - (mass Pb+ 20+ co] C0-0833908 + 0.0168 FO) to-03448858 - al mass w 2 t mass an 2 O- 35849 0.013448558 Ou2435d x 1007. z 12-849 +- 6 0168373g xion 0-4921 1004 13.566725 mass pk . mass sample xlooy. 0.0833818 - 0-4721 17.664 0-4921 to mass Pb a mass sn - 0 35843 x loot 75-924.7

at pH₂q, ayurz 5.40167 log kf 2 12-7 2) If 2 1012-7 25x10'2 Kpl 2 Lyk x kg a 5.4x152x5x102 2 2.7x108 - 3x10 equivalence pointy ut2 + yur y complex) moles of ut? maes of eata resent. added Calculate 0-0558 mx 28-54ml -0279m x una

2) the 57 of mL =stime 5-708 mit This so all added EDTA is in is before uth converted exces. equivalence to form point, complex. C complex) z mmdes ut? Tored volume lur) 0-0279 mx 5.708 mL 28.54mL +5.708 ml 0-00465m. CERTAD OLL 2 Toral - reached Tord roles Gella z moles uth of eqchalence Powe 2 0 -0274m [57.08 - 5.708 ml 28.54ml +5.708 ml 2 0.04185m. 2 Ut2 + y x 0.0185tX complex 0.00465-7 E Kpl 2 (complex] [u+2) Love Cedra 3x1011 2 (0.00465) Co-04185) Cor utz z Casca, 3-771073 2 4x1013 s pita – 10g (4x103) [ 12.4 Calwiate same for 22-83 mL to 56-97 mL (before eaunalence points)

57.08 met at equivalence point all edra reached, no excess Edra. Ccomplexa 2 0.0279 mx 57.08 ml 28.54ml +5t-of mi *2 20-0186 m. 2) kel? (complex] Cutz) cesta 2) 3x10 2 0-0186 2) x2 Cutz) ? 2x107 2 2:48+10 ~ 2x107 put2 a 10g (2-58107) 2.66 57-19mL7 after equulence point. uth as an excess Cutz 2 Cut2) Total modes - mmoles reached Total volume & 2 0.0279 m (stom - 57.08 mL) 28-54 mL & 5748 ml uth volume Edra volume - left 2 3-6 x 10 m 2 put2 2 4.44)