A 0.4405 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv

Question

A 0.4405 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2·4H2O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 300.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.26 mL of 0.001465 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 33.85 mL of the 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 27.15 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Explanation / Answer

0.4405 g sample of pewter

1) Titration1 (solution of lead, copper, and zinc )

V1=15.00ml (solution of lead, copper, and zinc )

C1=unknown

V2=volume of EDTA=35.26ml

C2=0.001465 M

Using formula,C1 V1=C2V2

                          C1=C2 V2/V1=0.001465 M*35.6 ml/15.00 ml=0.003477M or mol/L

2) titration2

V1=20.00ml(lead and zinc)

C1=unknown

V2=volume of EDTA=33.85ml

C2=0.001465 M

Using formula,C1 V1=C2V2

                          C1=C2 V2/V1=0.001465 M*33.85 ml/20.00 ml=0.002479 M

3) Titation 3

V1=25.00ml(lead only)

C1=unknown

V2=volume of EDTA=27.15 ml

C2=0.001465 M

Using formula,C1 V1=C2V2

                          C1=C2 V2/V1=0.001465 M*27.15ml/25.00 ml=0.001591 M

Let the concentrations of lead, copper, and zinc be denoted by X,Y,Z respectively(for convenience of calculations)

Now from the titration 3 results,

So lead in 300ml (acid solution initially prepared)=0.3 L*0.001591 mol/L=0.0004773 moles Pb=0.0004773 moles* molar mass of pb=0.0004773 moles *207.2g/mol=0.0989 g

(pb + Zn) X+Z=0.002479 mol/L

So moles of Zn and Pb in 300 ml=0.002479 mol/L* 0.3L=0.0007437 moles

But in (1) we found that moles of Pb in 300 ml metal solution=0.0004773 moles

Therefore moles of Zn=0.0007437 moles-0.0004773 moles=0.0002664 moles

Mass of Zn=moles*molar mass=0.0002664 moles*65.38 g/mol=0.01742g

3)from results of titration 1,

X+Y+Z=concentration of (pb+Zn+Cu)= 0.003477 mol/L

Total Moles of (pb+Zn+Cu) in 300 ml =0.003477 mol/L*0.3L=0.001043 moles

Moles of Cu alone=total moles-(moles of pb+ moles of Zn)= 0.001043 moles –(0.0002664 moles+0.0004773 moles)=0.0002993 moles

Mass of Cu=0.0002993 moles*63.55 g/mol=0.01902 g

4)Mass of Sn=total mass of pewter-(sum of masses of pb,zn,cu)=0.44050g-(0.0989 g +0.01742g +0.01902 g)=0.3052 g

5)Percent composition by mass-mass/total mass *100

Zn =0.01742g/0.44050g*100=4.0%

Pb=0.0989 g/0.44050g*100=22.4%

Cu=0.01902g/0.44050g*100=4.3%

Sn=0.3052 g/0.44050g*100=69.3%

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