A 0.4436 g sample of pewter, containing tin, lead, copper, and zinc, was dissolv

Question

A 0.4436 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 200.0 mL. A 20.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.52 mL of 0.001486 M EDTA. Thiosulfate was used to mask the copper in a second 25.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.26 mL of the EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 30.00 mL aliquot. Titration of the lead in this aliquot required 26.33 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample.

Explanation / Answer

we will calculate in backward direction:

first we wil calculate no. of moles of Pb,

second we will find out no .of moles of Zn

thirdly, we will calculate no. of moles of Cu

lastly , the remaining % wolud be weight % of Sn

1. 26.33 mL of 0.001486 M EDTA is used in titration of 30 mL Pb solution

M1= molarity of EDTA=0.001486M

V1 = volume of EDTA used to titrate Pb= 26.33 mL

M2 = Molarity of Pb

V2= volume of Pb solution = 30 mL

by formula: M1*V1=M2*V2

26.33*0.001486=30*M2

M2= 0.0013M

i.e. Molarity of Pb= 0.0013 M

i.e. no. of moles of Pb = n1 = 0.0013 mole

since, 1 mole of Pb= molar mass of Pb= 207 g

therefore, 0.0013 mole of Pb= 207*0.0013= 0.2691 g

thus, weight of Pb in sample= 0.2691g

since , weight of sample given = 0.4436 g

therfore, % weight of Pb in sample= (weight of Pb*100)weight of sample= (0.2691*100).4436 = 60.6%

2. 34.26 mL of 0.001486M of EDTA is used in titration of 25 mL of mixture of Pb & Zn

let, M3= molarity of EDTA= 0.001486M

V3 = Volume of EDTA used= 34.26 mL

M4= molarity of mixture of Pb & Zn

V4 = volume of mixture of Pb & Zn = 25 mL

by formula, M3*V3= M4*V4

0.001486*34.26= M4*25

M4= 0.002 M

therefore, M4 = molarity of mixture of Pb&Zn = 0.002 M

i.e. no. of moles of mixture of Pb& Zn= 0.002 mole

no.of moles of mixture= no.of moles of Pb + no.of moles of Zn

therefore no. of moles of Zn=n2= 0.002-0.0013= 0.0007 moles

since, 1mole of Zn = molar mass of zinc= 65 g

therefore, 0.0007 mole of Zn = 0.0007*65 = 0.0478 g Zn

  thus, weight of Zn in sample= 0.0478g

since , weight of sample given = 0.4436 g

% weight of Zn in sample= (weight of Zn in sample*100)weight of sample

= ( 0.0478*100).4436

= 10.7%

3. 35.52 mL of 0.001486 M EDTA is used in titration of 20 mL of solution of mixture of Pb,Cu,Zn

let M5= Molarity of EDTA= 0.001486 M

V5= VOLUME OF EDTA = 35.52 mL

M6= molarity of mixture of Pb,Cu,Zn

V5= volume of mixture of Pb,Cu,Zn = 20 mL   

by formula , M5*V5= M6*V6

0.001486*35.52 = M6*20

   M6 = 0.0026 M

therefore no. of moles of mixture ( Pb,Cu,Zn) = 0.0026 moles

no. of moles of mixture ( Pb,Cu,Zn) = no. of moles (Cu) +  no. of moles (Pb) + no. of moles(Zn)

0.0026 =  no. of moles (Cu)+ 0.0013+0.0007

no. of moles (Cu)= n3 = 0.0006 moles

1 mole of Cu = molar mass of Cu = 63 g

thus, 0.0006 moles of Cu = 63*0.0006 = 0.0378 g

therfore,

weight of Cu in sample = 0.0378 g

weight of sample = 0.4436 g

therfore, weight % of Cu = (weight of Cu in sample *100)weight of sample

= (0.0378*100) 0.4436 = 8.52 %

therfore, weight % of Cu = 8.52 %

  % weight of Zn in sample =10.7%

% weight of Pb in sample= 60.6 %

thus , remaining weight is weigth % of Sn = 100- (60.6+10.7+8.52)

= 20.18 %

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