Tungsten trioxide (WO3) has a rich yellow color and is often used as a pigment i

Question

Tungsten trioxide (WO3) has a rich yellow color and is often used as a pigment in ceramics and paints. In order to test a ceramic vase for its WO3 content, a 10.22 g sample of the vase was ground and reduced with Pb(Hg) to convert any WO3 to W3 . The resulting W3 was transferred to 500.0 mL of 1.00 M HCl. A 100.00 mL aliquot of the HCl solution required 14.66 mL of 0.08808 M potassium permanganate (KMnO4) to reach the purple endpoint. A blank required 0.26 mL. Balance the reaction below and determine the percent WO3 in the ceramic sample.

The balanced reaction is: 3 MnO4 + 5 W + 4 H ------> 3 Mn + 5 WO2 + 2 H2O

Explanation / Answer

Tungsten trioxide (WO3) has a rich yellow color and is often used as a pigment in ceramics and paints. In order to test a ceramic vase for its WO3 content, a 10.22 g sample of the vase was ground and reduced with Pb(Hg) to convert any WO3 to W3 . The resulting W3 was transferred to 500.0 mL of 1.00 M HCl. A 100.00 mL aliquot of the HCl solution required 14.66 mL of 0.08808 M potassium permanganate (KMnO4) to reach the purple endpoint. A blank required 0.26 mL. Balance the reaction below and determine the percent WO3 in the ceramic sample.

The Balanced reaction is : 3 MnO4 + 5 W + 4 H ------> 3 Mn + 5 WO2 + 2 H2O

sample taken=10.22g

solution prepared=500mL

sample taken=100mL=1/5 of original volume

3 MnO4 + 5 W + 4 H ------> 3 Mn + 5 WO2 + 2 H2O

volume of KMnO4sed for blank=0.26mL

Volume of KMnO4 used for sample=14.66-0.26=14.4mL

moles of KMnO4 used=0.08808moles/L*0.0144L=0.00127moles

from above equation 3mole of MnO4- reacts with =1.667 molesof W3

0.00127moles of KMnO4 reacts with=1.667/3*0.00127=0.00071 moles of W3

so 0.00071moles of W3 arepresent in(100mL) 1/5volume so

total of 0.00353moles of W3 present

Next Three moles of WO3 are produced from one moles of W3 hence

moles of WO3=3*moles of W3=3*0.00353moles=0.0106moles

mass of WO3=0.0106moles*231.84g/mole=2.454g

%age of WO3=2.454g/10.22g*100=24.01%

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