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 10.19 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 12.00 mL of 0.08454 M potassium permanganate (KMnO4) to reach the purple endpoint. A blank required 0.10 mL. Balance the reaction below and determine the percent WO3 in the ceramic sample.

MnO₄^-+ W³⁺--> Mn²⁺ + WO₂³⁺

___ % WO₃ ??

Answer 1

1. MnO₄^- + W³⁺ Mn²⁺ + WO₃

Oxidation half reaction (OHR):

W³⁺ WO₃

W³⁺ + 3H₂O WO₃ (balancing no. of oxygens)

W³⁺ + 3H₂O WO₃ + 6H⁺ (balancing no. of hydrogens)

W³⁺ + 3H₂O WO₃ + 6H⁺ + 3e^- (balancing no. of hydrogens)

Reduction half reaction (RHR):

MnO₄^- Mn²⁺

MnO₄^- Mn²⁺ + 4H₂O (Balancing no. of oxygens)

MnO₄^- + 8H⁺ Mn²⁺ + 4H₂O (Balancing no. of hydrogens)

MnO₄^- + 8H⁺ + 5e^- Mn²⁺ + 4H₂O (Balancing no. of electrons)

The complete balanced equation can be written by multiplying OHR with 5 and RHR with 3 and adding both of them

i.e. 3MnO₄^- + 5W³⁺ + 3H₂O 3Mn²⁺ + 5WO₃ + 6H⁺

No. of moles of KMnO₄ added to reach the purple end point = 12*10^-3 L * 0.08454 M, i.e. 1.0145*10^-3

No. of moles of W³⁺ used in the above reaction = (5/3) * 1.0145*10^-3, i.e. 1.6908*10^-3 (since according to the balanced equation, for 3 moles of MnO₄^-, 5 moles of W³⁺ is required).

Therefore, the no. of moles of WO₃ formed = no. of moles of W³⁺ added, i.e. 1.6908*10^-3 (according to the balanced equation).

The amount of WO₃ formed = 1.6908*10^-3 mol * 231.837 g.mol^-1, i.e. 0.392 g.

Hence, the % of WO₃ in the given ceramic sample = (0.392/10.19)*100, i.e ~ 3.85 %