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 10.29 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 11.43 ml of 0.09713 m potassium permanganate (kmno4) to reach the purple endpoint. a blank required 0.25 ml. balance the reaction below and determine the percent wo3 in the ceramic sample.

%WO₃ = 20.84%

Explanation:

The half-reaction for the reduction of WO₃ is:

a) W⁺⁶ +3e⁻ ⇒ W⁺³

The reaction that takes place in the analysis is

MnO₄⁻ + W⁺³ ⇒ Mn⁺² + WO₂⁺²

In order to balance this reaction, we write the two half-reactions:

W⁺³ ⇒ WO₂⁺²

2H₂O + W⁺³ ⇒ WO₂⁺²

2H₂O + W⁺³ ⇒ WO₂⁺² + 4H⁺

2H₂O + W⁺³ ⇒ WO₂⁺² + 4H⁺ + 3e⁻    b)

MnO₄⁻  ⇒ Mn⁺²

MnO₄⁻  ⇒ Mn⁺² + 4H₂O

MnO₄⁻ + 8H⁺  ⇒ Mn⁺² + 4H₂O

MnO₄⁻ + 8H⁺ + 5e⁻ ⇒ Mn⁺² + 4H₂O    c)

To balance charges, we multiply half-reaction b) by 5, and half-reaction c) by 3; and add them:

10H₂O + 5W⁺³ + 3MnO₄⁻ + 24H⁺ + 15e⁻  ⇒ 5WO₂⁺² + 20H⁺ + 15e⁻ + 3Mn⁺² + 12H₂O

We simplify the equation and we're left with the balanced reaction:

5W⁺³ + 3MnO₄⁻ + 4H⁺⇒ 5WO₂⁺² + 3Mn⁺² + 2H₂O d)

To determine the moles of W⁺³ in the aliquot, we make use of reaction d), using the known moles of MnO₄⁻ that reacted with the sample:

Now we determine the moles of W⁺³ in the sample of 500 mL:

0.00185 moles * 500.0 mL / 100.0 mL = 0.00925 moles

Due to half-reaction a), we know that the moles of W⁺³ are equal to the moles of WO₃ in the sample of the vase. Now we determine the mass of WO₃ in the sample, using the molecular weight:

0.00925 moles WO₃ * 231,84 g /mol= 2.145 g WO₃

All that's left now is to calculate the percent of WO₃ in the sample:

* 100% = 20.84%

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