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 10.29 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.

Answer 1

Answer:

%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:

• Oxidation:

W⁺³ ⇒ WO₂⁺²

2H₂O + W⁺³ ⇒ WO₂⁺²

2H₂O + W⁺³ ⇒ WO₂⁺² + 4H⁺

2H₂O + W⁺³ ⇒ WO₂⁺² + 4H⁺ + 3e⁻    b)

• Reduction:

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:

$0.01143 L * 0.09713 M *\frac{5molesW}{3molesMnO4} =0.00185 moles$

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:

$\frac{2.145 g}{10.29g}$ * 100% = 20.84%