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 10.20 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 12.96 mL of 0.08427 M potassium permanganate (KMnO4) to reach the purple endpoint. A blank required 0.29 mL. Balance the reaction below and determine the percent WO3 in the ceramic sample.
5 answers
I'm leaving this one for you. After the last two you should be able to do large chunks of this on your own. If you have trouble post the balanced equations and explain in detail what you don't understand about the problem/next step. I can help you through it but one step at a time.
Ok well I have this one however it keeps telling me to not forget to take into account that a blank sample required 0.29 mL of the potassium permanganate solution to reach the purple endpoint. To determine the volume of potassium permanganate solution that actually reacted with the W3 in the 100.0 mL aliquot, subtract the volume required for the blank from the volume required for the 100.0 mL sample.
What I did specifically is
(12.96ml)(0.08427M)(5e-)=(100mL)(?M)(3e-)
This gave me 0.01820232M
Then I took
(?M)(0.5L) = mol W which was 0.00910116
Then I took
(mol W)(231g/mol) = g W which was 2.10236796
Then to find weight percent I took this divided by the original and multiplied by 100%
2.10236796/10.2 = 0.2061145059 x 100% = 20.61%. However this is incorrect.
(12.96ml)(0.08427M)(5e-)=(100mL)(?M)(3e-)
This gave me 0.01820232M
Then I took
(?M)(0.5L) = mol W which was 0.00910116
Then I took
(mol W)(231g/mol) = g W which was 2.10236796
Then to find weight percent I took this divided by the original and multiplied by 100%
2.10236796/10.2 = 0.2061145059 x 100% = 20.61%. However this is incorrect.
I don't follow much of what you've done. That doesn't mean it isn't right. Apparently your database thinks you haven't corrected for the blank and I don't see that in your work either. Do you know how to do that? From your post I surmise that the data base told you exactly what to do (and did it very well). The titration took 12.96 mL, the blank was 0.29 ml, so the net volume used in the titration is 12.96-0.29 = 12.67 mL KMnO4 and that times M = mols KMnO4 used in titrating the W^3+. If you want to pursue this further I want to see those balanced equations.
Note that 12.67 is mL and must be converted to L before multilying by M to get mols.
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