When drying in air a suspension of WC particles and isohexane, the dried cake is soft. Switching isohexane with acetone and drying in the same conditions makes the cake harder. Using water makes the cake even harder.
Could anyone confirm or infirm my hypothesis that the polarity of the used solvent determines the properties of the dried cake. (Non polar - soft cake, the higher the polarity - the harder the cake)
Your question is interesting. It is related to the structure of carbide, which consists of metal (hydrophilic) and carbon (hydrophobic). Carbide has 2 allotropic modifications. Look what you have? Your hypothesis makes sense. But I would operate with the notion of adhesion, wetting, intermolecular interaction. Note that the carbide can be two-faced Janus. Can help you article
https://www.jim.or.jp/journal/e/pdf3/52/07/1471.pdf
Thank you for your response. I believe that, in my case, due to the fact that half of the powder is sub-micronic, it wouldn't sediment freely. In your proposed paper, can be seen that at low zeta potential, the WC particles are likely to agglomerate and form larger flocks. Also, the submicronic WC powder did not sediment, only the larger SiO2. In the case of larger particles, Stoke's law can apply.
Could this be the mechanism:
Non-polar -> agglomerates ->fast sedimentation ->soft cake
Polar solvents->disperse particle->slow sedimentation->hard and tight cake.
Did I understand correctly?
Non-polar ->disperse particle -> big agglomerates ->fast sedimentation ->soft cake
Polar solvents->disperse particles -> small agglomerates->low sedimentation->hard and tight cake.
First, the loss of aggregative stability, and then sedimentation . Interlayers of polar molecules better combine nanoparticles in cake .
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