The compound known as 12 CaO.7 Al
2O
3 corresponding to Ca
12Al
14O
33 has the possibility of anion substitution. The chemical formula of substitutional solid solution corresponds, to 11 CaO.7 Al
2O
3. CaX
2, where X is hydroxide ion or halogenide ion. In the present report, we investigated the substitution reaction of the fluoride ion between 12 CaO.7 Al
2O
3 and CaF
2. The phases in burned product were identified by mean of X-ray diffraction, and the amount of fluoride ion was measured with an ion-sensitive electrode., It was proved that 12 CaO.7 Al
2O
3 solid solution and CaO were formed by burning of 12 CaO.7 Al
2O
3 with CaF
2 at a temperature of 500°C. The compound 12 CaO.7 Al
2O
3 solid solution indicates 12 CaO.7 Al
2O
3 crystal occluding fluoride ions and the end member is 11CaO.7 Al
2O
3. CaF
2, that is, four fluoride ions are substituted for two oxide ions in the unit cell of 12 CaO.7 Al
2O, (Ca
24, Al
28O
66) as the maximum amount. Subsequently as the burning temperature was raised, 12 CaO.7 Al
2O
3 solid solution liberated fluoride ions gradually and was converted into 12 CaO.7 Al
2O
3. As the equimolecular mixture of 12 CaO.7 Al
2O
3 and CaF
2 was burned, the lattice constant (a
0) of 12 CaO.7 Al
2O
3 solid solution changed continuously. This showed that fluoride ions were occluded in 12 CaO.7 Al
2O
3 crystals and the value of a0 depended on the amount of fluoride ions. The value of a
0 decreased gradually with the rise of temperature between 400 and 1000°C, and then increased above 1000°C. Although 12 CaO.7 Al
2O
3, combined with CaO to form 3 CaO. Al
2O
3 at temperatures above 1050°C, 11CaO.7 Al
2O
3. CaF
2 formed 3 CaO. Al
2O
3 at above 1300°C. According to a quantitative analysis of the fluoride ions, 11CaO.7 Al
2O
3. CaF
2 reacted with CaO after the defluorination. These suggested that 11 CaO.7 Al
2O
3. CaF
2 was more thermostable than 12 CaO.7 Al
2O
3.
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