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Derive the equation of relation of solubility (S) of zirconium phosphate Zr_3(PO_4)_4 and K_(sp) solubility product. |
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Answer» SOLUTION :In concentrated salt solution the ionic equilibrium established between `Zr^(4+)` positive ion, `PO_4^(3-)`negative ion and insoluble MOLECULE `Zr_3(PO_4)_4`is as under `{:(Zr_3(PO_4)_"4(s)" HARR , 3Zr_((aq))^(4+) +, 4PO_(4(aq))^(3-)),(,3 S M, 4S M):}` `K=([Zr^(4+)]^3[PO_4^(3-)]^4)/([Zr_3(PO_4)_4])` But `[Zr_3(PO_4)_4]` = constant = `K^1` , because it is SOLID so concentration will be constant. `therefore K xx K^1` = Solubility product of sparingly soluble salt `K_(sp)` `therefore K_(sp)=[Zr^(4+)]^3 [PO_4^(3-)]^4` `=(3S)^3 (4S)^4` `=6912 S^7` `therefore S=(K_(sp)/6912)^(1/7)` |
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