`Ca_(3)(PO_(4))_(2)` is insoluble in in water. On adding a few drops of HCI to solid `Ca_(3)(PO_(4))_(2)` in contant with water,

`Ca_(3)(PO_(4))_(2)` is insoluble in in water. On adding a few drops of HCI to solid `Ca_(3)(PO_(4))_(2)` in contant with water, the solid dissolves. The reason is:
A. The solvent becomes more polar on adding HCl
B. `Ca_(3)(PO_(4))_(2)` combines with HCl to form soluble `CaCl_(2)` and `H_(3)PO_(4)`
C. `Ca(H_(2)PO_(4))_(2)` is formed, which dissolves
D. `H_(3)PO_(4)` a weak acid is formed and the solubility product of `Ca_(3)(PO_(4))_(2)` decrease

`Ca_(3)(PO_(4))_(2)` is insoluble in in water. On adding a few drops of HCI to solid `Ca_(3)(PO_(4))_(2)` in contant with water, the solid dissolves.

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

Please log in or register to add a comment.

Find MCQs & Mock Test

Related questions

Categories