1.

Calculate the solubility of A2xx3 pure water. Assuming that neither king of ion reacts with water. The solubility product of A2X3Ksp=1.1xx10-23 Explain with equation.

Answer»

Solution :`k_(sp)=[A^(3+)]^(2)[X^(-2)]^(3)=[2s]^(2).[3s]^(3)=108s^(5)`
`S^(5)=(1.1xx10^(-23))/108=1xx10^(-25)`
`S^(5)=1xx10.5"mol"DM^(-3)`


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