1.

The solubility of a compound MX_(2) at 25^(@)C is 5.0xx10^(-3) mol/L. Its solubility product at that temperature is :

Answer»

`25xx10^(-6)`
`5.0xx10^(-11)`
`5.0xx10^(-7)`
`1.25xx10^(-9)`

Solution :`MX_(2)hArr M^(2+)+2X^(-)`
`K_(SP)=(x)(2x)^(2)=4x^(3)`
`=4xx(5.0xx10^(-3))^(3)`
`=5.0xx10^(-7)`


Discussion

No Comment Found

Related InterviewSolutions