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The solubility product of Zr(OH)(2) is 4...

The solubility product of `Zr(OH)_(2)` is `4.5xx10^(-17) "mol"^(3)L^(-3)`. What is solubility?

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`Zn(OH)_(2) hArr Zn^(++)=2OH^(-)`
`:. K_(SP)=S xx 4S^(2)=4S^(3)`
`:. 4S^(3)=4.5xx10^(-17)`
`S^(3)=""^(3) sqrt(11.7)xx ""^(3) sqrt((10^(-18))`
`S=2.6xx10^(-6)`
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