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The solubility product of a sparingly so...

The solubility product of a sparingly soluble metal hydroxide `[M(OH)_(2)]` is `5xx10^(-16) mol^(3)dm^(-9)` at 298 K. Find the pH of its saturated aqueous solution.

A

5

B

9

C

11.5

D

2.5

Text Solution

Verified by Experts

The correct Answer is:
B

`M(OH)_(2) hArr underset(s)(M^(2+))+underset(2s)(2OH^(_))`
`K_(sp)=(s)(2s)^(2)=4s^(3):.s^(3)=(K_(sp))/(4)=(5 xx 10^(-16))/(4)`
`=1.25 xx 10^(-16)=125 xx10^(-18)`
`:. S=5 xx 10^(-6) M`
`i.e. |OH^(-)|=2 xx 5 xx 10^(-6)M =10^(-5)M`
`pOH =5.0 :. pH =14-5=9`
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