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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)`
` 5 xx10^(-16) = 4S^(3)`
`S^(3)= 1.25 xx10^(-16) = 125 xx10^(-18)`
`S = 5 xx 10^(-6)`
`[OH^(-)] = 2xx5 = 2xx 5 xx 10^(-6) = 10^(-5)`
`pOH = - log [OH^(-)] = - log (10^(-5)) = 5`
`pH = 14 - 5 = 9`
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