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Calculate The [H^(+)] ions of 0.008 M Ca...

Calculate The `[H^(+)]` ions of `0.008 M Ca(OH)_(2)` is

A

`2xx10^(-7)M`

B

`6.25xx10^(-13)M`

C

`1.4xx10^(-12)M`

D

`1.25xx10^(-11)M`

Text Solution

Verified by Experts

The correct Answer is:
B
`Ca(OH)_(2)toCa^(+2)+2OH`
`008-0.08" "0.08" "2xx0.08" "0.16M`
`[H^(+)][OH]=10^(-14)`
`[H^(+)]=(10^(-14))/([O^(-)H])=(10^(-14))/([0.16])=6.25xx10^(-13)M`
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