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2.5 mL of (2M)/(5) weak monoacidic base ...

2.5 mL of `(2M)/(5)` weak monoacidic base `(K_(b) = 1xx10^(-12) at 25^(@)C)` is titrated with `(2)/(15)`MHCl in water at `25^(@)C`. The concentration of that equivalence point is:

A

`3.2 xx 10 ^(-13) M`

B

`3.2 xx 10 ^(7) M`

C

`3.2 xx 10^(-2) M`

D

`2.7 xx 10 ^(-2) M`

Text Solution

Verified by Experts

The correct Answer is:
D
`BOH + HCl to BCl + H _(2) O`
Meq. Of BOH = Meq. Of HCl = Meq. Of BCl
`2.5 xx (2)/(5) xx 1 = V xx (2)/(15) xx 1 xx 1 therefore V = 7. 5 mL therefore` Total volume `= 2.5 + 7.5 = 10 mL`
Thus,`[BCl] = (1)/(10) = 0. 1 ,` Now for hydrolysis of `BCl , K _(H) = (CH ^(2))/( 1 - H ) = ( K _(w))/( K_(b))`
`therefore h = 0. 27 or [H^(+) ] = CH.H= 0. 27 xx 0.1 = 2. 7 xx 10 ^(-2) M`
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