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25 mL of 0.4 M of a weak base and 75 mL ...

`25 mL` of `0.4 M` of a weak base and `75 mL` of `0.2` M of its salt forms a buffer solution. The dissocation constant of base is `2 xx 10^(-5)` the `(log 1.5 = 0.176)`

A

`4.876`

B

`9.476`

C

`4.524`

D

`9.124`

Text Solution

Verified by Experts

The correct Answer is:
D
`pH = 14 - P_(kb) - "log" ([N_(s) xx V_(s)])/([N_(B) xx V_(B)])`
`= 14 - 4.6 - "log" (75 xx 0.2)/(25 xx 0.4)`
`14 - 4.6 - "log" (15)/(10) = 9.12`
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