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The pH of a buffer is 6.745. When 0.01 m...

The `pH` of a buffer is `6.745`. When `0.01` mole of `NaOH` is added to `1` litre of it, the `pH` changes to `6.832`. Its buffer capacity is

A

`0.187`

B

`0.115`

C

`0.076`

D

`0.896`

Text Solution

Verified by Experts

The correct Answer is:
B
Buffer capacity
`=` (No. of moles of base added//litre of buffer)/(Change in `pH`)
`=(0.01)/((6.832-6.745))=(0.01)/(0.087)=0.11`
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