Question

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`

Answer 1

Correct Answer - 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`