Solution of an acid and it's anion (that is it's conjugate base) or of a base and it's common cation are buffered. When we add a small amount of acid or base to any one of them, the pH of solution changes very little pH of buffer solution can be compoted
as , for , acidic , buffer
`pH = pK_(a) + "log"(["Conjugate base"])/(["Acid"])`
for basuc buffer
`pOH = pK_(b) + "log" (["Conjugated base"])/(["base"])`
it is generally accepteed that a solution has useful buffer capacity (`pH` change resistng power) provided that the value of [salt or conjugate base]/[acid] for acidic acid buffer lies within the range of `1:10` to `10:1`. Buffer capacity is max.
When [conjugate base]=[acid]
One litre of an aqueous solution contaion `0.15` mole of `CH_(3)COOH(pK_(a) = 4.8)` and `0.15` mole of `CH_(3)COONa`. After the addition of `0.05` moe of solid `NaOH` to this solution, the `pH` will be:

Options

A. `4.5`
B. `4.8`
C. `5.1`
D. `5.4`

Answer

C

Explanation

`{:(CH_(3)COOH,,CH_(3)COONa),(I:" "0.15,," "0.15):}`
after addition of `0.05` moles of `NaOH`
`{:(CH_(3)COOH,+,NaOH,rarr,CH_(3)COONa),(0.05,,0.05,,0.05):}`
left finally `{:(CH_(3)COOH,rarrCH_(3)COONa),(0.1 " "0.2,):}`
`:. pH = 4.8 + "log" (0.2)/(0.1) = 5.1`