Solution of `CH_3 COOH [Ka = 10^(-5) " at " 25^@ ] ` is been neutralized by using NaOH solution. pH of the solution is x at 20% neutralization and it is y at 80% neutralization then value of y-x will be `[ log_(10)^(2) = 0.30 ]`

Explanation

Incomplete Neutralization will provide an acidic buffer.
` pH = pKa + "log" (["salt"])/([W.A])`
` pH = x = pKa + " log " 20/80 ` [at 20% Neutralization]
`pH = y = pKa + " log " 80/90 ` [at 80% Neutralization]
So y - x = 2 log ` 80/20 = 2 log 4 = 1.20 `