`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2 M `acetic acid to give `70 mL` of the solution. What is the `pH` of this solution. Calculate the additional volume of `0.2M NaOh` required to make the `pH` of the solution `4.74`. (Ionisation constant of `CH_(3)COOh` is `1.8 xx 10^(-5))`

Explanation

mmol of NaOH = `20 xx 0.2` = 4
mmol of acetic acid = `50 xx 0.2` = 10
After neutralization, buffer solution is formed which contain 6 mmol `CH_(3)COOH` and 4 mmol `CH_(3)COONa`
`pH = pKa + log [CH_(3)COONa]/[CH_(3)COOH] = -log(1.8 xx 10^(-5)) + log 4/6 = 4.56`
Now, let x mmol of NaOH is further added so that pH of the resulting buffer solution is 4.74.
Now, the buffer solution contains (4 + x) mmol `CH_(3)COONa`
`4.74 = -log(1.8 xx 10^(-5)) + log4+x/6-x`
`implies 4+x/6-x implies x = 1.0 mmol = 0.2 xx V implies V = 5 mL of NaOH`