Answer:
pH after the addition of 10 ml NaOH = 4.81
pH after the addition of 20.1 ml NaOH = 8.76
pH after the addition of 25 ml NaOH = 8.78
Explanation:
(1)
Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles = 2 x 10⁻³ moles,
Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles
CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O
Initial conc. 2 x 10⁻³ 1 x 10⁻³ 0
Equilibrium 1 x 10⁻³ 0 1 x 10⁻³
Final volume = 20 + 10 = 30 ml = 0.03 lit
So final concentration of Acid = 
Final concentration of conjugate base [CH₃CH₂CH₂COONa]
Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .
Using Henderson Hasselbalch equation to find the pH
![pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%2Blog%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D%20%20%5C%5C%5C%5C%3D-log%281.54X10%5E%7B-5%7D%20%29%2Blog%5Cfrac%7B0.03%7D%7B0.03%7D%20%5C%5C%5C%5C%3D4.81)