(આપેલ : $\left.pK _{ a }\left( CH _{3} COOH \right)=4.76\right)$
$\log 2=0.30$
$\log 3=0.48$
$\log 5=0.69$
$\log 7=0.84$
$\log 11=1.04$
\(NaOH + CH _{3} COOH \longrightarrow CH _{3} COONa + H _{2} O\)
\(2.5\, m\, mole \quad\quad 2.5 \,m\,mole\)
\(0 \quad\quad\quad\quad\quad 2.5 m \text { mole } \quad\quad2.5 m \text { mole }\)
so buffer is formed
\(pH = pKa +\log \left(\frac{2.5 / 75}{2.5 / 75}\right)= pKa\)
\(pH =4.76\)
\(=476 \times 10^{-2}\)