b
\(\quad\quad\quad\quad\quad\quad2 {HA} \rightleftharpoons {H}_{2} {~A}_{2} \quad {HA} \rightleftharpoons {H}^{+}+{A}\)

\(\text { Initial moles } {a} \times \frac{50}{100} \quad\quad 0 \quad {a} \times \frac{50}{100} \quad0 \quad\quad 0\)

\(\text { Final moles }\quad 0 \quad 0.25 {a} \quad\quad\quad0 \quad\quad 0.5 {a} \quad\quad 0.5 {a}\)

Now, \(i=\frac{\text { final moles }}{\text { initial moles }}=\frac{0.25 a+0.5 a+0.5 a}{0.5 a+0.5 a}\)

\(=1.25=125 \times 10^{-2}\)