Explanation

(d)
We know that,
$\left\lbrack H^{+} \right\rbrack = 10^{- pH} = 10^{- 5}$
$\alpha = \frac{actual\ concentration}{molar\ concentration}$
$= \ \frac{10^{- 5}}{0.005} = 0.2 \times 10^{- 2}$
$\therefore$ Percentage ionisation = $0.2 \times 10^{- 2} \times 100$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.2\%$