Question

A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is ______.

पर्याय

• 1.0 × 10⁻³

• 1.0 × 10⁻⁶

• 1.0 × 10⁻⁸

• 1.0 × 10⁻¹¹

Answer 1

A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is 1.0 × 10⁻¹¹.

Explanation:

Monoprotic Acid (assume)

\[\ce{CH3COOH <=> CH3COO^- + H^+}\]
t = 0	C	0	0
t = ?	C − Cα	Cα	Cα

∵ K_a = `("C"^2α^2)/("C"(1 - α)) (1 - α)` = 1 for weak Acid

K_a = Cα² ....(α = very less)

= `0.1 xx ((0.001 xx 1)/100)^2`

= 1.0 × 10⁻¹¹