MCQ
A monoprotic acid in a $0.1\,\,M$ solution ionizes to $0.001\%$. Its ionisation constant is
  • A
    $1.0 \times {10^{ - 3}}$
  • B
    $1.0 \times {10^{ - 6}}$
  • C
    $1.0 \times {10^{ - 8}}$
  • $1.0 \times {10^{ - 11}}$

Answer

Correct option: D.
$1.0 \times {10^{ - 11}}$
(d) $\therefore$ Monoprotic acid $HA$

$HA$ $ \rightleftharpoons $ ${H^ + } + {A^ - }$

Ionisation constant = ?

$\alpha = 0.001\,\% = \frac{{0.001}}{{100}} = {10^{ - 5}}$

$K = \frac{{{\alpha ^2}}}{V} = \frac{{{{[{{10}^{ - 5}}]}^2}}}{{10}} = {10^{ - 11}}$.

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