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Chemistry
A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is
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Q. A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is
Equilibrium
A
$1.0 X 10^{-3}$
8%
B
$1.0 X 10^{-6}$
13%
C
$1.0 X 10^{-8}$
19%
D
$1.0 X 10^{-11}$
60%
Solution:
Ionisation constant, $K_{a}=\frac{a^{2} \times c}{1-a}$
$\alpha$= degree of dissociation $= 0.001 \%$
$=\frac{0.001}{100}=1\times10^{-5}$
$C$= concentration $= 0.1 \,M$
when $\alpha$ is very small
$K=\alpha^{2} \times C$
$=(1 \times 10^{-5})^{2} \times 0.1=1\times 10^{-11}$