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What is the molar concentration of acetic acid if value of its dissociation constant is $1.8 \times 10^{-5}$ and degree of dissociation is 0.02 ?
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The correct answer is:
$4.5 \times 10^{-2} \mathrm{M}$
$\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5} ; \alpha=0.02$
For a weak monobasic acid,
$\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}$
$\therefore \quad c=\frac{\mathrm{K}_{\mathrm{a}}}{\alpha^2}=\frac{1.8 \times 10^{-5}}{0.02^2}=0.045 \mathrm{M}=4.5 \times 10^{-2} \mathrm{M}$
For a weak monobasic acid,
$\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}$
$\therefore \quad c=\frac{\mathrm{K}_{\mathrm{a}}}{\alpha^2}=\frac{1.8 \times 10^{-5}}{0.02^2}=0.045 \mathrm{M}=4.5 \times 10^{-2} \mathrm{M}$
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