What is the molar concentration of acetic acid if value of its dissociation constant is $1.8 \times 10^{-5}$…

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 ?
  1. $4.6 \times 10^{-3} \mathrm{M}$
  2. $4.5 \times 10^{-2} \mathrm{M}$
  3. $4.0 \times 10^{-4} \mathrm{M}$
  4. $3.6 \times 10^{-2} \mathrm{M}$

Solution

$\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5} ; \alpha=0.02$ For a weak monobasic acid, $\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}$ $\therefore \quad c=\frac{\mathrm{K}_{\mathrm{a}}}{\alpha^2}=\frac{1.8 \times 10^{-5}}{0.02^2}=0.045 \mathrm{M}=4.5 \times 10^{-2} \mathrm{M}$

Asked in: MHT CET 2023 (13 May Shift 2)

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