MHT CET · Chemistry · Electrochemistry
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 ?
- A \(4.6 \times 10^{-3} \mathrm{M}\)
- B \(4.5 \times 10^{-2} \mathrm{M}\)
- C \(4.0 \times 10^{-4} \mathrm{M}\)
- D \(3.6 \times 10^{-2} \mathrm{M}\)
Answer & Solution
Correct Answer
(B) \(4.5 \times 10^{-2} \mathrm{M}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5} ; \alpha=0.02\)
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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