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Question

Chemistry Question on Ionic Equilibrium In Solution

The ionisation constant of acetic acid is 1.8×10−51.8 \times 10^{-5}. At what concentration will it be dissociated to 2% ?

A

1 M

B

0.018 M

C

0.18 M

D

0.045 M

Answer

0.045 M

Explanation

Solution

For weak acid Ka=Cα2K_{a}=C\alpha^{2} or C=Ka/α2C=K_{a}/\alpha^{2} =1.8×10−5.02×0.02=1.8×10−54×10−4=\frac{1.8\times10^{-5}}{.02\times0.02}=\frac{1.8\times10^{-5}}{4\times10^{-4}} =0.045 M=0.045\,M