Equilibrium Question 333

Question: A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is [BVP 2003]

Options:

A) $ 1\times {10^{-8}} $

B) $ 1\times {10^{-4}} $

C) $ 1\times {10^{-6}} $

D) $ {10^{-5}} $

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Answer:

Correct Answer: A

Solution:

$ K=\frac{{{\alpha }^{2}}C}{1-\alpha };\alpha =\frac{0.01}{100}\approx 1,\therefore K={{\alpha }^{2}}C={{[ \frac{0.01}{100} ]}^{2}}\times 1 $

$ =1\times {10^{-8}} $ .



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