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}} $
Show Answer
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}} $ .
