Given :
Molarity of acetic acid solution, M = 0.10 M.
pKa of acetic acid, pKa = 4.75 .
To Find :
Percentage dissociation of 0.10 M solution of acetic acid.
Solution :
We know, 
Taking antilog both side, we get :
Since, acetic acid has 1 hydrogen atom to loose , so it is a monoprotic acid.
Now, percentage dissociation of monoprotic acid is given by :
Hence, this is the required solution.
Circular motion. Hope this helps!
The Answer is C
<span> It can be separated by physical means and is uniform in composition.</span>
<span>From left to right:
1) starts at high pH, no buffer zone - strong base being titrated with strong acid
2) starts at low pH, no buffer zone - strong acid being titrated with strong base
3) starts at low pH, shows a buffer zone - weak acid titrated with strong base
4) starts at high pH, shows a buffer zone - weak base titrated with strong acid
5) polyprotic due to two equivalence points
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