How to calculate the sufficient ratio of acid and corresponding base of the indicator that allows to clearly observe the endpoint of a titration? Ram can see the purple colour of an acidic indicator below pH = 5.5 only and the blue colour only above pH = 8.5. How many times that of acidic form must the concentration of the basic form at least be for him to be able to see the end point of the titration of HCl and NaOH. 


Attempt at Solution

Since the indicator is acidic it is of the form $\ce{HIn}$ and dissociates as follows - $$\ce{HIn&lt;=&gt;H^+ + In^-}$$ 

where $\ce{HIn}$ is acidic form and $\ce{In^-}$ is basic form so we need to find the minimum value of $\frac{[\ce{In-}]}{[\ce{HIn}]}$. The colour change happens at ph 5.5 and 8.5

At the end of point of titration of HCl and NaOH ${pH = 7}$ which automatically falls within the pH range of indicator ( 5.5 - 8.5 ), so i don't know what to do.

Now from the Henderson Hasselbach equation we have, $${pH = pK_{In} + \log \frac{[\ce{In-}]}{[\ce{HIn}]}}$$ So I guess we put 7 in place of pH and then I again do not know what to do now.

Please help me out with this problem.

Answer in my book is 32
If you assume that the $pK_{a}$ of the indicator is 7.0 and you assume that you're going to the end point at pH 8.5 then the Henderson Hasselbach equation


  
    ${pH = pK_{a} + log \frac{[In^-]}{[HIn]}}$
  


can be rearranged to:


  
    $\frac{[HIn]}{[In^-]} = 10^{(pH - pK_{In})} = 31.62+ \approx 32$
  


However:

(1) There is no real statement in the problem that $pK_{a}$ of the indicator is 7.0 which was assumed from:


  
    $\dfrac{1}{2}(5.5 + 8.5) = 7.0$
  


There is little real chemical validity behind this assumption. 

(2) There isn't a real justification of why color at pH 8.5 is any better than color at pH 5.5. So the ratio is either 32 or 1/32. Flip a coin :-(