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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 - $$\ce{HIn<=>H^+ + In^-}$$

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

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  • $\begingroup$ Help guys... Please........... $\endgroup$ – user21540 Dec 22 '15 at 16:35
  • $\begingroup$ Frankly I don't understand the question - "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." ?!? // I'll point out that due to the nature of equilibriums the indicator is never ALL in the acid form or the base form. $\endgroup$ – MaxW Dec 22 '15 at 18:49
  • $\begingroup$ @max it is asking ratio of [In-]/[HIn] $\endgroup$ – user21540 Dec 22 '15 at 20:23
  • $\begingroup$ As indicator is acidic so HIn is the acidic form and In- must be the basic one $\endgroup$ – user21540 Dec 22 '15 at 20:25
  • $\begingroup$ re your assertion "so we need to find the minimum value of $\frac{[In^-]}{[HIn]}$." // That is clearly wrong. The ratio would be at a minimum in very, very acidic solution, not anywhere near pH 7. $\endgroup$ – MaxW Dec 22 '15 at 20:36
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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 :-(

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  • $\begingroup$ HCl and NaOH are both strong acids and bases so the solution should be neutral at end point. So isn't it that the pH at end poi t should have been 7? That is where I am confused now. $\endgroup$ – user21540 Dec 22 '15 at 23:02
  • $\begingroup$ Yes at pH 7 then [H+] = [OH-] so you're at the acid/base neutralization point. But pKa of the indicator is pH where [H+] = [In-] . The only way to get 7 for the pKa of the indicator is 1/2(5.5+8.5) which is really a stretch. $\endgroup$ – MaxW Dec 22 '15 at 23:08
  • $\begingroup$ No I understood how you got the pKa of the indicator as 7 but i do not understand why the pH of solution at end point is 8.5 or 5.5 .... I thought since the solution must be neutral at endpoint $\endgroup$ – user21540 Dec 22 '15 at 23:11
  • $\begingroup$ The question is really setup poorly so that you have to make some nebulous assumptions to get the right answer. You're literately flipping a coin to decide if pH 5.5 is better or worse than pH 8.5. $\endgroup$ – MaxW Dec 22 '15 at 23:11
  • $\begingroup$ The assumption in a titration is that the amount of acid (or base) to get from the neutralization point to the indicator end point is negligible. What this really means is that that a judicious choice of indicator has to be made. $\endgroup$ – MaxW Dec 22 '15 at 23:14

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