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enter image description here Hi, The question I have is above. Could someone explain why the answer is A? Thankyou, this is greatly appreciated (this is in preparation for an exam, it is not cheating)

What I think so far: -Does the sodium hydroxide turn the indicator blue because it is a strong base and hence it will react completely to form water, shifting equilibrium really far to the right? -Does the HCl turn the indicator red for a similar reason? -I am unsure of the sodium acetate effect

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  • $\begingroup$ Please note that the homework policy on this site requires you to show some personal effort. $\endgroup$ – Klaus-Dieter Warzecha Mar 7 '15 at 6:40
  • $\begingroup$ I've added some of my attempts on the question? Could you please help? $\endgroup$ – chemistryyo Mar 7 '15 at 6:45
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You're absolutely on the right track!

The keys to the question are

  1. to order $\ce{HCl}$, $\ce{NH4Cl}$, $\ce{NaOAc}$ and $\ce{NaOH}$ in terms of strong acid, weak acid, weak base and strong base

  2. to realize that in the presence of $\ce{HCl}$, the indicator exists as $\ce{H2Z}$, while it is fully deprotonated to $\ce{Z^{2-}}$ in the presence of $\ce{NaOH}$

  3. in the presence of weak acids or weak bases, the indicator exists in form of each two species. Consequently, a mixed colour is observed.

In the presence of $\ce{NaOAc}$, a weak base, $\ce{H2Y}$ is completely deprotonated, thus no red colour is observed. In addition, a part of $\ce{HY-}$ is deprotonated too: the second equilibrium isn't completely on the left side. Some amount of $\ce{Y^{2-}}$ is present too. As a result, the green colour (yellow + blue) is observed.

It is all about the position of the equilibria.

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  • $\begingroup$ Alright! I understand points 1 and 2, but why does the indicator exist as two species in the presence of weak base? I think I may have some sort of idea, but I would prefer it if you gave a definite explanation. Thankyou! $\endgroup$ – chemistryyo Mar 7 '15 at 6:57
  • $\begingroup$ @chemistryyo I've tried to address the last point too. $\endgroup$ – Klaus-Dieter Warzecha Mar 7 '15 at 7:06
  • $\begingroup$ Sorry, I just have another question: How do we know that H2Y is completely deprotonated. If this was a single equilibrium reaction, there should be some H2Y present $\endgroup$ – chemistryyo Mar 7 '15 at 7:13

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