examstudy
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( Why do indicators all have different end point ranges?).
Hi, I am struggling with understanding how pH indicators work.
If indicators change colour when their: [Hin] = [in-] ...why does this occur at a different pH value for every indicator. How can there be equal amounts of the weak acid and conjugate base present at different pH values for each indicator?!
Any help would be appreciated greatly.
Thanks
Last edited by examstudy; 1 year ago
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BobbJo
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different Ka

Ka = [H+][In-]/[HIn]
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examstudy
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(Original post by BobbJo)
different Ka

Ka = [H+][In-]/[HIn]
Sorry I still dont understand why doesn't pH = 7 for all when [Hin] = [in-]
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BobbJo
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(Original post by examstudy)
Sorry I still dont understand why doesn't pH = 7 for all when [Hin] = [in-]
[HIn]/[In-] = [H+]/Ka
if pKa = 7 then sure pH = 7
otherwise no
[H+] at which it changes colour depends on Ka

also pH is temperature dependent. 7 is pH of neutral water at 298K
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examstudy
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(Original post by BobbJo)
[HIn]/[In-] = [H+]/Ka
if pKa = 7 then sure pH = 7
otherwise no
[H+] at which it changes colour depends on Ka

also pH is temperature dependent. 7 is pH of neutral water at 298K
Thanks!
Essentially, different Ka so different dissociations therefore the different indicators require different amount of the base for the concs to be equal and therefore this occurs at different pH values?
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BobbJo
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(Original post by examstudy)
Thanks!
Essentially, different Ka so different dissociations therefore the different indicators require different amount of the base for the concs to be equal and therefore this occurs at different pH values?
Yes you basically got it
But I would rather say "different Ka so different degrees of dissociations" because the nature of dissociation may be considered the same since it is simply the release of a proton*, i.e, HA <-> H+ + A-.

*for a weak acid acting as indicator
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examstudy
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(Original post by BobbJo)
Yes you basically got it
But I would rather say "different Ka so different degrees of dissociations" because the nature of dissociation may be considered the same since it is simply the release of a proton, i.e, HA <-> H+ + A-.
Great thanks for your help!
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