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Ionic Equilibrum question

2004 · Q110
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Ionic Equilibrum question

2004 · Q110

NEETChemistryIonic EquilibrumMCQ+4 / −1
The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (HIn) and base (In−-−) forms of the indicator by the expression
  1. A
    log⁡[In−][HIn]=pKIn−pH\log {{\left[ {I{n^ - }} \right]} \over {\left[ {HIn} \right]}} = p{K_{In}} - pHlog[HIn][In−]​=pKIn​−pH
  2. B
    log⁡[HIn][In−]=pKIn−pH\log {{\left[ {HIn} \right]} \over {\left[ {I{n^ - }} \right]}} = p{K_{In}} - pHlog[In−][HIn]​=pKIn​−pH
  3. C
    log⁡[HIn][In−]=pH−pKIn\log {{\left[ {HIn} \right]} \over {\left[ {I{n^ - }} \right]}} = pH - p{K_{In}}log[In−][HIn]​=pH−pKIn​
  4. D
    log⁡[In−][HIn]=pH−pKIn\log {{\left[ {I{n^ - }} \right]} \over {\left[ {HIn} \right]}} = pH - p{K_{In}}log[HIn][In−]​=pH−pKIn​
View written solutionFree

Correct answer: D

For an acid-base indicator

HIn ⇌ H+ + In-

Kin = [H+][In−][Hin]{{\left[ {{H^ + }} \right]\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}[Hin​][H+][In−]​

⇒\Rightarrow⇒ [H+]=Kin[In−][Hin]\left[ {{H^ + }} \right] = {{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}[H+]=[Hin​]Kin​[In−]​

Take – log on both sides

−log⁡[H+]=−log⁡(Kin[In−][Hin]) - \log \left[ {{H^ + }} \right] = - \log \left( {{{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}} \right)−log[H+]=−log([Hin​]Kin​[In−]​)

⇒\Rightarrow⇒ pH = –log KIn + log⁡[In−][Hin]\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}log[Hin​][In−]​

⇒\Rightarrow⇒ pH = pKIn + log⁡[In−][Hin]\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}log[Hin​][In−]​

⇒\Rightarrow⇒ log⁡[In−][Hin]\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}log[Hin​][In−]​ = pH - pKIn

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