Question

An indicator \[{\text{HIn}}\] has a standard ionization constant of \[{\text{9}}{\text{.0}} \times {\text{1}}{{\text{0}}^{ - 9}}\]. The acid colour of the indicator is yellow and the alkaline colour is red. The yellow colour is visible when the ratio of yellow form to red form is 30 to 1 and the red colour is predominant when the ration of red form to yellow form is 2 to 1. What is the pH range of the indicator?
A.\[ < 6.568\]
B.\[6.568{\text{ to 8}}{\text{.346}}\]
C.\[{\text{ > 8}}{\text{.346}}\]
D.None of these

Answer 1

Hint: Inthe above question is based upon indicator theory. We have to calculate the pH for both the colour range separately using the formula of pH of indicators hence we will get the range in which an indicator will work.
Formula used:
 \[{\text{pH = p}}{{\text{K}}_{{{\text{I}}_{\text{n}}}}} + \log \dfrac{{\left[ {{\text{I}}_{\text{n}}^ - } \right]}}{{\left[ {{\text{H}}{{\text{I}}_{\text{n}}}} \right]}}\]
Where p is a operator that represent \[ - {\text{log}}\], \[{{\text{K}}_{{{\text{I}}_{\text{n}}}}}\] is ionization constant, \[\left[ {{\text{I}}_{\text{n}}^ - } \right],\left[ {{\text{H}}{{\text{I}}_{\text{n}}}} \right]\] are concentration of the respective species.

Complete step by step answer:
The ionization of indicator will be as follow:
\[{\text{HIn}} \rightleftharpoons {{\text{H}}^ + } + {{\text{I}}_{\text{n}}}^ - \]
Let us calculate the pH when the solution has yellow colour, that is it is in acidic solution, the ionization constant is given to us and the ratio of yellow form to red form or acid form to base form is given that is 30 to 1. Here \[{\text{HIn}}\] will act as acid form and \[{{\text{I}}_{\text{n}}}^ - \] is basic form so \[\dfrac{{{\text{HIn}}}}{{{{\text{I}}_{\text{n}}}^ - }} = \dfrac{{30}}{1}\] or we can reverse the equation \[\dfrac{{{{\text{I}}_{\text{n}}}^ - }}{{{\text{HIn}}}} = \dfrac{1}{{30}}\]. Substituting the values we will get the value of pH in acidic form
\[{\text{pH = }} - {\text{log(9}} \times {\text{1}}{{\text{0}}^{ - 9}}) + \log \dfrac{2}{1}\]
Solving the above equation we will get, \[{\text{pH = 6}}{\text{.568}}\]
Now we will calculate the pH when the solution has red colour, that is it is in the basic solution, the ionization constant is given to us and the ratio of red form to yellow form or base form to acid form is given that is 30 to 1. Here \[{\text{HIn}}\] will act as basic form and \[{{\text{I}}_{\text{n}}}^ - \] is acid form because indicator is in basic solution. So \[\dfrac{{{\text{HIn}}}}{{{{\text{I}}_{\text{n}}}^ - }} = \dfrac{2}{1}\] . Substituting the values we will get the value of pH in acidic form
\[{\text{pH = - log(9}} \times {\text{1}}{{\text{0}}^{ - 9}}) + \log \dfrac{2}{1}\]
Solving the above equation we will get, \[{\text{pH = 8}}{\text{.346}}\].

Hence the range of the indicator will be \[6.568{\text{ to 8}}{\text{.346}}\] , i.e. Option B.

Note:
Indicators are weak organic acid or weak organic bases, which show different colours in ionized and unionized form. The intensity of colour will depend upon the concentration of colour imparting species. The two colour imparting species are: \[{\text{HIn}}\] and \[{{\text{I}}_{\text{n}}}^ - \].