Question

Bromophenol is an acid indicator with a ${{K}_{a}}$ value of $6\times {{10}^{-5}}$. What % of this indicator is in its basic form at a pH of 5?

Answer 1

To solve this, use the Henderson-Hasselbalch equation. The equation is given as- $pH=p{{K}_{In}}+\log \left( \dfrac{\left[ I{{n}^{-}} \right]}{\left[ HIn \right]} \right)$. Use the ${{K}_{a}}$ to find out the $p{{K}_{In}}$ and pH is already given in the question. You can use this to solve the given problem.

Complete step by step solution:
We know that in chemistry, we use indicators for the determination of the pH of a solution. Different indicators work differently in acidic, basic and neutral solutions.
Due to the property of indicators having different colour in acidic and basic solutions, they are also known as acid – base indicators.
Generally, the indicators are weak acids that are one colour in its protonated form i.e. acidic form and a different colour in de-protonated i.e. basic form. Similar to other weak acids, they also dissociate according to its ${{K}_{a}}$ value.
For the indicator, we can write the dissociation reaction as -
$HIn(aq)\rightleftharpoons {{H}^{+}}(aq)+I{{n}^{-}}(aq)$
So, for the above reaction we can write ${{K}_{a}}$ as -
$\implies {{K}_{a}}=\dfrac{\left[ {{H}^{+}} \right]\left[ {{X}^{-}} \right]}{\left[ HX \right]}$
We can calculate the pH of a buffer solution by the Henderson-Hasselbalch equation. So, converting the above equation in this form, we will get -
$pH=p{{K}_{In}}+\log \left( \dfrac{\left[ I{{n}^{-}} \right]}{\left[ HIn \right]} \right)$
Now, in the question, pH is given to us as 5 and ${{K}_{a}}$ is given as $6\times {{10}^{-5}}$ .
$p{{K}_{In}}=-\log \text{ }{{\text{K}}_{a}}=-\log \left( 6\times {{10}^{-5}} \right)=4.22$
So, putting the value in the above equation, we will get-

& 5=4.22+\log \left( \dfrac{\left[ I{{n}^{-}} \right]}{\left[ HIn \right]} \right) \\\ & Or,\left( \dfrac{\left[ I{{n}^{-}} \right]}{\left[ HIn \right]} \right)=\dfrac{100}{16.6} \\\ \end{aligned}$$ The Henderson-Hasselbalch equation shows that the ratio of $I{{n}^{-}}$ and $HIn$ is indicated by pH and the $p{{K}_{a}}$ of the solution. As the two species have different colours, the ratio will determine the solution’s overall colour. So, % of indicator in its basic form will be $\dfrac{100}{100\times 16.6}\times 100=85.7%$ **Therefore, the correct answer is 85.7%.** **Note:** In chemistry we use indicators during titration to mark the end point of a titration. We see that in acid-base titrations, generally the end is marked by the change in colour of the solution. Depending upon the titration, it is very important for us to choose suitable indicators which will indicate the equivalence point.