Question

An acid type indicator, $\text{ HIn }$, differs in colour from its conjugate base $\text{I}{{\text{n}}^{-}}$. The human eye is sensitive to colour differences only when the ratio $\text{ }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ }$ is greater than 10 or smaller than $\text{ 0}\text{.1 }$. What should be the minimum change in $\text{ pH }$ of solution to observe a complete colour change? ($\text{ Ka = 1}{{\text{0}}^{\text{-5}}}\text{ }$)

Answer 1

Hint: The indicators are used to detect the change in the $\text{ pH }$of the solution. These have a very wide application in acid-base titration reaction. The acid indicators are expressed as $\text{ HIn }$dissociates into the $\text{I}{{\text{n}}^{-}}$. The range of the $\text{ pH }$ in which the complete change in the colour of the indicator can be observed is determined by the Henderson –Hasselbach equation. For Indicator, $\text{ HIn }$the equation is given as,
$\text{ pH = p}{{\text{K}}_{\text{In}}}\text{ + log }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}$
Where, $\text{ }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ }$ is the ratio of the concentration of dissociated to undissociated form and ${{\text{K}}_{\text{In}}}$ is the dissociation constant of the indicator.

Complete step by step answer:
Ostwald developed a theory of acid-base indicators which offers a simple explanation for the colour change with the variation of $\text{ pH }$. According to this theory, hydrogen ion indicators are either weak organic acid or base. The undissociated, molecules have one colour and the ion furnished by it, on dissociation has another the colour.
Let the indicator be the acid of the formula $\text{ HIn }$. Then, its dissociation in solution may be represented as,
$\text{ }\begin{matrix}
\text{HIn} & \rightleftharpoons & {{\text{H}}^{+}} & + & \text{I}{{\text{n}}^{-}} \\
\text{Colour A} & {} & {} & {} & \text{Colour B} \\
\end{matrix}\text{ }$
The undissociated molecule $\text{ HIn }$ has one colour say colour A and the dissociated ion $\text{I}{{\text{n}}^{-}}$ have colour B.
The equilibrium constant for the indicator may be written as,
$\text{ }{{\text{K}}_{\text{In}}}\text{ = }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ }$
We have to find the $\text{ pH }$ change associated with the indicator when the ratio $\text{ }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ }$ is varied.
Case a) when the ratio is $\text{ }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ = 10 }$
Applying the Henderson –Hasselbach equation the $\text{ }{{\text{p}}_{\text{1}}}\text{H }$ of the indicator is calculated as follows,
$\begin{align}
& \text{ }{{\text{p}}_{\text{1}}}\text{H = p}{{\text{K}}_{\text{In}}}\text{ + log }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ }\because {{\text{K}}_{\text{In}}}=1\times {{10}^{-5}}\text{ } \\
 & \Rightarrow {{\text{p}}_{\text{1}}}\text{H }=\text{ 5 + log 10 } \\
 & \therefore {{\text{p}}_{\text{1}}}\text{H }=\text{ 5 + 1} \\
\end{align}$
Case b) when the ratio is $\text{ }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ = 0}\text{.1 }$
Applying the Henderson –Hasselbach equation the $\text{ }{{\text{p}}_{2}}\text{H }$ of the indicator is calculated as follows,
$\begin{align}
& \text{ }{{\text{p}}_{2}}\text{H = p}{{\text{K}}_{\text{In}}}\text{ + log }\dfrac{\left[ \text{I}{{\text{n}}^{-}} \right]}{\left[ \text{HIn} \right]}\text{ }\because {{\text{K}}_{\text{In}}}=1\times {{10}^{-5}}\text{ } \\
 & \Rightarrow {{\text{p}}_{\text{1}}}\text{H }=\text{ 5 + log 0}\text{.1 } \\
 & \therefore {{\text{p}}_{\text{1}}}\text{H }=\text{ 5 }-\text{ 1} \\
\end{align}$
If we observe the case a) and case b), we see that the change in the $\text{ pH }$is equal to the $\text{ +1 }$ or $\text{ }-1\text{ }$ to that of the $\text{ p}{{\text{K}}_{\text{In}}}\text{ }$ value. That is,
$\text{ pH change = }\pm \text{ 1 }$

Therefore, the minimum change in $\text{ pH }$ of the indicator should be in the range of $\pm \text{ 1 }$ to the $\text{ p}{{\text{K}}_{\text{In}}}\text{ }$ value of the indicator to observe the visible and colour change.

Note: One of the most common acid-base indicators is phenolphthalein. It is a weak acid. In acidic medium i.e. $\text{ pH }<\text{7 }$, it is predominately in undissociated form. Thus colourless In basic medium i.e. $\text{ pH }>\text{7 }$, it loses its proton and exists in an undissociated form (anion). Due to extended conjugation, in the basic medium, it is pink in colour. Thus phenolphthalein is used for the acid-base titration reaction to determine the end point.