Question

The indicator phenol red is half in the ionic form when the pH is 7.2. The pH is altered if the ratio of the undissociated form to the ionic form becomes 1:5. At this new pH of solution, if the indicator is altered such that the ratio of undissociated form to dissociated form becomes 1:4, then find out the pH when 50% of the new indicator is in ionic form.
A ) 7.1
B ) 7.0
C ) 7.
D ) 7.3

Answer 1

Hint: Write the Henderson Hasselbalch equation for the acid buffer solution. Consider the undissociated form of the indicator as weak acid and the dissociated form as conjugate base.

Complete step by step answer:
Let \[{\text{HIn and I}}{{\text{n}}^ - }\]represent the undissociated and dissociated forms of the indicator. Consider the equilibrium for the dissociation of the indicator.
\[{\text{HIn }} \to {\text{ }}{{\text{H}}^ + }{\text{ + I}}{{\text{n}}^ - }\]
Write the expression for the pH of the indicator:
\[{\text{pH = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}}\]
When half of the indicator is dissociated and the remaining half is undissociated, so that \[\left[ {{\text{HIn}}} \right]{\text{ = }}\left[ {{\text{I}}{{\text{n}}^ - }} \right]\], the pH is 7.2.
\[{\text{pH = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}} \\
{\text{pH = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}} \\
{\text{7}}{\text{.2 = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}1 \\
{\text{7}}{\text{.2 = p}}{{\text{K}}_{{\text{In}}}}{\text{ + 0}} \\
{\text{p}}{{\text{K}}_{{\text{In}}}}{\text{ = }}7.2 \\\]
The ratio of undissociated to dissociated form is then changed to 1:5
\[{\text{pH = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}} \\
{\text{pH = 7}}{\text{.2 + lo}}{{\text{g}}_{10}}{\text{ }}\frac{5}{1} \\
{\text{pH = }}7.9 \\\]
The indicator is changed to make the ratio of undissociated form to dissociated form to 1:4.
\[{\text{pH = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}} \\
{\text{7}}{\text{.9 = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\frac{4}{1} \\
{\text{p}}{{\text{K}}_{{\text{In}}}}{\text{ = }}7.3 \\\]
Now, 50% of the new indicator is in dissociated form and the 50% is in undissociated form.
\[{\text{pH = p}}{{\text{K}}_{{\text{In}}}}{\text{ + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}} \\
{\text{pH = 7}}{\text{.3 + lo}}{{\text{g}}_{10}}{\text{ }}\dfrac{1}{1} \\
{\text{pH = }}7.3 \\\]

Hence, the option D ) 7.3 is the correct option.

Note: Use correct expression for the Henderson Hasselbalch equation. In this equation, natural logarithm is not used. Instead, logarithm to base 10 is used. Also, the 2.303 factor is not present in the equation. Also substitute the ratio as \[{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}} = \dfrac{1}{4}\]. Do not substitute the ratio as \[{\text{ }}\dfrac{{\left[ {{\text{I}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{HIn}}} \right]}} = \dfrac{4}{1}\] as it will lead to calculation mistake.