Question

The indicator constant for an acidic indicator, $\ln$ is $5 \times {10^{ - 6}}{\text{M}}$, this indicator appears only in the colour of acidic form when $\dfrac{{\left[ {{\text{l}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{Hln}}} \right]}} \leqslant \dfrac{1}{{20}}$ and it appears only in the colour as basic form when $\dfrac{{\left[ {{\text{Hl}}{{\text{n}}^ {-} }} \right]}}{{\left[ {{\text{i}}{{\text{n}}^ {-} }} \right]}} \leqslant 40$. The pH range of indicator is:
A) $4.3 - 6.3$
B) $4.0 - 6.6$
C) $4.0 - 6.9$
D) $3.3 - 6.6$

Answer 1

To answer this question, you should recall the concept of the equivalence point. The equivalence point of a titration is where we have mixed the two substances in exactly equation proportions. Use the Henderson Hasselbach equation to find the pH range.

Formulas used:
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{\left[ {{\text{conjugate base}}} \right]}}{{\left[ {{\text{acid}}} \right]}}$

Complete step by step solution:
An indicator is a chemical substance which when added to acid or base changes its colour up to a limit with variation in pH of the solution to which it is added. Indicators usually are either weak acids or weak bases with a special characteristic of exhibiting different colours in the ionized and unionized form e.g. phenolphthalein is a weak acid. Buffer Solution refers to a mixture containing a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base. They resist a change in pH upon dilution or upon the addition of small amounts of acid/alkali to them. The Henderson-Hasselbalch equation can be written as:
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{\left[ {{\text{conjugate base}}} \right]}}{{\left[ {{\text{acid}}} \right]}}$.
Conjugate base refers to $\left[ {{\text{l}}{{\text{n}}^ - }} \right]$and acid refers to ${\text{Hln}}$.
$\therefore {\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{\left[ {{\text{l}}{{\text{n}}^ - }} \right]}}{{\left[ {{\text{Hln}}} \right]}}$
Now substituting the values given in the question we have:
${\text{pH}} = 5.3 + \log \left( {\dfrac{1}{{20}}} \right)$.
After solving
${\text{pH}} = 3.99 \approx 4.0$
For the upper value, when the solution is basic.
$\Rightarrow {\text{pH}} = 5.3 + \log \left( {40} \right)$
We will get the final value as ${\text{pH}} = 6.9$.
Hence, the range is from $4.0 - 6.9$.

Thus, the correct answer to this question is option C.

Note:
The titration graphs help in determining the equivalence point:
1. Strong acid vs strong base: Neither indicator changes colour at the equivalence point. If we use phenolphthalein, we would titrate until it just becomes colourless (at pH $8.3$) because that point is the closest to the equivalence point. However, if we use methyl orange, the titration will be done until there is the very first trace of orange in the solution. If the solution becomes red, we are getting further from the equivalence point.
2. Strong acid vs weak base: Methyl orange starts to change from yellow towards orange very close to the equivalence point.
3. Weak acid vs strong base: Use of phenolphthalein is preferred as it changes colour exactly where you want it to.
4. Weak acid vs weak base: The curve is for a case where the acid and base are both equally weak. Neither indicator is of any use.