Question

In the titration of $C{{H}_{3}}COOH$ against NaOH, we cannot use the:
A. Methyl orange
B. Methyl red
C. Phenolphthalein
D. Bromothymol blue

Answer 1

Hint: An indicator is a substance that undergoes an observable change when the conditions in its solution change. This change could be a colour change, temperature change, precipitate formation or any other measurable quality. There are different types of indicators that have different ranges of pH.

Complete answer:
-Let us discuss a little bit about acetic acid first. As we know that acetic acid is a weak acid, which dissociates partially in water, we can see from the given equation:
\[C{{H}_{3}}COOH+{{H}_{2}}O\leftrightarrow C{{H}_{3}}CO{{O}^{-}}+{{H}_{3}}{{O}^{+}}\]
- Here we can see that acetic acid donates a ${{H}^{+}}$ion to water and forms the acetate ion. The thing to be Noticed here is that it is a reversible reaction, where the acetate ion accepts a ${{H}^{+}}$ ion from ${{H}_{3}}{{O}^{+}}$to form the acetic acid. It has the ability to accept a ${{H}^{+}}$, therefore it acts as a base. Acetate ion is known as the conjugate base of acetic acid.
- During the titration with sodium hydroxide, which is a strong base , when the end point is reached, all the acetic acid $C{{H}_{3}}COOH$molecules are been converted to acetate ions$C{{H}_{3}}CO{{O}^{-}}$.
- As we know that, $C{{H}_{3}}CO{{O}^{-}}$is the conjugate base of $C{{H}_{3}}COOH$.The acetate ions will then react with the water molecules that are present in the reaction flask, we can see it from the following reaction:
\[C{{H}_{3}}CO{{O}^{-}}+{{H}_{2}}O\leftrightarrow C{{H}_{3}}COOH+O{{H}^{-}}\]
- Here, we can find that the acetate ions act as a base that accepts the H+ ion from water, again forming the acetic acid molecule and in the process, generates OH- ions.
- We know that acetic acid $C{{H}_{3}}COOH$ is a weak acid with a pKa of 4.76. In the titration of acetic acid with a strong base like sodium hydroxide (NaOH). When acetic acid is titrated with NaOH, the pH will be 4.76 (which is higher than that needed for methyl orange to have changed to yellow)
- The pH at the end point will be greater than 7, which is slightly alkaline. As the pH at the end point is not at the range of methyl orange (3–5), methyl orange will not be able to act as an indicator in this titration as we will not be able to see a colour change when the reaction completes.
Hence, we can conclude that the correct option is (A)
That is in the titration of $C{{H}_{3}}COOH$ against NaOH, we cannot use the methyl orange.

Note:
- The choice of a suitable indicator for any titration depends on the nature of the acid and base involved during the process of titration and also on the working range of the indicator.