Question

Select the best indicator from the given table for titration of $20ml$ of $0.02M$ $C{H_3}COOH$ with $0.02M$ $NaOH$ . Given $p{K_a}$ of $C{H_3}COOH = 4.74$ .

| Indicator| $pH$ range
---|---|---
i.| Bromothymol blue| $6.0 - 7.6$
ii.| Thymolphthalein| $9.3 - 10.5$
iii.| Malachite Green| $11.4 - 13$
iv.| M-cresol purple| $7.4 - 9.0$

A. i.
B. ii.
C. iii.
D. iv.

Answer 1

Indicator is defined as a substance that gives a visible sign, generally by a colour change when brought in contact with acid or base. In order to select the best indicator we have to check the $pH$ of the solution and then we should use the indicator whose range lies more accurately to solve the $pH$ range.

Formula used:
1. $n = M \times V$
where, $n$ is the number of moles, $M$ is the Molarity and $V$ is the volume.
2. $pH = \dfrac{1}{2}(p{K_w} + p{K_a} + \log c)$
where, $pH$ is the measure of acidity, ${K_w}$ is the dissociation constant of water $p{K_a}$ is the negative base $- 10$ logarithm of an acid dissociation constant and $c$ is the concentration.

Complete step by step answer:
Let us first write the reaction
$C{H_3}COOH + NaOH \rightleftharpoons C{H_3}COONa + {H_2}O$
As we know, $n = M \times V$
Where, $n$ is the number of moles, $M$ is the Molarity and $V$ is the volume, In the case of above reaction:
$\mathop {C{H_3}COOH}\limits_{20 \times 0.02} + \mathop {NaOH}\limits_{20 \times 0.02} \rightleftharpoons \mathop {C{H_3}COONa}\limits_{20 \times 0.02} + {H_2}O$
Now, let us find the concentration of salt.
$[salt] = \dfrac{{mil\lim oles}}{{{\text{total volume}}}}$
Substituting the values in the above formula,
$[salt] = \dfrac{{20 \times 0.02}}{{20 + 20}} \Rightarrow 0.01$
This salt is a formation of weak acid and strong base.
Now, let us calculate the $pH$ and the formula is written as follows:
$pH = \dfrac{1}{2}(p{K_w} + p{K_a} + \log c)$
Substituting the above values in the above formula, we get,
$\ \Rightarrow pH = \dfrac{1}{2}(14 + 4.74 + \log 0.01) \\\ \Rightarrow pH = 8.37 \\\ \$
The best indicator will range at a $pH$ of $8.37$ . In this question, M – Cresol purple is the only indicator having the $pH$ range of $7.4 - 9$ . Therefore, the above calculated $pH$ , that is $8.37$ will lie between $7.4 - 9.0$ , which is a M – Cresol indicator.
Hence, the correct option is (iv), which is M – Cresol indicator.

So, the correct answer is Option D.

Note: $pH$ is defined as a scale which is used to measure acidity and basicity of an aqueous solution or liquid. The
$pH$ less than seven considers an acid concentration, whereas $pH$ more than seven considers a basic solution.
$pH = - \log [{H^ + }]$
where, ${H^ + }$ is the concentration of ${H^ + }$ ions.
${K_a}$ is defined as dissociation constant of an acid. If the value of dissociation constant of an acid is high then it is a strong acid, whereas if the value of dissociation constant of an acid is less, then it is a weak acid.
$p{K_a}$ is defined as the negative base $- 10$ logarithm of an acid dissociation constant.
M – Cresol purple is a triaryl methane dye and also used as an indicator. When it gets exposed to carbon dioxide, it changes its color from purple to yellow.
Moles is equal to the product of molarity and volume.
$n = M \times V$
$p{K_a}$ is used to measure the strength of an acid.