Question

Methyl red has ${K_a} = {10^{ - 5}}$ and the acid form $HIn$ is red and its conjugate base $In{d^ - }$ is yellow.

Reason:

pH	3	5	7
$\dfrac{{[In{d^ - }]}}{{[HIn]}}$	${\text{ 1}}{{\text{0}}^{{\text{ - 2}}}}$	${\text{1}}$	${{\text{0}}^{\text{2}}}$
Colour	Red	Orange	yellow

A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B) Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion
C) The assertion is correct and Reason is incorrect
D) Both Assertion and Reason are incorrect

Answer 1

An indicator is a weak organic acid or base which changes its colour within limits with variations in $pH{\text{ }}$value of the solution to which it is added. Indicator indicates the endpoint generally by a change of colour of the solution. The colour of the indicator is determined by the ratio of $\dfrac{{[In{d^ - }]}}{{[HIn]}}$.

Complete step-by-step answer:
Indicators are either weak organic acids or weak bases having different colours in the ionised form and unionised form. Methyl red is an indicator which turns red in acidic medium and turns yellow in basic medium. The acid form of methyl red indicator is $HIn$ and its conjugate base which is yellow in colour is denoted by $In{d^ - }$.

The $p{K_a}$ of methyl red is 5 so its ${K_a}$ value is ${10^{ - 5}}$. Hence, given assertion methyl red has ${K_a} = {10^{ - 5}}$ and the acid form $HIn$ is red and its conjugate base $In{d^ - }$ is yellow is correct.

Now we will verify the given reason as follows:
As we know that
$\dfrac{{[In{d^ - }]}}{{[HIn]}} = \dfrac{{{K_a}}}{{[{H^ + }]}}$

At $pH$3 the concentration of $[{H^ + }]$ is

pH = - \log [{H^ + }]$$ $$\Rightarrow [{H^ + }] = {\text{ antilog ( - }}pH) = {\text{antilog ( - 3}}) = {10^{ - 3}}$$ Now we will calculate the ratio of $$\dfrac{{[In{d^ - }]}}{{[HIn]}}$$ at $$pH$$3 as follows: $$\dfrac{{[In{d^ - }]}}{{[HIn]}} = \dfrac{{{K_a}}}{{[{H^ + }]}} = \dfrac{{{{10}^{ - 5}}}}{{{{10}^{ - 3}}}} = {10^{ - 2}}$$ From this ratio, we can say that at $$pH$$3 the concentration of $$HIn$$ is greater than the concentration of $$In{d^ - }$$ that is the solution will be acidic so the colour will be red. At $$pH$$5 the concentration of $$[{H^ + }]$$ is $$\Rightarrow pH = - \log [{H^ + }]$$ $$\Rightarrow [{H^ + }] = {\text{ antilog ( - }}pH) = {\text{antilog ( - 5}}) = {10^{ - 5}}$$ Now we will calculate the ratio of $$\dfrac{{[In{d^ - }]}}{{[HIn]}}$$ at $$pH$$5 as follows: $$\dfrac{{[In{d^ - }]}}{{[HIn]}} = \dfrac{{{K_a}}}{{[{H^ + }]}} = \dfrac{{{{10}^{ - 5}}}}{{{{10}^{ - 5}}}} = 1$$ From this ratio, we can say that at $$pH$$5 the concentration of $$HIn$$ is equal to the concentration of $$In{d^ - }$$ so the colour of the solution will be a mixture of red and yellow. At $$pH$$7 the concentration of $$[{H^ + }]$$ is $$\Rightarrow pH = - \log [{H^ + }]$$ $$\Rightarrow [{H^ + }] = {\text{ antilog ( - }}pH) = {\text{antilog ( - 7}}) = {10^{ - 7}}$$ Now we will calculate the ratio of $$\dfrac{{[In{d^ - }]}}{{[HIn]}}$$ at $$pH$$7 as follows: $$\dfrac{{[In{d^ - }]}}{{[HIn]}} = \dfrac{{{K_a}}}{{[{H^ + }]}} = \dfrac{{{{10}^{ - 5}}}}{{{{10}^{ - 7}}}} = {10^2}$$ From this ratio, we can say that at $$pH$$7 the concentration of $$In{d^ - }$$ is greater than the concentration of $$HIn$$ that is the solution will be basic so the colour will be yellow. So, we can say that Both Assertion and Reason are correct and Reason is the correct explanation for Assertion **Hence, the correct answer is an option (A).** **Note:** Each type of indicator shows colour change at a specific range of$$pH$$. The colour of unionised ($$HIn$$) indicator appears in acidic medium while colour of indicator anion ($$In{d^ - }$$) appears in basic medium.