An indicator $HIn$ has the value of ${{K}_{a}}=3\times {{10}^{-5}}$, the acid form is red and the basic form is blue. Which statement is correct?
A. $pH=5$ when indicator is $75%$red.
B. $pH=4.05$ when indicator is $75%$red.
C. $pH=5$ when indicator is $75%$blue.
D. $pH=4.05$ when indicator is $75%$blue.

Hint:An acid base indicator are those substances or compounds which have a tendency to change color with change in $pH$. They are generally weak acids or bases and when they dissolve in water, they dissociate slightly and form ions.

Complete step-by-step answer:Consider the indicator $HIn$ given in the question. This indicator is basically a weak acid and in equilibrium it undergoes partial dissociation as depicted below,
$HIn\left( aq \right){{H}^{+}}\left( aq \right)+I{{n}^{-}}\left( aq \right)$
Here $HIn$ is a weak acid and $I{{n}^{-}}$is its conjugate base. It is given in the question that acid form is red, this means that $HIn$ will be of red color. And it is also given that the basic form is blue, this means that $I{{n}^{-}}$is blue in color.
The value of dissociation constant for the acid is given as: ${{K}_{a}}=3\times {{10}^{-5}}$.
Now we know that the $pH$ of acid indicator can be calculated by the formula,
$pH=p{{K}_{a}}+\log \dfrac{[base]}{[acid]}$
For the dissociation, mentioned in the question this formula will be
$pH=p{{K}_{a}}+\log \dfrac{[I{{n}^{-}}]}{[HI]}$
Now firstly let us find out $p{{K}_{a}}$. We know that $p{{K}_{a}}=-\log {{K}_{a}}$
$\begin{align}
  & \Rightarrow p{{K}_{a}}=-\log \left( 3\times {{10}^{-5}} \right) \\
 & \Rightarrow p{{K}_{a}}=5-\log 3 \\
 & \text{we know log 3=0}\text{.47} \\
 & \Rightarrow p{{K}_{a}}=5-0.47=4.53 \\
 & \therefore p{{K}_{a}}=4.53 \\
\end{align}$
Let us take two cases.
A. when the indicator is $75%$red. This means that rest $25%$is the basic form which is blue.
$75%$red means that $[HIn]$= $75%$and $[I{{n}^{-}}]$= $25%$
Therefore, using formula $pH=p{{K}_{a}}+\log \dfrac{[I{{n}^{-}}]}{[HI]}$we get
$\begin{align}
  & pH=4.52+\log \dfrac{[25]}{[75]} \\
 & \Rightarrow pH=4.52+\log \dfrac{[1]}{[3]} \\
 & \text{we know that log }\dfrac{a}{b}=\log a-\log b \\
 & \Rightarrow pH=4.52+(\log 1-\log 3) \\
 & \text{we know that log1=0 and log 3= 0}\text{.47} \\
 & \Rightarrow pH=4.52+\left( 0-0.47 \right) \\
 & \Rightarrow pH=4.52-0.47=4.05 \\
\end{align}$
We get $pH$= $4.05$
 From this calculation we can interpret that statement of option B. which states that $pH=4.05$when indicator is $75%$red is correct.
Now let us see the other case.
When Indicator is $75%$blue. This means that rest $25%$is acid which is red.
Therefore, we can say that $[HIn]$= $25%$and $[I{{n}^{-}}]$= $75%$
Calculating, pH for this solution we will get,
$\begin{align}
  & pH=4.52+\log \dfrac{[75]}{[25]} \\
 & \Rightarrow pH=4.52+\log 3 \\
 & \text{we know that log 3= 0}\text{.47} \\
 & \Rightarrow pH=4.52+0.47 \\
 & \Rightarrow pH=5 \\
\end{align}$
Therefore, by this calculation we see that option c. which states that $pH=5$when indicator is $75%$blue is correct.

Hence, option B. and option C. are correct options.

Note:The acid and its conjugate base generally have different colors. At low pH values the hydronium ion concentration is high and hence the equilibrium shifts left. In this condition equilibrium solutions have a specific color (here red). When hydronium ion concentration is less than the equilibrium shifts in the right direction and at this position solution has different color (here blue).