${\mathrm{H}}_0\text{ is an acidic indicator }\left({\mathrm{K}}_{\ln }={10}^{-7}\right)\text{ which dissociates into }{\mathrm{H}}^{+}\text{and }{\mathrm{In}}^{-}\text{in }$aqueous solution. It is added to a solution of 30 ml of 0.15 M ${\mathrm{H}}_3{\mathrm{PO}}_4$

$({\mathrm{K}}_1=1\times {10}^{-3},{ \mathrm{K}}_2=1\times {10}^{-7}\stackrel{-1}{=}{\mathrm{K}}_3=1\times {10}^{-13})\text{ . If }\mathrm{Hln}\text{ and }{\mathrm{In}}^{-}\text{posses colour }\mathrm{P}$and Q respectively and color P predominates over color Q, when concentration of HIn is 120 times than that of $I{n}^{-}$and color Q predominates over P, and when concentration of $I{n}^{-}$is 127 times of HIn­. If this solution treated with 30 ml KOH, 

then correct statements among the following is/are :  

 

$\begin{array}{llllll} & {\text{H}}_3{\text{PO}}_4& \stackrel{{\text{K}}_1={10}^{-3}}{\rightleftharpoons }& {\text{H}}^{+}& +& {\text{H}}_2{\text{PO}}_4^{-}\\ \text{t}=0& \text{C}& & 0& & 0\\ \text{t}=\text{t}& \text{C}(1-\alpha )& & \text{C}\alpha & & \text{C}\alpha \end{array}$

$H^{+}=C\alpha =\sqrt{Ka.C}=\sqrt{1.5\times {10}^{-3}}=1.23\times {10}^{-2}$

$K_{In}=\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[H_{In}\right]}$

${10}^{-7}=1.23\times {10}^{-2}\times \frac{\left[I{n}^{-}\right]}{\left[H_{In}\right]}$

$\frac{\left[H_{In}\right]}{\left[I{n}^{-}\right]}=\frac{1.23\times {10}^{-2}}{{10}^{-7}}=1.23\times {10}^5$

Now, when P colour predominates over Q colour, then

$pH=p{K}_{In}+\log \frac{1}{120}$

$pH=7-\log 120=4.93$

When Q colour predominates over P colour, then

$pH=p{K}_{In}+\log 127=9.1$

\ pH range of the indicator is 4.93 to 9.1

$\text{ At }{1}^{\text{st }}\text{ neutralization point, the acid }{\mathrm{H}}_3{\mathrm{PO}}_4\text{ will be convert into }{\mathrm{KH}}_2{\mathrm{PO}}_4\text{ , here }\mathrm{pH}$ of the solution

$=\frac{p{K}_1+p{K}_2}{2}=\frac{3+7}{2}=5$

$\text{ At the }{2}^{\text{nd }}\text{ step of neutralization, the acid }{\mathrm{H}}_3{\mathrm{PO}}_4\text{ will be convert into }{\mathrm{K}}_2{\mathrm{HPO}}_4\text{ , }$

here the pH of the solution $=\frac{p{K}_2+p{K}_3}{2}=\frac{7+13}{2}=10$

The equivalence point will be obtained when acid $H_{3}P{O}_4$will change into $K_{2}HP{O}_4$

\n- factor of the acid = 2

\normality of $KOH=\frac{30\times 0.15\times 2}{30}N$

Molarity of KOH = 0.3M