${\mathrm{H}}_0\text{ is an acidic indicator }\left({\mathrm{K}}_{\ln }={10}^{-7}\right)\text{ which dissociates into }{\mathrm{H}}^{+}\text{and }{\mathrm{In}}^{-}\text{in }$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 :  

 

detailed solution

Correct option is A

H3PO4⇌K1=10−3H++H2PO4−t=0C 0 0t=tC(1−α) Cα CαH+=Cα=Ka.C=1.5×10−3=1.23×10−2KIn=H+In−HIn10−7=1.23×10−2×In−HInHInIn−=1.23×10−210−7=1.23×105Now, when P colour predominates over Q colour, thenpH=pKIn+log1120pH=7−log120=4.93When Q colour predominates over P colour, thenpH=pKIn+log127=9.1\ pH range of the indicator is 4.93 to 9.1 At 1st  neutralization point, the acid H3PO4 will be convert into KH2PO4 , here pH of the solution=pK1+pK22=3+72=5 At the 2nd  step of neutralization, the acid H3PO4 will be convert into K2HPO4 , here the pH of the solution =pK2+pK32=7+132=10The equivalence point will be obtained when acid H3PO4will change into K2HPO4\n- factor of the acid = 2\normality of KOH=30×0.15×230NMolarity of KOH = 0.3M