H0 is an acidic indicator Kln=10-7 which dissociates into H+and In-in aqueous solution. It is added to a solution of 30 ml of 0.15 M H3PO4(K1=1×10-3, K2=1×10-7=-1K3=1×10-13) . If Hln and In-posses colour Pand Q respectively and color P predominates over color Q, when concentration of HIn is 120 times than that of In−and color Q predominates over P, and when concentration of In−is 127 times of HIn­. If this solution treated with 30 ml KOH, then correct statements among the following is/are :

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