`Hin` is an acidic indicator `(K_(Ind) =10^(-7))` which dissociates into aqueous acidic solution of `30mL` of `0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13))`
Calculate the `[(Ind^(Theta))/(Hin)]`

Hin is an acidic indicator (K_(Ind) =10^(-7)) which dissociates into aqueous acidic solution of 30mL of 0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13)) If Hin and Ind^(Theta) posses colour P and Q , respectively, and concentration of HIn is 120 times than that of Ind^(Theta) . colour Q predominates over P when concnetration of Ind^(Theta) is 127 times of HIn . What is the pH range of the indicator.

Hin is an acidic indicator (K_(Ind) =10^(-7)) which dissociates into aqueous acidic solution of 30mL of 0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13)) If Hin and Ind^(Theta) posses colour P and Q , respectively, and concentration of HIn is 120 times than that of Ind^(Theta) . colour Q predominates over P when concnetration of Ind^(Theta) is 127 times of HIn . What is the pH range of the indicator.

Hin is an acidic indicator (K_(Ind) =10^(-7)) which dissociates into aqueous acidic solution of 30mL of 0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13)) If this solution is treated with 30mL of NaOH solution, then what molarity of NaOH is needed to reach the equivalence point with indicator?

Hin is an acidic indicator (K_(Ind) =10^(-7)) which dissociates into aqueous acidic solution of 30mL of 0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13)) If this solution is treated with 30mL of NaOH solution, then what molarity of NaOH is needed to reach the equivalence point with indicator?

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) pH of solution is

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) Concentration of H_(2)PO_(4)^(-) is

In a solution of 0.1M H_(3)PO_(4) acid Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) Concentration of HPO_(4)^(2-) is

Hydrocyanic acid (K_(a) = 4.8xx10^(-10)) is dissociated to an extent of .... In 1.0 M solution . F