An acid base indicator (HIn) is `50%` ionized at `pH=6.0`.What will be the dissociation constant of HIn? `10^(-6)` , `10^(-5)` , `10^(-4)` , `0.5times10^(-6)`

The dissociation constant of a weak acid is 10^(-6) . Then the P^(H) of 0.01 Nof that acid is

Calculate the pH of 0.10 M solution of NH_4CI . The dissociation constant (K_b) "of" NH_3 "is" 1.8 xx 10^(-5) .

Calculate the pH of 0.01 M Solution of CH_3 COOH . The dissociation constant of the acid is 1.8 xx 10 ^(-5)

An acid HA ionizes as HAhArrH^(+)+A^(-) The pH of 1.0 M solution is 5 . Its dissociation constant would be

An acid HA ionizes as HAhArrH^(+)+A^(-) The pH of 1.0 M solution is 5 . Its dissociation constant would be

Calculate the pH of 0.10 M solution of NH_(4)Cl. The dissociation constant (K_(b)) of NH_(3) is 1.8 X 10^(-5)

The pH of a 0.1M solution of NH_(4)OH (having dissociation constant K_(b) = 1.0 xx 10^(-5)) is equal to

The first and second dissociation constant of an acid H_2A are 1.0 xx 10 ^(-5) and 5.0 xx 10^(-10) respectively . The overall dissociation constant of the acid will be : 5.0 xx 10^(-5) 5.0 xx 10 ^(-15) 5.0 xx 10 ^(-15) 0.2 xx 10 ^(5)

The dissociation constant of an acid, HA is 1 x 10^-5 The pH of 0.1 M solution of the acid will be

Solve with due regards to significant figures 4.0 xx (10^-4) - 2.5 xx 10^(-6) .