At `25^(@)C K_(b)` for `BOH=1.0xx10^(-12).0.01M` solution of `BOH` has `[OH^(-)]`:

At 25^@C, K_b for BOH= 1.0xx10^(-12) .0.01 M solution of BOH has [OH^-] :

K_(b) for NH_(4)OH is 1.81xx10^(-5) . The pH of 0.01 M NH_(4)Cl solution at 25^(@)C is :

Total number of solutions from the following which has pHlt7 at 25^(@)C . (K_(a)(CH_(3)COOH)=1.8xx10^(-5),K_(b)(NH_(4)OH)=1.8xx10^(-5)) (i) 10^(-8) MHCl (ii) 0.01 M solution of NH_(4)Cl (iii) 0.01 M solution of CH_(3)COONa (iv) 0.01M solution is B(OH)_(3) (v) 0.01M solution of CH_(3)COONH_(4)

In the equilibrium A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4)) . The degree of hydrolysis of 0.01 M solution of the salt is

In the equilibrium A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4)) . The degree of hydrolysis of 0.01 M solution of the salt is

K_(sp) of Mg(OH)_(2) is 1xx10^(-12).0.01 M MgCl_(2) will show precipitation is a solution of pH greater than :

At 25^(@) C dissociation constants of acid HA and base BOH in aqueous solution are same. The pH of 0.01 M solution of HA is 5. The pOH of 10^(-4) M solution of BOH at the same temperature is :

At 25^(@) C dissociation constants of acid HA and base BOH in aqueous solution are same. The pH of 0.01 M solution of HA is 5. The pOH of 10^(-4) M solution of BOH at the same temperature is :

At 25^(@) C dissociation constants of acid HA and base BOH in aqueous solution are same. The pH of 0.01 M solution of HA is 5. The pOH of 10^(-4) M solution of BOH at the same temperature is :