Calculate pOH of 0.1 M aq. Solution of weak base BOH `(K_(b)=10^(-7))` at `25^(@)C.`

Calculate the pH of 500mL Solution of 1 M BOH (K_(b)=2.5xx10^(-5))

Calculate pH of a resultant solution of 0.1 M HA (K_(a)=10^(-6)) and 0.5 M HB (K_(a)=2xx10^(-6)) at 25^(@)C.

The vapour pressur of 0.01molal solution of weak base BOH in water at 20^(@)C is 17.536mm . Calculate K_(b) for base. Aqueous tension at 20^(@)C is 17.540mm . Assume molatilly and molarity same.

For 10^(-4) M BOH (weak base) , K_(b)=5xx10^(-5)) :

For a solution obtained by mixing equal volumes of 0.02M KOH solution &0.2MB (weak diacidic base), K_(b_(1))=10^(-7) & K_(b_(2))=10^(-14)) solution:

The freezing point depression of a 0.1 M aq. Solution of weak acid (HX) is -0.20^(@)"C". What is the value of equilibrium constant for the reaction: "HX(aq)"iff"H"^(+)"(aq)"+"X"^(-1)"(aq)" [Given: "K"_("f")" for water=1.8 Kg mol"^(-1)"K. & Molality=Molarity" ]

Calculate the pH of 0.1M ammonium acetate at 25^@C . K_a=K_b=1.8xx10^(-10)

What is the pOH of 0.1 M KB (salt of weak acid and strong base) at 25^(@)C ? (Given : pK_(b) of B^(-) =7)