For a weak base `BOH,K_(b)=10^(-4)` .Calculate `pH` of `10^(-4)BOH` solution. (Take `log6.2=0.79)`

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

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

The pKb of weak base BOH [Kb( BOH) = 1 xx 10^-5 ] will be

The pK_a of a weak acid HA and pk_b of a weak base BOH are 4.60 and 4.80 respectively. The pH of 0.1 M solution of the salt, BA, formed from the acid HA and base BOH is

HA is a weak acid [K_(a) = 10^(-4)] and BOH is a weak base [K_(b) = 10^(-5)] . BA is the salt formed from them, In the aquoeus solution of BA

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

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