Question

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:
A. `pH=12`
B. `[BH^(+)]=10^(-6)M`
C. `[BH_(2)^(2+)]=10^(-18)M`
D. All of these

Answer 1

`[OH^(-)]=(0.02)/(2)=0.01M:.pH=12`
`10^(-7)=(0.01[BH^(+)])/(0.1)rArr [BH^(+)]=10^(-6)M`
`10^(-7)=((0.01)^(2)[BH_(2)^(2+)])/(0.1)rArr [BH^(+)]=10^(-18)M`