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))` solut

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))` solut

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

Monjur Alahi

16 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`[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`

16 views Answered 2023-02-05 15:29