Question

A certain ion $B ^{-}$ has an Arrhenius constant for basic character (eq. constant $2.8 \times 10^{-7}$ ). The equilibrium constant for Lowry' Bronsted basic character is _____ $\times 10^7$

Answer 1

$B ^{-}+ H ^{+} \rightleftharpoons HB ; K _{ b }=2.8 \times 10^{-7}=\frac{[ HB ]}{\left[ B ^{-}\right]\left[ H ^{+}\right]}\, \dots(i)$

= Bronsted base

$B ^{-}+ H _{2} O \rightleftharpoons HB + OH ; K _{ a }=\frac{[ HB ]\left[ OH ^{-}\right]}{\left[ B ^{-}\right]}\, \dots(ii)$

Arrhenius base

By (i) and (ii)

$\frac{ K _{ a }}{ K _{ b }}=\frac{[ HB ]\left[ OH ^{-}\right]}{\left[ B ^{-}\right]} \times \frac{\left[ B ^{-}\right]\left[ H ^{+}\right]}{[ HB ]}$

$=\left[ H ^{+}\right][ OH ]= K _{ w }\, \dots(iii)$

$\therefore K _{ b }=\frac{ K _{ a }}{ K _{ w }}=\frac{2.8 \times 10^{-7}}{10^{-14}}=2.8 \times 10^{7}$