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

Answer: 2.8

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

= Bronsted base

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

Arrhenius base

By (i) and (ii)

$\frac{K_a}{K_b}=\frac{[HB]\left[O{H}^{-}\right]}{\left[B^{-}\right]}\times \frac{\left[B^{-}\right]\left[H^{+}\right]}{[HB]}$

$=\left[H^{+}\right][OH]=K_w\dots (iii)$

$\therefore K_b=\frac{K_a}{K_w}=\frac{2.8\times 10^{-7}}{10^{-14}}=2.8\times 10^7$