Question

A certain ion, $B^{-}$ has a dissociation constant for Arrhenius basic character ($2.8\times {10}^{-7}$). The equilibrium constant for Lowry-Bronsted basic character is:

• A. $2.8\times {10}^{-7}$
• B. $3.57\times {10}^{-8}$
• C. $3.57\times {10}^8$
• D. $2.8\times {10}^7$

Answer 1

Answer: (D)

$2.8\times {10}^7$

$\underset{Bronstedbase}{B^{-}}+H^{+}\rightleftharpoons HB;K_b=\frac{\left[HB\right]}{\left[B^{-}\right]\left[H^{+}\right]}......\left(i\right)$
$\underset{Arrhneiusbase}{B^{-}}+H^{+}\rightleftharpoons HB+{OH}^{-};K_a=\frac{\left[HB\right]\left[{OH}^{-}\right]}{\left[B^{-}\right]}......\left(ii\right)$
By Eqs. $\left(i\right)$ and $\left(ii\right)$
$\frac{K_a}{K_b}=\frac{\left[HB\right]\left[{OH}^{-}\right]}{\left[B^{-}\right]}\times \frac{\left[B^{-}\right]\left[H^{+}\right]}{\left[HB\right]}=\left[H^{+}\right]\left[{OH}^{-}\right]=K_w........\left(iii\right)$
By Eqs. $\left(i\right)$ and $\left(ii\right)$ and $\left(iii\right)$,
$K_b=\frac{K_a}{K_w}=\frac{2.8\times {10}^{-7}}{{10}^{-14}}=2.8\times {10}^7$