$\left[\mathrm{OH}^{-}\right]=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{H}^{+}\right]}=\frac{7.1 \times 10^{-15}}{5.37 \times 10^{-12}}=\frac{7.1}{5.37} \times 10^{-3}\, \mathrm{M}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}$
$\left(\frac{7.1}{5.37}\right)^{2} \times 10^{-6}=\mathrm{K}_{\mathrm{b}} \times 0.1$
$\boxed{{K_b} = 1.747 \times {{10}^{ - 5}}}$
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$A$. $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$
$B$. $\mathrm{N} \rightarrow \mathrm{N}^{-}$
$C$. $\mathrm{O} \rightarrow \mathrm{O}^{2-}$
$D$. $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$
$E$. $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$
Choose the most appropriate answer from the options given below :