Question

OH- ( From KOH) vs C2H5O- (C2H5OH) Which is More strong ?

Answer 1

Answer to Question #39038, Chemistry, Organic Chemistry

Question

OH⁻ (From KOH) vs C₂H₅O⁻ (C₂H₅OH) Which is More strong?

Answer

Strength of the base can be considered as its affinity to proton.

Consider the equilibrium

$$\mathrm{B}^{\cdot} + \mathrm{H}^{+} = \mathrm{BH},$$

where B⁻ denotes the base. More stronger base will have higher value of the equilibrium constant. The latter is about $10^{14}$ for water:

$$\begin{array}{l} \mathrm{OH}^{-} + \mathrm{H}^{+} \xleftarrow{K} \mathrm{H}_{2}\mathrm{O} \\ K = \frac{[\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{OH}^{-}][\mathrm{H}^{+}]} = 10^{14} \mathrm{mol}^{2} \mathrm{l}^{-2} \end{array}$$

and about $10^{16}$ for ethanol

$$\begin{array}{l} \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{O}^{-} + \mathrm{H}^{+} \xleftarrow{K} \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH} \\ K = \frac{[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}]}{[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{O}^{-}][\mathrm{H}^{+}]} = 10^{16} \mathrm{mol}^{2} \mathrm{l}^{-2} \end{array}$$

(two orders higher), hence $\mathrm{C}_2\mathrm{H}_5\mathrm{O}^{-}$ is a stronger base.