× – , thus, x = . × – [H + ] = . × – M. Therefore, Percent dissociation = {[HOCl] dissociated / [HOCl] initial }× = .
Ionization of Weak Bases The ionization of base MOH can be represented by equation: MOH(aq) M + (aq) + OH – (aq) In a weak base there is partial ionization of MOH into M + and OH – , the case is similar to that of acid-dissociation equilibrium. The equilibrium constant for base ionization is called base ionization constant and is represented by K b . It can be expressed in terms of concentration in molarity of various species in equilibrium by the following equation: K b = [M + ][OH – ] / [MOH] ( . ) Alternatively, if c = initial concentration of base and α = degree of ionization of base i.e.
the extent to which the base ionizes. When equilibrium is reached, the equilibrium constant can be written as: K b = (c α ) / c ( - α ) = c α / ( - α ) The values of the ionization constants of some selected weak bases, K b are given in Table . . Many organic compounds like amines are weak bases.
Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by another group. For example, methylamine, codeine, quinine and nicotine all behave as very weak bases due to their very small K b . Ammonia produces OH – in aqueous solution: NH (aq) + H O(l) NH + (aq) + OH – (aq) The pH scale for the hydrogen ion concentration has been extended to get: p K b = –log ( K b ) ( . ) Problem .
The pH of .004M hydrazine solution is . . Calculate its ionization constant K b and p K b . NH NH + H O NH NH + +