w = K a × K b , can also be obtained by considering the base-dissociation equilibrium reaction: B(aq) + H O(l) BH + (aq) + OH – (aq) K b = [BH + ][OH – ] / [B] As the concentration of water remains constant it has been omitted from the denominator and incorporated within the dissociation constant. Then multiplying and dividing the above expression by [H + ], we get: K b = [BH + ][OH – ][H + ] / [B][H + ] ={[ OH – ][H + ]}{[BH + ] / [B][H + ]} = K w / K a or K a × K b = K w It may be noted that if we take negative logarithm of both sides of the equation, then p K values of the conjugate acid and base are related to each other by the equation: p K a + p K b = p K w = (at 298K) Problem . Determine the degree of ionization and pH of a .05M of ammonia solution. The ionization constant of ammonia can be taken from Table .
. Also, calculate the ionization constant of the conjugate acid of ammonia. The ionization of NH in water is represented by equation: NH + H O NH + + OH – We use equation ( . ) to calculate hydroxyl ion concentration, [OH – ] = c α = .
α K b = . α / ( – α ) The value of α is small, therefore the quadratic equation can be simplified by neglecting α in comparison to in the denominator on right hand side of the equation, Thus, K b = c α or α = √ ( . × – / . ) = .