pH of a salt solution of wak acid `(pK_(a) = 4)` & weak base `(pK_(b) = 5)` at `25^(@)C` is :

pH of a salt solution of weak acid (pK_(a) = 4) & weak base (pK_(b) = 5) at 25^(@)C is :

Define pK_(b) for a weak base.

The pH of a buffer solution containing a weak acid (pK_(a)) and its salt is :

pH of a solution of salt of weak acid and weak base is : pH=1/2pK_w+1/2pK_a-1/2pK_b and that of weak acid and strong base is pH=1/2pK_w+1/2pK_a+1/2logc For a salt of weak acid and weak base having K_a=K_b , the pH at 25^@C will be

For a salt of weak acid and weak base [pK_a- pK_b] would be equal to –

pH of a solution of salt of weak acid and weak base is : pH=1/2pK_w+1/2pK_a-pK_b and that of weak acid and strong base is pH=1/2pK_w+1/2pK_a+1/2logc The pH of 0.1 M sodium acetate ( K_a for CH_3COOH=1.8xx10^(-5) ) is

pH of a solution of salt of weak acid and weak base is : pH=1/2pK_w+1/2pK_a-1/2pK_b and that of weak acid and strong base is pH=1/2pK_w+1/2pK_a+1/2logc pH of 0.1 M solution of ammonium cyanide ( pK_a =9.02 and pK_b =4.76 ) is

Match Column -I with Column -II : Column -I Column-II (A) Ph of water ( p ) ( 1)/( 2) p K_(w) ( B ) Ph of a salt of strong acid and strong base ( q) pH = (1)/(2) [ pK_(w) + pK_(a) + log c ] (C ) Ph of a salt of weak acid and strong base ( r ) pH = ( 1)/( 2) [ pK _(w) + pK_(a) - pK_(b)] (D) Ph of a salt of weak acid and weak base (s) 7 where K_(a) , K_(b) are dissociation constants of weak acid and weak base and K_(w) = Ionic product of water.