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

A salt of weak acid and weak base undergoes

Salts Of Weak Acid And Weak Bases

Hydrolysis constant for a salt of weak acid and weak base would be

Hydrolysis constant for a salt of weak acid and weak base would be

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

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

When a salt reacts with water resulting into formation of acidic or basic solution, the process is referred to as salt hydrolysis. The pH of salt solution can be calculated using the following equations. pH = ( 1)/(2) ( p K_(w) + pK_(a) + log C) for salt of weak acid and strong base. pH = (1)/(2) ( pK_(w)- pK_(b) - log C ) for salt of weak base and strong acid. pH = (1)/(2) ( pK_(w) + pK_(a) - pK_(b)) for salt of weak acid and weak base equal volume of 0.1M solution of weak acid HA is titrated with 0.1M NaOH solution till the end point pK_(a) for acid is 6 and degree of hydroglysis is less compared to 1.The pH of the resultant solution at the end point is