If `K_(a)` and `K_(b)` are the dissociation constants of weak acid and its conjugate base , `pK_(a) + pK_(b)`

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.

A solution containing a weak acid and its conjugate base acts as an acidic buffer. In a buffer the dissociation of the weak acid is suppressed by its conjugate base. (K_(1),K_(2), and K_(3) of H_(3)PO_(4) , are 10^(-4), 10^(-8), 10^(-13) respectively What is the pH of a solution obtained by mixing 100ml of 0.1 MH_(3) PO_(4) , and 150ml of 0.IM NaOH

pK_(a) of a weak acid is defined as :

How are pK_(a) values related to acid strength of an acid?

If ionisation constant of an acid (HA) at equlibrium is 1.0xx10^(-8) then calculate the value of pK_(a) and pK_(b) (for its conjugate base)

If degree of dissociation is 0.01 of decimolar solution of weak acid HA then pK_(a) of acid is :

Pure water is netural in nature [H^(+)]=[OH^(-)]. When this condition is disturbed by changing the concentration of H^(+)or OH^(-), the natural solution changes to acidic {[H^(+)]gt[OH^(-)]} or basic {[H^(+)]lt[OH^(-)]}. This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation (i) Salt of weak acid and strong base pH =1/2[pK_(w)+pK_(a)+logC] (ii) Salt of weak base and strong acid pH=1/2[pK_(w)-pK_(b)-logC] (iii) For salt of weak base and weak acid pH=1/2[pK_(w)+pK_(a)-pK_(b)] The pH of buffer can be calculated using t he following formula pH=pK_(a)+log""(["Salt"])/(["Acid"]) pOH=pK_(b)=log""(["Salt"])/(["Base"]) Answer t he following questions when pK_(a)=4.7447 pK_(b)=4.7447 ltb rgt pK_(w)=14 When 50 ml of 0.1 m NaOH is added of 50 ml of 0.1 MCH_(3)COOH solution the pH will be

Pure water is netural in nature [H^(+)]=[OH^(-)]. When this condition is disturbed by changing the concentration of H^(+)or OH^(-), the natural solution changes to acidic {[H^(+)]gt[OH^(-)]} or basic {[H^(+)]lt[OH^(-)]}. This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation (i) Salt of weak acid and strong base pH =1/2[pK_(w)+pK_(a)+logC] (ii) Salt of weak base and strong acid pH=1/2[pK_(w)-pK_(b)-logC] (iii) For salt of weak base and strong acid pH=1/2[pK_(w)+pK_(a)-pK_(b)] The pH of buffer can be calculated using t he following formula pH=pK_(a)+log""(["Salt"])/(["Acid"]) pOH=pK_(b)=log""(["Salt"])/(["Base"]) Answer t he following questions when pK_(a)=4.7447 pK_(b)=4.75 lt rgt pK_(w)=14 When 50 ml of 0.1 M NH_(4)OH is added to 50 ml of 0.05 M HCl solution, the pH is nearly