What is the pOH of 0.1 M KB (salt of weak acid and strong base) at `25^(@)C` ? (Given : `pK_(b) of B^(-)`=7)

Degree hydrolysis (h) of a salt of weak acid and a strong base is given by

pH of a salt of a strong base with weak acid

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 50 mL 0.1 M NaOH is added to 50 mL of 0.1 M CH_(3)COOH solution , the pH will be

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 When 100 mL of 0.1 M NH_(4)OH is added to 50 mL of 0.1M HCl solution , the pH is

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 0.001 M NH_(4)Cl aqueous solution has pH :

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.

Calculate pOH of 0.1 M aq. Solution of weak base BOH (K_(b)=10^(-7)) at 25^(@)C.

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

The extent to which hydrolysis proceeds is expressed as degree of hydrolysis and is defined as fraction of ne mole of the salt that is hydrolysed when equilibrium has been attained . The nature of solution also depends upon the extent upto which the salt has been hydrolysed . Degree of hydrolysis of a salt ofdepends upon concentration of salt at a particular temperature and varies inversely . Consequently , nature of the solution whether acidic or alkaline can be determined . pH of the solution of salt of weak acid and strong base or weak base and strong acid depends upon its concentration . So by varying the concentration of salts their pH could be adjusted . 0.1 M of which salt will have pH close to 7 , if pK_(a) HA = 5, pK_(a)HB = 6, pK_(a)HC = 7 ,pK_(a)HD = 8 ?

The extent to which hydrolysis proceeds is expressed as degree of hydrolysis and is defined as fraction of ne mole of the salt that is hydrolysed when equilibrium has been attained . The nature of solution also depends upon the extent upto which the salt has been hydrolysed . Degree of hydrolysis of a salt ofdepends upon concentration of salt at a particular temperature and varies inversely . Consequently , nature of the solution whether acidic or alkaline can be determined . pH of the solution of salt of weak acid and strong base or weak base and strong acid depends upon its concentration . So by varying the concentration of salts their pH could be adjusted . For strong acid and weak base salt solution , the [H_(3)O^(+)] at equilibrium will be