For anionic hydrolysis "pH" is given by (A) pH=`(1)/(2)pK_(w)` - `(1)/(2)pK_(b)`-`(1)/(2)log C`
(B) pH=`(1)/(2)pK_(w)`+`(1)/(2)pK_(a)`-`(1)/(2)pK_(b)`
(C) pH=`(1)/(2)pK_(w)`+`(1)/(2)pK_(a)`+`(1)/(2)`log C (D) None of the above

pH of some solution is given by pH = (pK_(a_(1))+pK_(a_(2)))/(2) . This formula is valid for the compound NaH_(2)BO_(3) .

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 The pH of 1M PO_(4(aq))^(3-) soluiton will be ( given pK_(b) of PO_(4)^(3-) = 1.62 )

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

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)=4.7447 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)=4.7447 ltb rgt pK_(w)=14 1 mole of CH_(3)COOH is dessolved in water to from 1 litre aqueous solution. The pH of resulting solution 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

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 50 mL 0.1 M NaOH is added to 50 mL of 0.1 M CH_(3)COOH solution , the pH will be