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What is the pOH of 0.1 M KB (salt of wea...

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)

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Salts Of Weak Acids And Strong Bases

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

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

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 :

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

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