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Pure water is netural in nature [H^(+)]=...

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

A

`9.2253`

B

`2.3723`

C

14

D

7

Text Solution

Verified by Experts

The correct Answer is:
B
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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

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

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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 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

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