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Assertion : pH of the buffer solution is...

Assertion : pH of the buffer solution is not affected by dilution.
Reason : `pH=pK_(a)+"log"(["Conjugate base"])/(["Acid"])`

A

If both assertion and reason are true and reason is the correct explanation of assertion.

B

If both assertion and reason are true but reason is not the correct explanation of assertion.

C

If assertion is true but reason is false.

D

If both assertion and reason are false.

Text Solution

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The correct Answer is:
A
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pH of which solution is not affected by dilution ?

Solution of a weak acid and its anion (that is,its conjugate base) or of a base and its common cation are buffered. When we add a small amount of acid or base to any one of the, the pH of solution change very little. pH of buffer solution can be computed as for acidic buffer : pH=pK_(a)+ log.(["Conjugate base"])/(["Acid"]) for basic buffer : pOH=pK_(b)+log.(["Conjugate acid"])/([Base]) It is generly accepted that a has useful buffer cpacity (pH change resisting power) provided that the value of [salt or conjugate base] /[acid] for acidic buffer lies within the range of 1 : 10 to 1. Buffer capacity is maximum when [conjugate base] = [acid] Useful correct statement :

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 :

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

On diluting a buffer solution, its pH

The pH of an acidic buffer solution & a basic buffer solution at 25^(@)C is :

STATEMENT-1: Buffer capacity is maximum when concentrtion of salt is equal concentrtion of acid. STATEMEN T-2: pH of the buffer is given by pH= pK_(a)+log""(["salt"])/(["acid"]).

Assertion: The pH of a buffer solution does not change appreciably on additions of a small amount of an acid or a base. Reason: A buffer solution consists of either a weak acid and its salt with a strong base or a weak base its salt with a strong acid.

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