Home
Class 12
CHEMISTRY
STATEMENT-1: Buffer capacity is maximum ...

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"]).`

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-11

B

Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-11

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both statements regarding buffer solutions and their properties. ### Step-by-Step Solution: 1. **Understanding Buffer Solutions:** - A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in pH when small amounts of acid or base are added. 2. **Analyzing Statement 1:** - **Statement 1:** "Buffer capacity is maximum when the concentration of salt is equal to the concentration of acid." - When the concentrations of the weak acid and its conjugate base (salt) are equal, the buffer can effectively neutralize added acids or bases. This is because there are equal amounts of acid and base to react with any added species, maximizing the buffer's capacity to resist pH changes. 3. **Analyzing Statement 2:** - **Statement 2:** "pH of the buffer is given by pH = pKa + log([salt]/[acid])." - This is the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the salt (conjugate base) to the acid. 4. **Connecting the Statements:** - When the concentrations of the salt and acid are equal, the ratio [salt]/[acid] equals 1. Therefore, log(1) = 0, which means pH = pKa. - This indicates that at maximum buffer capacity (when concentrations are equal), the pH of the buffer is equal to the pKa of the weak acid. 5. **Conclusion:** - Both statements are true. Statement 1 is correct because buffer capacity is indeed maximum when the concentrations of salt and acid are equal. Statement 2 is also true as it accurately describes how to calculate the pH of a buffer solution. Furthermore, Statement 2 provides a correct explanation for Statement 1. ### Final Answer: Both Statement 1 and Statement 2 are true, and Statement 2 is a correct explanation for Statement 1. ---
Promotional Banner

Topper's Solved these Questions

  • EQUILIBRIUM

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION-F) (MATRIX-MATCH TYPE QUESTIONS)|3 Videos
  • EQUILIBRIUM

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION-G) (INTEGER ANSWER TYPE QUESTIONS)|5 Videos
  • EQUILIBRIUM

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION-D) (LINKED COMPREHENSION TYPE QUESTIONS)|8 Videos
  • ENVIRONMENTAL CHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D) (Assertion - Reason Type Questions)|4 Videos
  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section-D) Assertion-Reason Type Question|15 Videos

Similar Questions

Explore conceptually related problems

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

The buffering action of an acidic buffer is maximum when its pH equals to

An acidic buffer contains equal concentrations of acid and salt. The dissociation constant of acid is 10^(-5) . The P^(H) of the buffer solution is

In a mixture of weak acid and its salt, the ratio of concentration of salt to acid is increased ten fold. The pH of the solution

The Ph of basic buffer mixtures is given by : Ph=Pk_(a)+ log (["Base"])/(["Salt"]) whereas Ph of acidic buffer mixtures is given by : Ph = pK_(a)+"log"(["Salt"])/(["Acid"]) . Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. Calculate the ratio of pH of a solution containing 1 mole of CH_3COONa + 1 mole of HCl per litre and of other solution containing 1 mole CH_3COONa + 1 mole of acetic acid per litre

The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. A solution containing 0.2 mole of dichloroacetic acid (K_(a)=5xx10^(-2)) and 0.1 mole sodium dichloroacetate in one litre solution has [H^(+)] :

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

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 :

STATEMENT-1 : When a small amount of strong acid is added to a buffer solution, its pH value does not change significantly STATEMENT-2 : Buffer action of the buffer sloution resist the changee in pH when small amount of acid is added to it