229723
A buffer solution has equal volumes of $0.2 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{OH}$ and $0.02 \mathrm{M} \mathrm{NH} \mathrm{NC}_4$. The $\mathrm{pK}_{\mathrm{b}}$ of the base is 5 . The $\mathrm{pH}$ is
1 10
2 9
3 4
4 7
Explanation:
Given thatvolume of $\mathrm{NH}_4 \mathrm{OH}=0.2 \mathrm{M}$ and volume of $\mathrm{NH}_4 \mathrm{Cl}=0.02 \mathrm{M}$ $\mathrm{pK}_{\mathrm{b}}=5$ According to the Henderson's equation. $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{[\text { base }]} \\ & =5+\log \frac{0.02}{0.2}=5+\log \frac{1}{10}=5+(-1)=4 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-4=10 \end{aligned}$
COMEDK-2017**
Ionic Equilibrium
229724
When 0.1 mole of an acid is added to $2 \mathrm{~L}$ of a buffer solution, the $\mathrm{pH}$ of the buffer decreases by 0.5 . The buffer capacity of the solution is
1 0.6
2 0.4
3 0.2
4 0.1
Explanation:
Given data: Number of moles of acid added $=\frac{0.1}{2}$ per litre Change of $\mathrm{pH}=0.5$ Now, Buffer capacity $=\frac{\text { number of moles of acid added/L }}{\text { change in } \mathrm{pH}}$ Buffer capacity $=\frac{0.1}{0.5 \times 2}$ Buffer capacity $=0.1$
2008
Ionic Equilibrium
229726
Assertion : Mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and **Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 In the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the assertion is incorrect but the Reason is correct.
Explanation:
Acidic buffer contains equimolar mixture of weak acid and its salt with strong base. $\mathrm{CH}_3 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COONH}_4$ is not an example of acidic buffer.
AIIMS-(2007)
Ionic Equilibrium
229730
For a buffer of a mixture of $0.12 \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{CH}_3 \mathrm{COOH}$ and $0.12 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{CH}_3 \mathrm{COONa}$, the buffer capacity is
1 1.38
2 0.130
3 0.06
4 0.60
Explanation:
Buffer capacity $(\beta)=\frac{\Delta \mathrm{C}_{\mathrm{a}}}{\Delta \mathrm{pH}}$ Where $C_a=$ concentration of acid for acidic buffer Concentration of acid = concentration of salt $\therefore \Delta \mathrm{C}_{\mathrm{a}}=0.12 \mathrm{molL}^{-1}$ Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mathrm{pH}_{\mathrm{pH}}^{\mathrm{pH}}-\log \left[\mathrm{H}^{+}\right]=-\log (0.12)=0.92$ Buffer capacity $(\beta)=\frac{0.12}{0.92}=\frac{12}{92}=\frac{6}{46}=\frac{3}{23}=0.130$
229723
A buffer solution has equal volumes of $0.2 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{OH}$ and $0.02 \mathrm{M} \mathrm{NH} \mathrm{NC}_4$. The $\mathrm{pK}_{\mathrm{b}}$ of the base is 5 . The $\mathrm{pH}$ is
1 10
2 9
3 4
4 7
Explanation:
Given thatvolume of $\mathrm{NH}_4 \mathrm{OH}=0.2 \mathrm{M}$ and volume of $\mathrm{NH}_4 \mathrm{Cl}=0.02 \mathrm{M}$ $\mathrm{pK}_{\mathrm{b}}=5$ According to the Henderson's equation. $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{[\text { base }]} \\ & =5+\log \frac{0.02}{0.2}=5+\log \frac{1}{10}=5+(-1)=4 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-4=10 \end{aligned}$
COMEDK-2017**
Ionic Equilibrium
229724
When 0.1 mole of an acid is added to $2 \mathrm{~L}$ of a buffer solution, the $\mathrm{pH}$ of the buffer decreases by 0.5 . The buffer capacity of the solution is
1 0.6
2 0.4
3 0.2
4 0.1
Explanation:
Given data: Number of moles of acid added $=\frac{0.1}{2}$ per litre Change of $\mathrm{pH}=0.5$ Now, Buffer capacity $=\frac{\text { number of moles of acid added/L }}{\text { change in } \mathrm{pH}}$ Buffer capacity $=\frac{0.1}{0.5 \times 2}$ Buffer capacity $=0.1$
2008
Ionic Equilibrium
229726
Assertion : Mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and **Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 In the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the assertion is incorrect but the Reason is correct.
Explanation:
Acidic buffer contains equimolar mixture of weak acid and its salt with strong base. $\mathrm{CH}_3 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COONH}_4$ is not an example of acidic buffer.
AIIMS-(2007)
Ionic Equilibrium
229730
For a buffer of a mixture of $0.12 \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{CH}_3 \mathrm{COOH}$ and $0.12 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{CH}_3 \mathrm{COONa}$, the buffer capacity is
1 1.38
2 0.130
3 0.06
4 0.60
Explanation:
Buffer capacity $(\beta)=\frac{\Delta \mathrm{C}_{\mathrm{a}}}{\Delta \mathrm{pH}}$ Where $C_a=$ concentration of acid for acidic buffer Concentration of acid = concentration of salt $\therefore \Delta \mathrm{C}_{\mathrm{a}}=0.12 \mathrm{molL}^{-1}$ Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mathrm{pH}_{\mathrm{pH}}^{\mathrm{pH}}-\log \left[\mathrm{H}^{+}\right]=-\log (0.12)=0.92$ Buffer capacity $(\beta)=\frac{0.12}{0.92}=\frac{12}{92}=\frac{6}{46}=\frac{3}{23}=0.130$
229723
A buffer solution has equal volumes of $0.2 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{OH}$ and $0.02 \mathrm{M} \mathrm{NH} \mathrm{NC}_4$. The $\mathrm{pK}_{\mathrm{b}}$ of the base is 5 . The $\mathrm{pH}$ is
1 10
2 9
3 4
4 7
Explanation:
Given thatvolume of $\mathrm{NH}_4 \mathrm{OH}=0.2 \mathrm{M}$ and volume of $\mathrm{NH}_4 \mathrm{Cl}=0.02 \mathrm{M}$ $\mathrm{pK}_{\mathrm{b}}=5$ According to the Henderson's equation. $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{[\text { base }]} \\ & =5+\log \frac{0.02}{0.2}=5+\log \frac{1}{10}=5+(-1)=4 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-4=10 \end{aligned}$
COMEDK-2017**
Ionic Equilibrium
229724
When 0.1 mole of an acid is added to $2 \mathrm{~L}$ of a buffer solution, the $\mathrm{pH}$ of the buffer decreases by 0.5 . The buffer capacity of the solution is
1 0.6
2 0.4
3 0.2
4 0.1
Explanation:
Given data: Number of moles of acid added $=\frac{0.1}{2}$ per litre Change of $\mathrm{pH}=0.5$ Now, Buffer capacity $=\frac{\text { number of moles of acid added/L }}{\text { change in } \mathrm{pH}}$ Buffer capacity $=\frac{0.1}{0.5 \times 2}$ Buffer capacity $=0.1$
2008
Ionic Equilibrium
229726
Assertion : Mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and **Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 In the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the assertion is incorrect but the Reason is correct.
Explanation:
Acidic buffer contains equimolar mixture of weak acid and its salt with strong base. $\mathrm{CH}_3 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COONH}_4$ is not an example of acidic buffer.
AIIMS-(2007)
Ionic Equilibrium
229730
For a buffer of a mixture of $0.12 \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{CH}_3 \mathrm{COOH}$ and $0.12 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{CH}_3 \mathrm{COONa}$, the buffer capacity is
1 1.38
2 0.130
3 0.06
4 0.60
Explanation:
Buffer capacity $(\beta)=\frac{\Delta \mathrm{C}_{\mathrm{a}}}{\Delta \mathrm{pH}}$ Where $C_a=$ concentration of acid for acidic buffer Concentration of acid = concentration of salt $\therefore \Delta \mathrm{C}_{\mathrm{a}}=0.12 \mathrm{molL}^{-1}$ Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mathrm{pH}_{\mathrm{pH}}^{\mathrm{pH}}-\log \left[\mathrm{H}^{+}\right]=-\log (0.12)=0.92$ Buffer capacity $(\beta)=\frac{0.12}{0.92}=\frac{12}{92}=\frac{6}{46}=\frac{3}{23}=0.130$
229723
A buffer solution has equal volumes of $0.2 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{OH}$ and $0.02 \mathrm{M} \mathrm{NH} \mathrm{NC}_4$. The $\mathrm{pK}_{\mathrm{b}}$ of the base is 5 . The $\mathrm{pH}$ is
1 10
2 9
3 4
4 7
Explanation:
Given thatvolume of $\mathrm{NH}_4 \mathrm{OH}=0.2 \mathrm{M}$ and volume of $\mathrm{NH}_4 \mathrm{Cl}=0.02 \mathrm{M}$ $\mathrm{pK}_{\mathrm{b}}=5$ According to the Henderson's equation. $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{[\text { base }]} \\ & =5+\log \frac{0.02}{0.2}=5+\log \frac{1}{10}=5+(-1)=4 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-4=10 \end{aligned}$
COMEDK-2017**
Ionic Equilibrium
229724
When 0.1 mole of an acid is added to $2 \mathrm{~L}$ of a buffer solution, the $\mathrm{pH}$ of the buffer decreases by 0.5 . The buffer capacity of the solution is
1 0.6
2 0.4
3 0.2
4 0.1
Explanation:
Given data: Number of moles of acid added $=\frac{0.1}{2}$ per litre Change of $\mathrm{pH}=0.5$ Now, Buffer capacity $=\frac{\text { number of moles of acid added/L }}{\text { change in } \mathrm{pH}}$ Buffer capacity $=\frac{0.1}{0.5 \times 2}$ Buffer capacity $=0.1$
2008
Ionic Equilibrium
229726
Assertion : Mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and **Reason :** Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 In the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the assertion is incorrect but the Reason is correct.
Explanation:
Acidic buffer contains equimolar mixture of weak acid and its salt with strong base. $\mathrm{CH}_3 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COONH}_4$ is not an example of acidic buffer.
AIIMS-(2007)
Ionic Equilibrium
229730
For a buffer of a mixture of $0.12 \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{CH}_3 \mathrm{COOH}$ and $0.12 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{CH}_3 \mathrm{COONa}$, the buffer capacity is
1 1.38
2 0.130
3 0.06
4 0.60
Explanation:
Buffer capacity $(\beta)=\frac{\Delta \mathrm{C}_{\mathrm{a}}}{\Delta \mathrm{pH}}$ Where $C_a=$ concentration of acid for acidic buffer Concentration of acid = concentration of salt $\therefore \Delta \mathrm{C}_{\mathrm{a}}=0.12 \mathrm{molL}^{-1}$ Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mathrm{pH}_{\mathrm{pH}}^{\mathrm{pH}}-\log \left[\mathrm{H}^{+}\right]=-\log (0.12)=0.92$ Buffer capacity $(\beta)=\frac{0.12}{0.92}=\frac{12}{92}=\frac{6}{46}=\frac{3}{23}=0.130$
