Buffer solution are those solutions which resist the change in $\mathrm{pH}$ on addition of small amount of acid or base. Buffer solutions are mixture of either weak acid and its salt with strong base or a weak base and its salt with strong acid. (a) $\mathrm{NaCl}+\mathrm{NaOH}$ is not buffer solution because $\mathrm{NaOH}$ is strong base. (c) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONH}_4$ is not buffer solution because $\mathrm{CH}_3 \mathrm{COONH}_4$ is salt with weak base. (d) $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{CaSO}_4$ is not buffer solution because $\mathrm{H}_2 \mathrm{SO}_4$ is strong acid. (b) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$ is buffer solution. Because $\mathrm{CH}_3 \mathrm{COOH}$ is weak acid and $\mathrm{CH}_3 \mathrm{COONa}$ is its salt with strong base.
UP CPMT-2003
Ionic Equilibrium
229763
A buffer solution is prepared by mixing- 0.1M ammonia and $1.0 \mathrm{M}$ ammonium chloride. At $298 \mathrm{~K}$ the $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$ is 5.0 . The $\mathrm{PH}$ of the buffer is
229764
The ratio of volumes of $\mathrm{CH}_3 \mathrm{COOH} 0.1$ (N) to $\mathrm{CH}_3 \mathrm{COONa} 0.1(\mathrm{~N})$ required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{CH}_3 \mathrm{COOH}$ is 4.74)
1 $10: 1$
2 $5: 1$
3 $1: 5$
4 $1: 10$
Explanation:
The volumes of acetic acid and sodium acetate be VL and V'L respectively. Given that normality of $\mathrm{CH}_3 \mathrm{COOH}[$ Acetic acid] $=0.1 \mathrm{~N}$ Normality of $\mathrm{CH}_3 \mathrm{COONa}=0.1 \mathrm{~N}$ $\mathrm{pH}=5.74$ According to the Henderson's equation- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_3 \mathrm{COONa}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ 5.74=4.74+\log \frac{0.1 \times \mathrm{V}^{\prime}}{0.1 \times \mathrm{V}} \\ 5.74-4.74=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \\ 1=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \end{gathered}$ $\begin{aligned} & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10^1 \quad\left(\begin{array}{c} \because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \\ \mathrm{a}=\mathrm{b}^{\mathrm{x}} \end{array}\right) \\ & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10 \\ & \frac{\mathrm{V}}{\mathrm{V}^{\prime}}=1: 10 \end{aligned}$
WB-JEE-2015
Ionic Equilibrium
229713
Which of the following is correct for acid buffer? $[$ salt $=\mathrm{S}$, acid $=\mathrm{A}]$ ?
Buffer solution are those solutions which resist the change in $\mathrm{pH}$ on addition of small amount of acid or base. Buffer solutions are mixture of either weak acid and its salt with strong base or a weak base and its salt with strong acid. (a) $\mathrm{NaCl}+\mathrm{NaOH}$ is not buffer solution because $\mathrm{NaOH}$ is strong base. (c) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONH}_4$ is not buffer solution because $\mathrm{CH}_3 \mathrm{COONH}_4$ is salt with weak base. (d) $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{CaSO}_4$ is not buffer solution because $\mathrm{H}_2 \mathrm{SO}_4$ is strong acid. (b) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$ is buffer solution. Because $\mathrm{CH}_3 \mathrm{COOH}$ is weak acid and $\mathrm{CH}_3 \mathrm{COONa}$ is its salt with strong base.
UP CPMT-2003
Ionic Equilibrium
229763
A buffer solution is prepared by mixing- 0.1M ammonia and $1.0 \mathrm{M}$ ammonium chloride. At $298 \mathrm{~K}$ the $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$ is 5.0 . The $\mathrm{PH}$ of the buffer is
229764
The ratio of volumes of $\mathrm{CH}_3 \mathrm{COOH} 0.1$ (N) to $\mathrm{CH}_3 \mathrm{COONa} 0.1(\mathrm{~N})$ required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{CH}_3 \mathrm{COOH}$ is 4.74)
1 $10: 1$
2 $5: 1$
3 $1: 5$
4 $1: 10$
Explanation:
The volumes of acetic acid and sodium acetate be VL and V'L respectively. Given that normality of $\mathrm{CH}_3 \mathrm{COOH}[$ Acetic acid] $=0.1 \mathrm{~N}$ Normality of $\mathrm{CH}_3 \mathrm{COONa}=0.1 \mathrm{~N}$ $\mathrm{pH}=5.74$ According to the Henderson's equation- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_3 \mathrm{COONa}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ 5.74=4.74+\log \frac{0.1 \times \mathrm{V}^{\prime}}{0.1 \times \mathrm{V}} \\ 5.74-4.74=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \\ 1=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \end{gathered}$ $\begin{aligned} & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10^1 \quad\left(\begin{array}{c} \because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \\ \mathrm{a}=\mathrm{b}^{\mathrm{x}} \end{array}\right) \\ & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10 \\ & \frac{\mathrm{V}}{\mathrm{V}^{\prime}}=1: 10 \end{aligned}$
WB-JEE-2015
Ionic Equilibrium
229713
Which of the following is correct for acid buffer? $[$ salt $=\mathrm{S}$, acid $=\mathrm{A}]$ ?
Buffer solution are those solutions which resist the change in $\mathrm{pH}$ on addition of small amount of acid or base. Buffer solutions are mixture of either weak acid and its salt with strong base or a weak base and its salt with strong acid. (a) $\mathrm{NaCl}+\mathrm{NaOH}$ is not buffer solution because $\mathrm{NaOH}$ is strong base. (c) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONH}_4$ is not buffer solution because $\mathrm{CH}_3 \mathrm{COONH}_4$ is salt with weak base. (d) $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{CaSO}_4$ is not buffer solution because $\mathrm{H}_2 \mathrm{SO}_4$ is strong acid. (b) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$ is buffer solution. Because $\mathrm{CH}_3 \mathrm{COOH}$ is weak acid and $\mathrm{CH}_3 \mathrm{COONa}$ is its salt with strong base.
UP CPMT-2003
Ionic Equilibrium
229763
A buffer solution is prepared by mixing- 0.1M ammonia and $1.0 \mathrm{M}$ ammonium chloride. At $298 \mathrm{~K}$ the $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$ is 5.0 . The $\mathrm{PH}$ of the buffer is
229764
The ratio of volumes of $\mathrm{CH}_3 \mathrm{COOH} 0.1$ (N) to $\mathrm{CH}_3 \mathrm{COONa} 0.1(\mathrm{~N})$ required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{CH}_3 \mathrm{COOH}$ is 4.74)
1 $10: 1$
2 $5: 1$
3 $1: 5$
4 $1: 10$
Explanation:
The volumes of acetic acid and sodium acetate be VL and V'L respectively. Given that normality of $\mathrm{CH}_3 \mathrm{COOH}[$ Acetic acid] $=0.1 \mathrm{~N}$ Normality of $\mathrm{CH}_3 \mathrm{COONa}=0.1 \mathrm{~N}$ $\mathrm{pH}=5.74$ According to the Henderson's equation- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_3 \mathrm{COONa}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ 5.74=4.74+\log \frac{0.1 \times \mathrm{V}^{\prime}}{0.1 \times \mathrm{V}} \\ 5.74-4.74=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \\ 1=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \end{gathered}$ $\begin{aligned} & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10^1 \quad\left(\begin{array}{c} \because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \\ \mathrm{a}=\mathrm{b}^{\mathrm{x}} \end{array}\right) \\ & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10 \\ & \frac{\mathrm{V}}{\mathrm{V}^{\prime}}=1: 10 \end{aligned}$
WB-JEE-2015
Ionic Equilibrium
229713
Which of the following is correct for acid buffer? $[$ salt $=\mathrm{S}$, acid $=\mathrm{A}]$ ?
Buffer solution are those solutions which resist the change in $\mathrm{pH}$ on addition of small amount of acid or base. Buffer solutions are mixture of either weak acid and its salt with strong base or a weak base and its salt with strong acid. (a) $\mathrm{NaCl}+\mathrm{NaOH}$ is not buffer solution because $\mathrm{NaOH}$ is strong base. (c) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONH}_4$ is not buffer solution because $\mathrm{CH}_3 \mathrm{COONH}_4$ is salt with weak base. (d) $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{CaSO}_4$ is not buffer solution because $\mathrm{H}_2 \mathrm{SO}_4$ is strong acid. (b) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$ is buffer solution. Because $\mathrm{CH}_3 \mathrm{COOH}$ is weak acid and $\mathrm{CH}_3 \mathrm{COONa}$ is its salt with strong base.
UP CPMT-2003
Ionic Equilibrium
229763
A buffer solution is prepared by mixing- 0.1M ammonia and $1.0 \mathrm{M}$ ammonium chloride. At $298 \mathrm{~K}$ the $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$ is 5.0 . The $\mathrm{PH}$ of the buffer is
229764
The ratio of volumes of $\mathrm{CH}_3 \mathrm{COOH} 0.1$ (N) to $\mathrm{CH}_3 \mathrm{COONa} 0.1(\mathrm{~N})$ required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{CH}_3 \mathrm{COOH}$ is 4.74)
1 $10: 1$
2 $5: 1$
3 $1: 5$
4 $1: 10$
Explanation:
The volumes of acetic acid and sodium acetate be VL and V'L respectively. Given that normality of $\mathrm{CH}_3 \mathrm{COOH}[$ Acetic acid] $=0.1 \mathrm{~N}$ Normality of $\mathrm{CH}_3 \mathrm{COONa}=0.1 \mathrm{~N}$ $\mathrm{pH}=5.74$ According to the Henderson's equation- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_3 \mathrm{COONa}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ 5.74=4.74+\log \frac{0.1 \times \mathrm{V}^{\prime}}{0.1 \times \mathrm{V}} \\ 5.74-4.74=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \\ 1=\log \frac{\mathrm{V}^{\prime}}{\mathrm{V}} \end{gathered}$ $\begin{aligned} & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10^1 \quad\left(\begin{array}{c} \because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \\ \mathrm{a}=\mathrm{b}^{\mathrm{x}} \end{array}\right) \\ & \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=10 \\ & \frac{\mathrm{V}}{\mathrm{V}^{\prime}}=1: 10 \end{aligned}$
WB-JEE-2015
Ionic Equilibrium
229713
Which of the following is correct for acid buffer? $[$ salt $=\mathrm{S}$, acid $=\mathrm{A}]$ ?
