NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229731
A buffer contains equal concentrations of $\mathrm{X}^{-}$ and HX. The $K_a$ for $H X$ is $10^{-8}$. The $p H$ of the buffer is-
1 3
2 8
3 7
4 11
Explanation:
A buffer consist of a weak acid and its conjugate base or a weak base and its conjugate acid. $\mathrm{k}_{\mathrm{a}}$ is given to us it is an indication that the given buffer is an example of an acidic buffer. Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mid \begin{aligned} & \mathrm{pH}=-\log \mathrm{k}_{\mathrm{W}}+\log \frac{\text { [conjugate base }]}{[\text { Acid }]} \\ & \mathrm{pH}=\mathrm{pk}_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]} \\ & \mathrm{pH}=-\log _{10} 10^{-8}+\log _{10} 1=8 \end{aligned}$ As the concentrations of $\mathrm{X}^{-}$and $\mathrm{HX}$ are equal
BCECE-2009
Ionic Equilibrium
229732
A buffer solution contains 0.1 mole of sodium acetate in $1000 \mathrm{~cm}^3$ of $0.1 \mathrm{M}$ acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The $\mathrm{pH}$ of the resulting buffer is equal to-
229727
During titration of acetic acid with aq. $\mathrm{NaOH}$ solution, the neutralisation graph has a vertical line. This line indicates
1 alkaline nature of equivalence.
2 acidic nature of equivalence.
3 neutral nature of equivalence.
4 depends on experimental proceeding.
Explanation:
In the titration of a weak acid $\mathrm{CH}_3 \mathrm{COOH}$ with a strong base $(\mathrm{NaOH})$, there is a change in the $\mathrm{pH}$ value at the end point. But it is not sharp due to weak ionisation of $\mathrm{CH}_3 \mathrm{COOH}$. Due to the excess of free base beyond the equivalence point, there is a step rise in $\mathrm{pH}$ which is indicated by the vertical line. Hence, the ventical line in the graph indicates alkaline nature of equivalence.
AIIMS- 2007
Ionic Equilibrium
229735
Which solution is buffer?
1 Acetic acid $+\mathrm{NaOH}$ (equimolar ratio)
2 Acetic acid+ $\mathrm{NaOH}(1: 2$ molar ratio
3 Acetic acid $+\mathrm{NaOH}$ (2:1 molar ratio)
4 $\mathrm{HCl}+\mathrm{NaOH}$ (equimolar ratio)
Explanation:
A buffer solution can be described as a solution, which will resist changes in $\mathrm{pH}$ when a small amount of a strong acid or base is added. An acidic buffer is compound of a weak acid and its conjugated base or a weak acid and the salt of the weak acid and a strong base. e.g., $\left(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}\right)$. $\mathrm{CH}_3 \mathrm{COOH}$ ? $\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$ (feebly ionised) \(\mathrm{CH}_3 \mathrm{COONa}\) \(\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}\) (completely ionised) How to make a acidic buffer (i) start with taking weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and strong base \((\mathrm{NaOH})\) in 2:1 molar concentration ratio, so that a sufficient number of moles of the \(\mathrm{CH}_3 \mathrm{COOH}\) neutralize completely the same number of moles of \(\mathrm{NaOH}\). \(\underset{1 \mathrm{~mol}}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{1 \mathrm{~mol}}{\mathrm{NaOH}} \longrightarrow \underset{\substack{\text { Sodiumacetate } \\ \text { 1mmol }}}{\mathrm{CH}_3 \mathrm{COONa}}+\mathrm{H}_2 \mathrm{O}\) How to make a acidic buffer. Now, the resulting solution will contain the salt and acid in equimolar concentration and water. \(\underset{\text { 1limiting(mol) }}{\mathrm{OH}^{-}(\mathrm{aq})}+\underset{\text { (excess 2mol) }}{\mathrm{H}^{+}(\mathrm{aq})} \longrightarrow \underset{\substack{\text { salt(1mol) buffer } \\ \text { solution }}}{\mathrm{A}^{+}(\mathrm{aq})}+\underset{\text { remaining(mol) }}{\mathrm{H}^{+}(\mathrm{aq})}\) In the buffer solution, \([\mathrm{HA}]=\left[\mathrm{A}^{\top}\right]\) i.e., equimolar concentrations (ii) The acidic buffer can also prepared by taking equimolar concentrations and volume of weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and the salt of the weak acid and strong base \(\left(\mathrm{CH}_3 \mathrm{COONa}\right)\).
229731
A buffer contains equal concentrations of $\mathrm{X}^{-}$ and HX. The $K_a$ for $H X$ is $10^{-8}$. The $p H$ of the buffer is-
1 3
2 8
3 7
4 11
Explanation:
A buffer consist of a weak acid and its conjugate base or a weak base and its conjugate acid. $\mathrm{k}_{\mathrm{a}}$ is given to us it is an indication that the given buffer is an example of an acidic buffer. Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mid \begin{aligned} & \mathrm{pH}=-\log \mathrm{k}_{\mathrm{W}}+\log \frac{\text { [conjugate base }]}{[\text { Acid }]} \\ & \mathrm{pH}=\mathrm{pk}_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]} \\ & \mathrm{pH}=-\log _{10} 10^{-8}+\log _{10} 1=8 \end{aligned}$ As the concentrations of $\mathrm{X}^{-}$and $\mathrm{HX}$ are equal
BCECE-2009
Ionic Equilibrium
229732
A buffer solution contains 0.1 mole of sodium acetate in $1000 \mathrm{~cm}^3$ of $0.1 \mathrm{M}$ acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The $\mathrm{pH}$ of the resulting buffer is equal to-
229727
During titration of acetic acid with aq. $\mathrm{NaOH}$ solution, the neutralisation graph has a vertical line. This line indicates
1 alkaline nature of equivalence.
2 acidic nature of equivalence.
3 neutral nature of equivalence.
4 depends on experimental proceeding.
Explanation:
In the titration of a weak acid $\mathrm{CH}_3 \mathrm{COOH}$ with a strong base $(\mathrm{NaOH})$, there is a change in the $\mathrm{pH}$ value at the end point. But it is not sharp due to weak ionisation of $\mathrm{CH}_3 \mathrm{COOH}$. Due to the excess of free base beyond the equivalence point, there is a step rise in $\mathrm{pH}$ which is indicated by the vertical line. Hence, the ventical line in the graph indicates alkaline nature of equivalence.
AIIMS- 2007
Ionic Equilibrium
229735
Which solution is buffer?
1 Acetic acid $+\mathrm{NaOH}$ (equimolar ratio)
2 Acetic acid+ $\mathrm{NaOH}(1: 2$ molar ratio
3 Acetic acid $+\mathrm{NaOH}$ (2:1 molar ratio)
4 $\mathrm{HCl}+\mathrm{NaOH}$ (equimolar ratio)
Explanation:
A buffer solution can be described as a solution, which will resist changes in $\mathrm{pH}$ when a small amount of a strong acid or base is added. An acidic buffer is compound of a weak acid and its conjugated base or a weak acid and the salt of the weak acid and a strong base. e.g., $\left(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}\right)$. $\mathrm{CH}_3 \mathrm{COOH}$ ? $\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$ (feebly ionised) \(\mathrm{CH}_3 \mathrm{COONa}\) \(\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}\) (completely ionised) How to make a acidic buffer (i) start with taking weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and strong base \((\mathrm{NaOH})\) in 2:1 molar concentration ratio, so that a sufficient number of moles of the \(\mathrm{CH}_3 \mathrm{COOH}\) neutralize completely the same number of moles of \(\mathrm{NaOH}\). \(\underset{1 \mathrm{~mol}}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{1 \mathrm{~mol}}{\mathrm{NaOH}} \longrightarrow \underset{\substack{\text { Sodiumacetate } \\ \text { 1mmol }}}{\mathrm{CH}_3 \mathrm{COONa}}+\mathrm{H}_2 \mathrm{O}\) How to make a acidic buffer. Now, the resulting solution will contain the salt and acid in equimolar concentration and water. \(\underset{\text { 1limiting(mol) }}{\mathrm{OH}^{-}(\mathrm{aq})}+\underset{\text { (excess 2mol) }}{\mathrm{H}^{+}(\mathrm{aq})} \longrightarrow \underset{\substack{\text { salt(1mol) buffer } \\ \text { solution }}}{\mathrm{A}^{+}(\mathrm{aq})}+\underset{\text { remaining(mol) }}{\mathrm{H}^{+}(\mathrm{aq})}\) In the buffer solution, \([\mathrm{HA}]=\left[\mathrm{A}^{\top}\right]\) i.e., equimolar concentrations (ii) The acidic buffer can also prepared by taking equimolar concentrations and volume of weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and the salt of the weak acid and strong base \(\left(\mathrm{CH}_3 \mathrm{COONa}\right)\).
229731
A buffer contains equal concentrations of $\mathrm{X}^{-}$ and HX. The $K_a$ for $H X$ is $10^{-8}$. The $p H$ of the buffer is-
1 3
2 8
3 7
4 11
Explanation:
A buffer consist of a weak acid and its conjugate base or a weak base and its conjugate acid. $\mathrm{k}_{\mathrm{a}}$ is given to us it is an indication that the given buffer is an example of an acidic buffer. Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mid \begin{aligned} & \mathrm{pH}=-\log \mathrm{k}_{\mathrm{W}}+\log \frac{\text { [conjugate base }]}{[\text { Acid }]} \\ & \mathrm{pH}=\mathrm{pk}_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]} \\ & \mathrm{pH}=-\log _{10} 10^{-8}+\log _{10} 1=8 \end{aligned}$ As the concentrations of $\mathrm{X}^{-}$and $\mathrm{HX}$ are equal
BCECE-2009
Ionic Equilibrium
229732
A buffer solution contains 0.1 mole of sodium acetate in $1000 \mathrm{~cm}^3$ of $0.1 \mathrm{M}$ acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The $\mathrm{pH}$ of the resulting buffer is equal to-
229727
During titration of acetic acid with aq. $\mathrm{NaOH}$ solution, the neutralisation graph has a vertical line. This line indicates
1 alkaline nature of equivalence.
2 acidic nature of equivalence.
3 neutral nature of equivalence.
4 depends on experimental proceeding.
Explanation:
In the titration of a weak acid $\mathrm{CH}_3 \mathrm{COOH}$ with a strong base $(\mathrm{NaOH})$, there is a change in the $\mathrm{pH}$ value at the end point. But it is not sharp due to weak ionisation of $\mathrm{CH}_3 \mathrm{COOH}$. Due to the excess of free base beyond the equivalence point, there is a step rise in $\mathrm{pH}$ which is indicated by the vertical line. Hence, the ventical line in the graph indicates alkaline nature of equivalence.
AIIMS- 2007
Ionic Equilibrium
229735
Which solution is buffer?
1 Acetic acid $+\mathrm{NaOH}$ (equimolar ratio)
2 Acetic acid+ $\mathrm{NaOH}(1: 2$ molar ratio
3 Acetic acid $+\mathrm{NaOH}$ (2:1 molar ratio)
4 $\mathrm{HCl}+\mathrm{NaOH}$ (equimolar ratio)
Explanation:
A buffer solution can be described as a solution, which will resist changes in $\mathrm{pH}$ when a small amount of a strong acid or base is added. An acidic buffer is compound of a weak acid and its conjugated base or a weak acid and the salt of the weak acid and a strong base. e.g., $\left(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}\right)$. $\mathrm{CH}_3 \mathrm{COOH}$ ? $\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$ (feebly ionised) \(\mathrm{CH}_3 \mathrm{COONa}\) \(\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}\) (completely ionised) How to make a acidic buffer (i) start with taking weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and strong base \((\mathrm{NaOH})\) in 2:1 molar concentration ratio, so that a sufficient number of moles of the \(\mathrm{CH}_3 \mathrm{COOH}\) neutralize completely the same number of moles of \(\mathrm{NaOH}\). \(\underset{1 \mathrm{~mol}}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{1 \mathrm{~mol}}{\mathrm{NaOH}} \longrightarrow \underset{\substack{\text { Sodiumacetate } \\ \text { 1mmol }}}{\mathrm{CH}_3 \mathrm{COONa}}+\mathrm{H}_2 \mathrm{O}\) How to make a acidic buffer. Now, the resulting solution will contain the salt and acid in equimolar concentration and water. \(\underset{\text { 1limiting(mol) }}{\mathrm{OH}^{-}(\mathrm{aq})}+\underset{\text { (excess 2mol) }}{\mathrm{H}^{+}(\mathrm{aq})} \longrightarrow \underset{\substack{\text { salt(1mol) buffer } \\ \text { solution }}}{\mathrm{A}^{+}(\mathrm{aq})}+\underset{\text { remaining(mol) }}{\mathrm{H}^{+}(\mathrm{aq})}\) In the buffer solution, \([\mathrm{HA}]=\left[\mathrm{A}^{\top}\right]\) i.e., equimolar concentrations (ii) The acidic buffer can also prepared by taking equimolar concentrations and volume of weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and the salt of the weak acid and strong base \(\left(\mathrm{CH}_3 \mathrm{COONa}\right)\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229731
A buffer contains equal concentrations of $\mathrm{X}^{-}$ and HX. The $K_a$ for $H X$ is $10^{-8}$. The $p H$ of the buffer is-
1 3
2 8
3 7
4 11
Explanation:
A buffer consist of a weak acid and its conjugate base or a weak base and its conjugate acid. $\mathrm{k}_{\mathrm{a}}$ is given to us it is an indication that the given buffer is an example of an acidic buffer. Hence, we use the following formula for calculating its $\mathrm{pH}$. $\mid \begin{aligned} & \mathrm{pH}=-\log \mathrm{k}_{\mathrm{W}}+\log \frac{\text { [conjugate base }]}{[\text { Acid }]} \\ & \mathrm{pH}=\mathrm{pk}_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]} \\ & \mathrm{pH}=-\log _{10} 10^{-8}+\log _{10} 1=8 \end{aligned}$ As the concentrations of $\mathrm{X}^{-}$and $\mathrm{HX}$ are equal
BCECE-2009
Ionic Equilibrium
229732
A buffer solution contains 0.1 mole of sodium acetate in $1000 \mathrm{~cm}^3$ of $0.1 \mathrm{M}$ acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The $\mathrm{pH}$ of the resulting buffer is equal to-
229727
During titration of acetic acid with aq. $\mathrm{NaOH}$ solution, the neutralisation graph has a vertical line. This line indicates
1 alkaline nature of equivalence.
2 acidic nature of equivalence.
3 neutral nature of equivalence.
4 depends on experimental proceeding.
Explanation:
In the titration of a weak acid $\mathrm{CH}_3 \mathrm{COOH}$ with a strong base $(\mathrm{NaOH})$, there is a change in the $\mathrm{pH}$ value at the end point. But it is not sharp due to weak ionisation of $\mathrm{CH}_3 \mathrm{COOH}$. Due to the excess of free base beyond the equivalence point, there is a step rise in $\mathrm{pH}$ which is indicated by the vertical line. Hence, the ventical line in the graph indicates alkaline nature of equivalence.
AIIMS- 2007
Ionic Equilibrium
229735
Which solution is buffer?
1 Acetic acid $+\mathrm{NaOH}$ (equimolar ratio)
2 Acetic acid+ $\mathrm{NaOH}(1: 2$ molar ratio
3 Acetic acid $+\mathrm{NaOH}$ (2:1 molar ratio)
4 $\mathrm{HCl}+\mathrm{NaOH}$ (equimolar ratio)
Explanation:
A buffer solution can be described as a solution, which will resist changes in $\mathrm{pH}$ when a small amount of a strong acid or base is added. An acidic buffer is compound of a weak acid and its conjugated base or a weak acid and the salt of the weak acid and a strong base. e.g., $\left(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}\right)$. $\mathrm{CH}_3 \mathrm{COOH}$ ? $\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$ (feebly ionised) \(\mathrm{CH}_3 \mathrm{COONa}\) \(\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}\) (completely ionised) How to make a acidic buffer (i) start with taking weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and strong base \((\mathrm{NaOH})\) in 2:1 molar concentration ratio, so that a sufficient number of moles of the \(\mathrm{CH}_3 \mathrm{COOH}\) neutralize completely the same number of moles of \(\mathrm{NaOH}\). \(\underset{1 \mathrm{~mol}}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{1 \mathrm{~mol}}{\mathrm{NaOH}} \longrightarrow \underset{\substack{\text { Sodiumacetate } \\ \text { 1mmol }}}{\mathrm{CH}_3 \mathrm{COONa}}+\mathrm{H}_2 \mathrm{O}\) How to make a acidic buffer. Now, the resulting solution will contain the salt and acid in equimolar concentration and water. \(\underset{\text { 1limiting(mol) }}{\mathrm{OH}^{-}(\mathrm{aq})}+\underset{\text { (excess 2mol) }}{\mathrm{H}^{+}(\mathrm{aq})} \longrightarrow \underset{\substack{\text { salt(1mol) buffer } \\ \text { solution }}}{\mathrm{A}^{+}(\mathrm{aq})}+\underset{\text { remaining(mol) }}{\mathrm{H}^{+}(\mathrm{aq})}\) In the buffer solution, \([\mathrm{HA}]=\left[\mathrm{A}^{\top}\right]\) i.e., equimolar concentrations (ii) The acidic buffer can also prepared by taking equimolar concentrations and volume of weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and the salt of the weak acid and strong base \(\left(\mathrm{CH}_3 \mathrm{COONa}\right)\).
