NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI07:EQUILIBRIUM
314779
What is the percent ionization \((\alpha)\) of \({\text{0}}{\text{.01}}\,\,{\text{M}}\) HA solution? \(\left(\mathrm{K}_{\mathrm{a}}=10^{-4}\right)\)
314780
Calculate the degree of ionisation of \({\text{0}}{\text{.05}}\,\,{\text{M}}\) acetic acid in \({\text{0}}{\text{.01}}\,\,{\text{M}}\) \({\mkern 1mu} {\text{HCl}}\) if its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) value is 4.74.
1 \(1.82 \times 10^{-2}\)
2 \(1.82 \times 10^{-3}\)
3 \(1.82 \times 10^{-4}\)
4 \(1.82 \times 10^{-5}\)
Explanation:
\(\mathrm {\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {4.74=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {\log \mathrm{K}_{\mathrm{a}}=-4.74=\overline{5} .26}\) \({{\text{K}}_{\text{a}}}{\text{ = antilog}}{\mkern 1mu} {\mkern 1mu} \overline {\text{5}} {\text{.26 = 1}}{\text{.82 }} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}\) In the presence of \({\text{0}}.{\text{01 M }}{{\text{H}}^{\text{ + }}}\) \[\begin{array}{*{20}{l}} {}&{{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}}& \rightleftharpoons &{{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}\;{\text{ + }}}&{{{\text{H}}^{\text{ + }}}} \\ {{\text{Initial}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.005}}\,{\text{M}}}&{}&{\text{0}}&{\text{0}} \\ {{\text{Equili}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.05-C} \alpha}}&{}&{{\text{C}\alpha}}&{{\text{C} \alpha}} \end{array}\] \(\rm {(\mathrm{C} \alpha+0.01) \approx 0.05 \approx 0.01}\) \(\mathrm {\left[\mathrm{CH}_{3} \mathrm{COOH}\right.}\) is a weak acid and \(\mathrm {\mathrm{HCl}}\) is a strong acid so we can assume that \(\mathrm {(\mathrm{C} \alpha+0.01) \approx 0.01]}\) \(\rm {\therefore \mathrm{K}_{\mathrm{a}}=\dfrac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}}\) \(\mathrm {1.82 \times 10^{-5}=\dfrac{\mathrm{C} \alpha \times 0.01}{0.05}}\) \(\mathrm {\mathrm{C} \alpha=\dfrac{1.82 \times 10^{-5} \times 0.05}{0.01}=9.1 \times 10^{-5}}\) \(\mathrm {\mathrm{C} \alpha=9.1 \times 10^{-5}}\) \(\mathrm {\alpha=\dfrac{9.1 \times 10^{-5}}{0.05}=1.82 \times 10^{-3}}\) In the presence of strong acid, dissociation of weak acid i.e., \(\mathrm {\mathrm{CH}_{3} \mathrm{COOH}}\) decreases due to common ion effect.
CHXI07:EQUILIBRIUM
314781
The dissociation constants of formic acid and acetic acid are \(1.77 \times 10^{-4}\) and \(1.77 \times 10^{-5}\) respectively. The relative strength of the two acids is
1 3.18
2 100
3 6.36
4 5.0
Explanation:
Relative strength of acids \(=\sqrt{\dfrac{\mathrm{K}_{\mathrm{a}_{2}}}{\mathrm{~K}_{\mathrm{a}_{1}}}}\) \(=\sqrt{\dfrac{1.77 \times 10^{-4}}{1.77 \times 10^{-5}}}=3.18\)
CHXI07:EQUILIBRIUM
314782
For a weak acid, the incorrect statement is
1 its dissociation constant is low
2 its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) is very low
3 it is partially dissociated
4 solution of its sodium salt is alkaline in water
Explanation:
A weak acid dissociates partially when dissolved in water. Thus, it has very low value of \(\mathrm{K}_{\mathrm{a}}\) i.e. dissociation constant. Since, \(\mathrm{pK}_{\mathrm{a}}\) is the negative logarithm of \(\mathrm{K}_{\mathrm{a}}\left(\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right)\) thus, weak acid has a very high value of \(\mathrm{pK}_{\mathrm{a}}\) Also, the solution of its sodium salt is alkaline in water, e.g. \(\mathrm{CH}_{3} \mathrm{COONa}\) is alkaline in aqueous solution is incouse. Thus, option (v) is micousect while rest the statement are correct.
314779
What is the percent ionization \((\alpha)\) of \({\text{0}}{\text{.01}}\,\,{\text{M}}\) HA solution? \(\left(\mathrm{K}_{\mathrm{a}}=10^{-4}\right)\)
314780
Calculate the degree of ionisation of \({\text{0}}{\text{.05}}\,\,{\text{M}}\) acetic acid in \({\text{0}}{\text{.01}}\,\,{\text{M}}\) \({\mkern 1mu} {\text{HCl}}\) if its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) value is 4.74.
1 \(1.82 \times 10^{-2}\)
2 \(1.82 \times 10^{-3}\)
3 \(1.82 \times 10^{-4}\)
4 \(1.82 \times 10^{-5}\)
Explanation:
\(\mathrm {\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {4.74=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {\log \mathrm{K}_{\mathrm{a}}=-4.74=\overline{5} .26}\) \({{\text{K}}_{\text{a}}}{\text{ = antilog}}{\mkern 1mu} {\mkern 1mu} \overline {\text{5}} {\text{.26 = 1}}{\text{.82 }} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}\) In the presence of \({\text{0}}.{\text{01 M }}{{\text{H}}^{\text{ + }}}\) \[\begin{array}{*{20}{l}} {}&{{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}}& \rightleftharpoons &{{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}\;{\text{ + }}}&{{{\text{H}}^{\text{ + }}}} \\ {{\text{Initial}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.005}}\,{\text{M}}}&{}&{\text{0}}&{\text{0}} \\ {{\text{Equili}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.05-C} \alpha}}&{}&{{\text{C}\alpha}}&{{\text{C} \alpha}} \end{array}\] \(\rm {(\mathrm{C} \alpha+0.01) \approx 0.05 \approx 0.01}\) \(\mathrm {\left[\mathrm{CH}_{3} \mathrm{COOH}\right.}\) is a weak acid and \(\mathrm {\mathrm{HCl}}\) is a strong acid so we can assume that \(\mathrm {(\mathrm{C} \alpha+0.01) \approx 0.01]}\) \(\rm {\therefore \mathrm{K}_{\mathrm{a}}=\dfrac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}}\) \(\mathrm {1.82 \times 10^{-5}=\dfrac{\mathrm{C} \alpha \times 0.01}{0.05}}\) \(\mathrm {\mathrm{C} \alpha=\dfrac{1.82 \times 10^{-5} \times 0.05}{0.01}=9.1 \times 10^{-5}}\) \(\mathrm {\mathrm{C} \alpha=9.1 \times 10^{-5}}\) \(\mathrm {\alpha=\dfrac{9.1 \times 10^{-5}}{0.05}=1.82 \times 10^{-3}}\) In the presence of strong acid, dissociation of weak acid i.e., \(\mathrm {\mathrm{CH}_{3} \mathrm{COOH}}\) decreases due to common ion effect.
CHXI07:EQUILIBRIUM
314781
The dissociation constants of formic acid and acetic acid are \(1.77 \times 10^{-4}\) and \(1.77 \times 10^{-5}\) respectively. The relative strength of the two acids is
1 3.18
2 100
3 6.36
4 5.0
Explanation:
Relative strength of acids \(=\sqrt{\dfrac{\mathrm{K}_{\mathrm{a}_{2}}}{\mathrm{~K}_{\mathrm{a}_{1}}}}\) \(=\sqrt{\dfrac{1.77 \times 10^{-4}}{1.77 \times 10^{-5}}}=3.18\)
CHXI07:EQUILIBRIUM
314782
For a weak acid, the incorrect statement is
1 its dissociation constant is low
2 its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) is very low
3 it is partially dissociated
4 solution of its sodium salt is alkaline in water
Explanation:
A weak acid dissociates partially when dissolved in water. Thus, it has very low value of \(\mathrm{K}_{\mathrm{a}}\) i.e. dissociation constant. Since, \(\mathrm{pK}_{\mathrm{a}}\) is the negative logarithm of \(\mathrm{K}_{\mathrm{a}}\left(\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right)\) thus, weak acid has a very high value of \(\mathrm{pK}_{\mathrm{a}}\) Also, the solution of its sodium salt is alkaline in water, e.g. \(\mathrm{CH}_{3} \mathrm{COONa}\) is alkaline in aqueous solution is incouse. Thus, option (v) is micousect while rest the statement are correct.
314779
What is the percent ionization \((\alpha)\) of \({\text{0}}{\text{.01}}\,\,{\text{M}}\) HA solution? \(\left(\mathrm{K}_{\mathrm{a}}=10^{-4}\right)\)
314780
Calculate the degree of ionisation of \({\text{0}}{\text{.05}}\,\,{\text{M}}\) acetic acid in \({\text{0}}{\text{.01}}\,\,{\text{M}}\) \({\mkern 1mu} {\text{HCl}}\) if its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) value is 4.74.
1 \(1.82 \times 10^{-2}\)
2 \(1.82 \times 10^{-3}\)
3 \(1.82 \times 10^{-4}\)
4 \(1.82 \times 10^{-5}\)
Explanation:
\(\mathrm {\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {4.74=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {\log \mathrm{K}_{\mathrm{a}}=-4.74=\overline{5} .26}\) \({{\text{K}}_{\text{a}}}{\text{ = antilog}}{\mkern 1mu} {\mkern 1mu} \overline {\text{5}} {\text{.26 = 1}}{\text{.82 }} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}\) In the presence of \({\text{0}}.{\text{01 M }}{{\text{H}}^{\text{ + }}}\) \[\begin{array}{*{20}{l}} {}&{{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}}& \rightleftharpoons &{{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}\;{\text{ + }}}&{{{\text{H}}^{\text{ + }}}} \\ {{\text{Initial}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.005}}\,{\text{M}}}&{}&{\text{0}}&{\text{0}} \\ {{\text{Equili}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.05-C} \alpha}}&{}&{{\text{C}\alpha}}&{{\text{C} \alpha}} \end{array}\] \(\rm {(\mathrm{C} \alpha+0.01) \approx 0.05 \approx 0.01}\) \(\mathrm {\left[\mathrm{CH}_{3} \mathrm{COOH}\right.}\) is a weak acid and \(\mathrm {\mathrm{HCl}}\) is a strong acid so we can assume that \(\mathrm {(\mathrm{C} \alpha+0.01) \approx 0.01]}\) \(\rm {\therefore \mathrm{K}_{\mathrm{a}}=\dfrac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}}\) \(\mathrm {1.82 \times 10^{-5}=\dfrac{\mathrm{C} \alpha \times 0.01}{0.05}}\) \(\mathrm {\mathrm{C} \alpha=\dfrac{1.82 \times 10^{-5} \times 0.05}{0.01}=9.1 \times 10^{-5}}\) \(\mathrm {\mathrm{C} \alpha=9.1 \times 10^{-5}}\) \(\mathrm {\alpha=\dfrac{9.1 \times 10^{-5}}{0.05}=1.82 \times 10^{-3}}\) In the presence of strong acid, dissociation of weak acid i.e., \(\mathrm {\mathrm{CH}_{3} \mathrm{COOH}}\) decreases due to common ion effect.
CHXI07:EQUILIBRIUM
314781
The dissociation constants of formic acid and acetic acid are \(1.77 \times 10^{-4}\) and \(1.77 \times 10^{-5}\) respectively. The relative strength of the two acids is
1 3.18
2 100
3 6.36
4 5.0
Explanation:
Relative strength of acids \(=\sqrt{\dfrac{\mathrm{K}_{\mathrm{a}_{2}}}{\mathrm{~K}_{\mathrm{a}_{1}}}}\) \(=\sqrt{\dfrac{1.77 \times 10^{-4}}{1.77 \times 10^{-5}}}=3.18\)
CHXI07:EQUILIBRIUM
314782
For a weak acid, the incorrect statement is
1 its dissociation constant is low
2 its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) is very low
3 it is partially dissociated
4 solution of its sodium salt is alkaline in water
Explanation:
A weak acid dissociates partially when dissolved in water. Thus, it has very low value of \(\mathrm{K}_{\mathrm{a}}\) i.e. dissociation constant. Since, \(\mathrm{pK}_{\mathrm{a}}\) is the negative logarithm of \(\mathrm{K}_{\mathrm{a}}\left(\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right)\) thus, weak acid has a very high value of \(\mathrm{pK}_{\mathrm{a}}\) Also, the solution of its sodium salt is alkaline in water, e.g. \(\mathrm{CH}_{3} \mathrm{COONa}\) is alkaline in aqueous solution is incouse. Thus, option (v) is micousect while rest the statement are correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXI07:EQUILIBRIUM
314779
What is the percent ionization \((\alpha)\) of \({\text{0}}{\text{.01}}\,\,{\text{M}}\) HA solution? \(\left(\mathrm{K}_{\mathrm{a}}=10^{-4}\right)\)
314780
Calculate the degree of ionisation of \({\text{0}}{\text{.05}}\,\,{\text{M}}\) acetic acid in \({\text{0}}{\text{.01}}\,\,{\text{M}}\) \({\mkern 1mu} {\text{HCl}}\) if its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) value is 4.74.
1 \(1.82 \times 10^{-2}\)
2 \(1.82 \times 10^{-3}\)
3 \(1.82 \times 10^{-4}\)
4 \(1.82 \times 10^{-5}\)
Explanation:
\(\mathrm {\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {4.74=-\log \mathrm{K}_{\mathrm{a}}}\) \(\mathrm {\log \mathrm{K}_{\mathrm{a}}=-4.74=\overline{5} .26}\) \({{\text{K}}_{\text{a}}}{\text{ = antilog}}{\mkern 1mu} {\mkern 1mu} \overline {\text{5}} {\text{.26 = 1}}{\text{.82 }} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}\) In the presence of \({\text{0}}.{\text{01 M }}{{\text{H}}^{\text{ + }}}\) \[\begin{array}{*{20}{l}} {}&{{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}}& \rightleftharpoons &{{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}\;{\text{ + }}}&{{{\text{H}}^{\text{ + }}}} \\ {{\text{Initial}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.005}}\,{\text{M}}}&{}&{\text{0}}&{\text{0}} \\ {{\text{Equili}}\;{\text{conc}}{\text{.}}}&{{\text{0}}{\text{.05-C} \alpha}}&{}&{{\text{C}\alpha}}&{{\text{C} \alpha}} \end{array}\] \(\rm {(\mathrm{C} \alpha+0.01) \approx 0.05 \approx 0.01}\) \(\mathrm {\left[\mathrm{CH}_{3} \mathrm{COOH}\right.}\) is a weak acid and \(\mathrm {\mathrm{HCl}}\) is a strong acid so we can assume that \(\mathrm {(\mathrm{C} \alpha+0.01) \approx 0.01]}\) \(\rm {\therefore \mathrm{K}_{\mathrm{a}}=\dfrac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}}\) \(\mathrm {1.82 \times 10^{-5}=\dfrac{\mathrm{C} \alpha \times 0.01}{0.05}}\) \(\mathrm {\mathrm{C} \alpha=\dfrac{1.82 \times 10^{-5} \times 0.05}{0.01}=9.1 \times 10^{-5}}\) \(\mathrm {\mathrm{C} \alpha=9.1 \times 10^{-5}}\) \(\mathrm {\alpha=\dfrac{9.1 \times 10^{-5}}{0.05}=1.82 \times 10^{-3}}\) In the presence of strong acid, dissociation of weak acid i.e., \(\mathrm {\mathrm{CH}_{3} \mathrm{COOH}}\) decreases due to common ion effect.
CHXI07:EQUILIBRIUM
314781
The dissociation constants of formic acid and acetic acid are \(1.77 \times 10^{-4}\) and \(1.77 \times 10^{-5}\) respectively. The relative strength of the two acids is
1 3.18
2 100
3 6.36
4 5.0
Explanation:
Relative strength of acids \(=\sqrt{\dfrac{\mathrm{K}_{\mathrm{a}_{2}}}{\mathrm{~K}_{\mathrm{a}_{1}}}}\) \(=\sqrt{\dfrac{1.77 \times 10^{-4}}{1.77 \times 10^{-5}}}=3.18\)
CHXI07:EQUILIBRIUM
314782
For a weak acid, the incorrect statement is
1 its dissociation constant is low
2 its \({\rm{p}}{{\rm{K}}_{\rm{a}}}\) is very low
3 it is partially dissociated
4 solution of its sodium salt is alkaline in water
Explanation:
A weak acid dissociates partially when dissolved in water. Thus, it has very low value of \(\mathrm{K}_{\mathrm{a}}\) i.e. dissociation constant. Since, \(\mathrm{pK}_{\mathrm{a}}\) is the negative logarithm of \(\mathrm{K}_{\mathrm{a}}\left(\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right)\) thus, weak acid has a very high value of \(\mathrm{pK}_{\mathrm{a}}\) Also, the solution of its sodium salt is alkaline in water, e.g. \(\mathrm{CH}_{3} \mathrm{COONa}\) is alkaline in aqueous solution is incouse. Thus, option (v) is micousect while rest the statement are correct.