314446
Among the following is only Bronsted Lowry acid but not an Arrhenius acid?
1 \(\mathrm{AlCl}_{3}\)
2 \(\mathrm{NH}_{4}^{+}\)
3 \(\mathrm{BF}_{3}\)
4 \(\mathrm{CH}_{3} \mathrm{COOH}\)
Explanation:
Ammonium ion \(\mathrm {\left(\mathrm{NH}_{4}^{+}\right)}\)behaves as Bronsted Lowry acid as it has a capacity to donate \(\mathrm {\mathrm{H}^{+}}\)ion.
CHXI07:EQUILIBRIUM
314447
Which of the following can act as both Bronsted acid and Bronsted base?i) \(\mathrm{HCOO}^{-}\)ii) \(\mathrm{NH}_{3}\)iii) \(\mathrm{O}^{-2}\)iv) \(\mathrm{HSO}_{4}^{-}\)
1 (i) and (ii)
2 (ii) and (iii)
3 (ii) and (iv)
4 (i) and (iv)
Explanation:
Both \(\mathrm{NH}_{3}\) and \(\mathrm{HSO}_{4}^{-}\)have a tendency to accept and donate protons.
CHXI07:EQUILIBRIUM
314448
The conjugate base of hydrazoic acid is:
1 \({{\text{N}}^{{\text{ -3}}}}\)
2 \({\text{N}}_{\text{3}}^{\text{ - }}\)
3 \({\text{N}}_{\text{2}}^{\text{ - }}\)
4 \({\text{HN}}_{\text{3}}^{\text{ - }}\)
Explanation:
\(\mathrm {\underset{\text { Hydrazoicacid }}{\mathrm{N}_{3} H} \rightleftharpoons N_{3}^{-}+H^{+}}\) i.e., conjugate base of hydrazoic acid is \(\mathrm {N_{3}}\)
JEE - 2014
CHXI07:EQUILIBRIUM
314769
A 0.1 molar solution of weak base \(\mathrm{BOH}\) is \(1 \%\) dissociated. If 0.2 mole of \(\mathrm{BCl}\) is added in 1 litre solution of \(\mathrm{BOH}\), the degree of dissociation of \(\mathrm{BOH}\) will become
1 \(2 \times 10^{-3}\)
2 \(5 \times 10^{-5}\)
3 0.05
4 0.02
Explanation:
\(\mathrm {K_{b}=C \alpha^{2}=1 \times 10^{-5}}\) In the presence of 0.2 moles \(\mathrm {\mathrm{BCl}}\) \(\rm {\underset{0.1-0.1 \alpha}{\mathrm{BOH}} \rightleftharpoons \underset{0.1 \alpha+0.2}{\mathrm{~B}^{+}}+\underset{0.1 \alpha}{\mathrm{OH}^{-}}}\) \(\mathrm {0.1 \alpha+0.2 \approx 0.2}\). Due to common ion, \(\mathrm {\mathrm{B}^{+}}\) ion concentration due to \(\mathrm {\mathrm{BOH}}\) is negligible. \(\mathrm {0.1-0.1 \alpha=0.1}\) Since \(\mathrm {\alpha}\) is very less. \(\rm {K_{b}=\dfrac{0.2 \times 0.1 \alpha}{0.1} \Rightarrow 10^{-5}=2 \times 10^{-1} \times \alpha \\ \therefore \quad \alpha=5 \times 10^{-5}}\)
314446
Among the following is only Bronsted Lowry acid but not an Arrhenius acid?
1 \(\mathrm{AlCl}_{3}\)
2 \(\mathrm{NH}_{4}^{+}\)
3 \(\mathrm{BF}_{3}\)
4 \(\mathrm{CH}_{3} \mathrm{COOH}\)
Explanation:
Ammonium ion \(\mathrm {\left(\mathrm{NH}_{4}^{+}\right)}\)behaves as Bronsted Lowry acid as it has a capacity to donate \(\mathrm {\mathrm{H}^{+}}\)ion.
CHXI07:EQUILIBRIUM
314447
Which of the following can act as both Bronsted acid and Bronsted base?i) \(\mathrm{HCOO}^{-}\)ii) \(\mathrm{NH}_{3}\)iii) \(\mathrm{O}^{-2}\)iv) \(\mathrm{HSO}_{4}^{-}\)
1 (i) and (ii)
2 (ii) and (iii)
3 (ii) and (iv)
4 (i) and (iv)
Explanation:
Both \(\mathrm{NH}_{3}\) and \(\mathrm{HSO}_{4}^{-}\)have a tendency to accept and donate protons.
CHXI07:EQUILIBRIUM
314448
The conjugate base of hydrazoic acid is:
1 \({{\text{N}}^{{\text{ -3}}}}\)
2 \({\text{N}}_{\text{3}}^{\text{ - }}\)
3 \({\text{N}}_{\text{2}}^{\text{ - }}\)
4 \({\text{HN}}_{\text{3}}^{\text{ - }}\)
Explanation:
\(\mathrm {\underset{\text { Hydrazoicacid }}{\mathrm{N}_{3} H} \rightleftharpoons N_{3}^{-}+H^{+}}\) i.e., conjugate base of hydrazoic acid is \(\mathrm {N_{3}}\)
JEE - 2014
CHXI07:EQUILIBRIUM
314769
A 0.1 molar solution of weak base \(\mathrm{BOH}\) is \(1 \%\) dissociated. If 0.2 mole of \(\mathrm{BCl}\) is added in 1 litre solution of \(\mathrm{BOH}\), the degree of dissociation of \(\mathrm{BOH}\) will become
1 \(2 \times 10^{-3}\)
2 \(5 \times 10^{-5}\)
3 0.05
4 0.02
Explanation:
\(\mathrm {K_{b}=C \alpha^{2}=1 \times 10^{-5}}\) In the presence of 0.2 moles \(\mathrm {\mathrm{BCl}}\) \(\rm {\underset{0.1-0.1 \alpha}{\mathrm{BOH}} \rightleftharpoons \underset{0.1 \alpha+0.2}{\mathrm{~B}^{+}}+\underset{0.1 \alpha}{\mathrm{OH}^{-}}}\) \(\mathrm {0.1 \alpha+0.2 \approx 0.2}\). Due to common ion, \(\mathrm {\mathrm{B}^{+}}\) ion concentration due to \(\mathrm {\mathrm{BOH}}\) is negligible. \(\mathrm {0.1-0.1 \alpha=0.1}\) Since \(\mathrm {\alpha}\) is very less. \(\rm {K_{b}=\dfrac{0.2 \times 0.1 \alpha}{0.1} \Rightarrow 10^{-5}=2 \times 10^{-1} \times \alpha \\ \therefore \quad \alpha=5 \times 10^{-5}}\)
314446
Among the following is only Bronsted Lowry acid but not an Arrhenius acid?
1 \(\mathrm{AlCl}_{3}\)
2 \(\mathrm{NH}_{4}^{+}\)
3 \(\mathrm{BF}_{3}\)
4 \(\mathrm{CH}_{3} \mathrm{COOH}\)
Explanation:
Ammonium ion \(\mathrm {\left(\mathrm{NH}_{4}^{+}\right)}\)behaves as Bronsted Lowry acid as it has a capacity to donate \(\mathrm {\mathrm{H}^{+}}\)ion.
CHXI07:EQUILIBRIUM
314447
Which of the following can act as both Bronsted acid and Bronsted base?i) \(\mathrm{HCOO}^{-}\)ii) \(\mathrm{NH}_{3}\)iii) \(\mathrm{O}^{-2}\)iv) \(\mathrm{HSO}_{4}^{-}\)
1 (i) and (ii)
2 (ii) and (iii)
3 (ii) and (iv)
4 (i) and (iv)
Explanation:
Both \(\mathrm{NH}_{3}\) and \(\mathrm{HSO}_{4}^{-}\)have a tendency to accept and donate protons.
CHXI07:EQUILIBRIUM
314448
The conjugate base of hydrazoic acid is:
1 \({{\text{N}}^{{\text{ -3}}}}\)
2 \({\text{N}}_{\text{3}}^{\text{ - }}\)
3 \({\text{N}}_{\text{2}}^{\text{ - }}\)
4 \({\text{HN}}_{\text{3}}^{\text{ - }}\)
Explanation:
\(\mathrm {\underset{\text { Hydrazoicacid }}{\mathrm{N}_{3} H} \rightleftharpoons N_{3}^{-}+H^{+}}\) i.e., conjugate base of hydrazoic acid is \(\mathrm {N_{3}}\)
JEE - 2014
CHXI07:EQUILIBRIUM
314769
A 0.1 molar solution of weak base \(\mathrm{BOH}\) is \(1 \%\) dissociated. If 0.2 mole of \(\mathrm{BCl}\) is added in 1 litre solution of \(\mathrm{BOH}\), the degree of dissociation of \(\mathrm{BOH}\) will become
1 \(2 \times 10^{-3}\)
2 \(5 \times 10^{-5}\)
3 0.05
4 0.02
Explanation:
\(\mathrm {K_{b}=C \alpha^{2}=1 \times 10^{-5}}\) In the presence of 0.2 moles \(\mathrm {\mathrm{BCl}}\) \(\rm {\underset{0.1-0.1 \alpha}{\mathrm{BOH}} \rightleftharpoons \underset{0.1 \alpha+0.2}{\mathrm{~B}^{+}}+\underset{0.1 \alpha}{\mathrm{OH}^{-}}}\) \(\mathrm {0.1 \alpha+0.2 \approx 0.2}\). Due to common ion, \(\mathrm {\mathrm{B}^{+}}\) ion concentration due to \(\mathrm {\mathrm{BOH}}\) is negligible. \(\mathrm {0.1-0.1 \alpha=0.1}\) Since \(\mathrm {\alpha}\) is very less. \(\rm {K_{b}=\dfrac{0.2 \times 0.1 \alpha}{0.1} \Rightarrow 10^{-5}=2 \times 10^{-1} \times \alpha \\ \therefore \quad \alpha=5 \times 10^{-5}}\)
314446
Among the following is only Bronsted Lowry acid but not an Arrhenius acid?
1 \(\mathrm{AlCl}_{3}\)
2 \(\mathrm{NH}_{4}^{+}\)
3 \(\mathrm{BF}_{3}\)
4 \(\mathrm{CH}_{3} \mathrm{COOH}\)
Explanation:
Ammonium ion \(\mathrm {\left(\mathrm{NH}_{4}^{+}\right)}\)behaves as Bronsted Lowry acid as it has a capacity to donate \(\mathrm {\mathrm{H}^{+}}\)ion.
CHXI07:EQUILIBRIUM
314447
Which of the following can act as both Bronsted acid and Bronsted base?i) \(\mathrm{HCOO}^{-}\)ii) \(\mathrm{NH}_{3}\)iii) \(\mathrm{O}^{-2}\)iv) \(\mathrm{HSO}_{4}^{-}\)
1 (i) and (ii)
2 (ii) and (iii)
3 (ii) and (iv)
4 (i) and (iv)
Explanation:
Both \(\mathrm{NH}_{3}\) and \(\mathrm{HSO}_{4}^{-}\)have a tendency to accept and donate protons.
CHXI07:EQUILIBRIUM
314448
The conjugate base of hydrazoic acid is:
1 \({{\text{N}}^{{\text{ -3}}}}\)
2 \({\text{N}}_{\text{3}}^{\text{ - }}\)
3 \({\text{N}}_{\text{2}}^{\text{ - }}\)
4 \({\text{HN}}_{\text{3}}^{\text{ - }}\)
Explanation:
\(\mathrm {\underset{\text { Hydrazoicacid }}{\mathrm{N}_{3} H} \rightleftharpoons N_{3}^{-}+H^{+}}\) i.e., conjugate base of hydrazoic acid is \(\mathrm {N_{3}}\)
JEE - 2014
CHXI07:EQUILIBRIUM
314769
A 0.1 molar solution of weak base \(\mathrm{BOH}\) is \(1 \%\) dissociated. If 0.2 mole of \(\mathrm{BCl}\) is added in 1 litre solution of \(\mathrm{BOH}\), the degree of dissociation of \(\mathrm{BOH}\) will become
1 \(2 \times 10^{-3}\)
2 \(5 \times 10^{-5}\)
3 0.05
4 0.02
Explanation:
\(\mathrm {K_{b}=C \alpha^{2}=1 \times 10^{-5}}\) In the presence of 0.2 moles \(\mathrm {\mathrm{BCl}}\) \(\rm {\underset{0.1-0.1 \alpha}{\mathrm{BOH}} \rightleftharpoons \underset{0.1 \alpha+0.2}{\mathrm{~B}^{+}}+\underset{0.1 \alpha}{\mathrm{OH}^{-}}}\) \(\mathrm {0.1 \alpha+0.2 \approx 0.2}\). Due to common ion, \(\mathrm {\mathrm{B}^{+}}\) ion concentration due to \(\mathrm {\mathrm{BOH}}\) is negligible. \(\mathrm {0.1-0.1 \alpha=0.1}\) Since \(\mathrm {\alpha}\) is very less. \(\rm {K_{b}=\dfrac{0.2 \times 0.1 \alpha}{0.1} \Rightarrow 10^{-5}=2 \times 10^{-1} \times \alpha \\ \therefore \quad \alpha=5 \times 10^{-5}}\)
