The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of\(\rm {{{\text{K}}_{\text{a}}}}\) Since HI is a stronger acid than HF which is a stronger acid than \(\mathrm {\mathrm{H}_{2} \mathrm{~S}}\), a partial order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < H}}{{\text{S}}^{\text{ - }}}}\) Since \(\mathrm {\mathrm{NH}_{3}}\) is a very weak acid, \(\mathrm {\mathrm{NH}_{2}^{-}}\)must be a very strong base. Therefore the correct order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < HS < NH}}_{\text{2}}^{\text{ - }}}\)
JEE - 2013
CHXI07:EQUILIBRIUM
314445
Which from the following compounds accepts proton from water molecule according to Bronsted - Lowry theory?
1 \(\mathrm{NaOH}_{(\text {aq })}\)
2 \(\mathrm{HCl}_{(\mathrm{aq})}\)
3 \(\mathrm{NH}_{3(\mathrm{aq})}\)
4 \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
Explanation:
Ammonia accepts proton and becomes ammonium. \(\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of\(\rm {{{\text{K}}_{\text{a}}}}\) Since HI is a stronger acid than HF which is a stronger acid than \(\mathrm {\mathrm{H}_{2} \mathrm{~S}}\), a partial order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < H}}{{\text{S}}^{\text{ - }}}}\) Since \(\mathrm {\mathrm{NH}_{3}}\) is a very weak acid, \(\mathrm {\mathrm{NH}_{2}^{-}}\)must be a very strong base. Therefore the correct order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < HS < NH}}_{\text{2}}^{\text{ - }}}\)
JEE - 2013
CHXI07:EQUILIBRIUM
314445
Which from the following compounds accepts proton from water molecule according to Bronsted - Lowry theory?
1 \(\mathrm{NaOH}_{(\text {aq })}\)
2 \(\mathrm{HCl}_{(\mathrm{aq})}\)
3 \(\mathrm{NH}_{3(\mathrm{aq})}\)
4 \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
Explanation:
Ammonia accepts proton and becomes ammonium. \(\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of\(\rm {{{\text{K}}_{\text{a}}}}\) Since HI is a stronger acid than HF which is a stronger acid than \(\mathrm {\mathrm{H}_{2} \mathrm{~S}}\), a partial order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < H}}{{\text{S}}^{\text{ - }}}}\) Since \(\mathrm {\mathrm{NH}_{3}}\) is a very weak acid, \(\mathrm {\mathrm{NH}_{2}^{-}}\)must be a very strong base. Therefore the correct order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < HS < NH}}_{\text{2}}^{\text{ - }}}\)
JEE - 2013
CHXI07:EQUILIBRIUM
314445
Which from the following compounds accepts proton from water molecule according to Bronsted - Lowry theory?
1 \(\mathrm{NaOH}_{(\text {aq })}\)
2 \(\mathrm{HCl}_{(\mathrm{aq})}\)
3 \(\mathrm{NH}_{3(\mathrm{aq})}\)
4 \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
Explanation:
Ammonia accepts proton and becomes ammonium. \(\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of\(\rm {{{\text{K}}_{\text{a}}}}\) Since HI is a stronger acid than HF which is a stronger acid than \(\mathrm {\mathrm{H}_{2} \mathrm{~S}}\), a partial order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < H}}{{\text{S}}^{\text{ - }}}}\) Since \(\mathrm {\mathrm{NH}_{3}}\) is a very weak acid, \(\mathrm {\mathrm{NH}_{2}^{-}}\)must be a very strong base. Therefore the correct order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < HS < NH}}_{\text{2}}^{\text{ - }}}\)
JEE - 2013
CHXI07:EQUILIBRIUM
314445
Which from the following compounds accepts proton from water molecule according to Bronsted - Lowry theory?
1 \(\mathrm{NaOH}_{(\text {aq })}\)
2 \(\mathrm{HCl}_{(\mathrm{aq})}\)
3 \(\mathrm{NH}_{3(\mathrm{aq})}\)
4 \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
Explanation:
Ammonia accepts proton and becomes ammonium. \(\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of\(\rm {{{\text{K}}_{\text{a}}}}\) Since HI is a stronger acid than HF which is a stronger acid than \(\mathrm {\mathrm{H}_{2} \mathrm{~S}}\), a partial order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < H}}{{\text{S}}^{\text{ - }}}}\) Since \(\mathrm {\mathrm{NH}_{3}}\) is a very weak acid, \(\mathrm {\mathrm{NH}_{2}^{-}}\)must be a very strong base. Therefore the correct order of proton affinity is \(\rm {{{\text{I}}^{\text{ - }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < HS < NH}}_{\text{2}}^{\text{ - }}}\)
JEE - 2013
CHXI07:EQUILIBRIUM
314445
Which from the following compounds accepts proton from water molecule according to Bronsted - Lowry theory?
1 \(\mathrm{NaOH}_{(\text {aq })}\)
2 \(\mathrm{HCl}_{(\mathrm{aq})}\)
3 \(\mathrm{NH}_{3(\mathrm{aq})}\)
4 \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
Explanation:
Ammonia accepts proton and becomes ammonium. \(\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)