
All of the following can act as Bronsted- Lowry acids (proton donors) in aqueous solution except :
A.$HI$
B.$N{H_4}^ + $
C.$HC{O_3}^ - $
D.${H_2}S$
E.$N{H_3}$
Answer
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Hint:A bronsted lowry acid is any species that's capable of giving up or donating one or more hydrogen ions during a chemical action. Every bronsted lowry acid donates its proton to a species which is usually its conjugate base.
Complete step by step answer:
-A Brønsted-Lowry acid may be a proton (hydrogen ion), donor.
-A Brønsted-Lowry base could be a proton (hydrogen ion), the acceptor.
-In this theory, an acid could be a substance which will release a proton (like within the Arrhenius theory) and a base may be a substance which will accept a proton. When a Brønsted acid dissociates, it increases the concentration of hydrogen ions $\left[ {{H^ + }} \right]$within the solution, conversely, Brønsted bases dissociate by taking a proton from the solvent (water) to come up with hydroxide ions $\left[ {O{H^ - }} \right]$ within the solution.
Acid dissociation:
$HA\underset {} \leftrightarrows {A^ - } + {H_{(aq)}}^ + $
Base dissociation:
${B_{(aq)}} + {H_2}O\underset {} \leftrightarrows H{B^ + } + O{H^ - }_{(aq)}^{}$
A.$HI$ dissociates to form ${H_3}{O^ + }$ ions in the solution hence it is a Brønsted-Lowry acid.
B.$NH_4^ + + {H_2}O\underset {} \leftrightarrows {H_3}{O^ + } + N{H_3}$
Ammonium ion dissociates to give ${H^ + }$ions in the solution hence it is a Brønsted-Lowry acid.
C.$HC{O_3}^ - $is the conjugate base of carbonic acid and the conjugate acid of the carbonate ion. Hence it can act as both acid and base.
D. ${H_2}S$$ + {H_2}O\underset {} \leftrightarrows {H_3}{O^+ } + S{H^-}$
${H_{2}S}$ is a Bronsted-Lowry acid because it is donating a proton to water.
E. $N{H_3} + {H_2}O \to NH_4^+ + O{H^ - }$
Since Ammonia is accepting a proton from ${H_2}O$, it is behaving like a Brønsted-Lowry base.
Thus the correct option is E.
Note:
-When a Bronsted-Lowry acid donates a proton to a different compound, the particle that's missing the proton becomes its conjugate base. Together they're called a conjugate acid-base pair.
-When an acid $HA$ dissolves in water, it reacts reversibly with the water to supply hydronium ions and ${A^ - }$ ions.
$HA + {H_2}O\underset {} \leftrightarrows {H_3}O + {A^ - }$
-In this, $HA$ has donated a proton to water thus HA is the Bronsted-Lowry
acid and water is the Bronsted-Lowry base.
-A conjugate base is whatever is left behind after the proton has left. So ${A^ - }$ is the conjugate base of $HA$
-A conjugate acid is whatever is formed after the proton has been accepted. So, ${H_3}{O^ + }$ is the conjugate acid of water.
Complete step by step answer:
-A Brønsted-Lowry acid may be a proton (hydrogen ion), donor.
-A Brønsted-Lowry base could be a proton (hydrogen ion), the acceptor.
-In this theory, an acid could be a substance which will release a proton (like within the Arrhenius theory) and a base may be a substance which will accept a proton. When a Brønsted acid dissociates, it increases the concentration of hydrogen ions $\left[ {{H^ + }} \right]$within the solution, conversely, Brønsted bases dissociate by taking a proton from the solvent (water) to come up with hydroxide ions $\left[ {O{H^ - }} \right]$ within the solution.
Acid dissociation:
$HA\underset {} \leftrightarrows {A^ - } + {H_{(aq)}}^ + $
Base dissociation:
${B_{(aq)}} + {H_2}O\underset {} \leftrightarrows H{B^ + } + O{H^ - }_{(aq)}^{}$
A.$HI$ dissociates to form ${H_3}{O^ + }$ ions in the solution hence it is a Brønsted-Lowry acid.
B.$NH_4^ + + {H_2}O\underset {} \leftrightarrows {H_3}{O^ + } + N{H_3}$
Ammonium ion dissociates to give ${H^ + }$ions in the solution hence it is a Brønsted-Lowry acid.
C.$HC{O_3}^ - $is the conjugate base of carbonic acid and the conjugate acid of the carbonate ion. Hence it can act as both acid and base.
D. ${H_2}S$$ + {H_2}O\underset {} \leftrightarrows {H_3}{O^+ } + S{H^-}$
${H_{2}S}$ is a Bronsted-Lowry acid because it is donating a proton to water.
E. $N{H_3} + {H_2}O \to NH_4^+ + O{H^ - }$
Since Ammonia is accepting a proton from ${H_2}O$, it is behaving like a Brønsted-Lowry base.
Thus the correct option is E.
Note:
-When a Bronsted-Lowry acid donates a proton to a different compound, the particle that's missing the proton becomes its conjugate base. Together they're called a conjugate acid-base pair.
-When an acid $HA$ dissolves in water, it reacts reversibly with the water to supply hydronium ions and ${A^ - }$ ions.
$HA + {H_2}O\underset {} \leftrightarrows {H_3}O + {A^ - }$
-In this, $HA$ has donated a proton to water thus HA is the Bronsted-Lowry
acid and water is the Bronsted-Lowry base.
-A conjugate base is whatever is left behind after the proton has left. So ${A^ - }$ is the conjugate base of $HA$
-A conjugate acid is whatever is formed after the proton has been accepted. So, ${H_3}{O^ + }$ is the conjugate acid of water.
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