Chapter 15, Problem 15.109QA

Interpretation Introduction

To find:

Net ionic equations and identify Bronsted-Lowry acids and bases

Answer to Problem 15.109QA

Solution:

a) $H^{+}\left(aq\right)+{OH}^{-}\left(aq\right)\to H_{2}O\left(l\right)$, acid = $HN{O}_3$, base = $Ca{\left(OH\right)}_2$

b) $C{O}_3^{2-}\left(aq\right)+ 2H^{+}\left(aq\right)\to C{O}_2\left(g\right)+ H_{2}O\left(l\right)$, acid = $H_{2}S{O}_4$, base = $N{a}_{2}C{O}_3$

c) $C{H}_{3}N{H}_2\left(aq\right)+ H^{+}\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)$, acid =$HBr$, base = $C{H}_{3}N{H}_2$

d) $C{H}_{3}COOH\left(aq\right)+ O{H}^{-}\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+ H_{2}O\left(l\right)$, acid = $C{H}_{3}COOH$, base = $Mg{\left(OH\right)}_2$

e) $CaO\left(s\right)+ H_{2}O\left(l\right)\to {Ca}^{2+}\left(aq\right)+ 2O{H}^{-}\left(aq\right)$, acid =$H_{2}O$, base = CaO

f) $LiH\left(aq\right)+ H_{2}O\left(l\right)\to {Li}^{+}\left(aq\right)+ O{H}^{-}\left(aq\right)+ H_2\left(g\right)$, acid = $H_{2}O$, base = $LiH$

g) $H^{+}\left(aq\right)+{OH}^{-}\left(aq\right)\to H_{2}O\left(l\right)$, acid = $H_{2}S{O}_4$, base = $Ba{\left(OH\right)}_2$

h) $S{H}^{-}\left(aq\right)+ H^{+}\left(aq\right)\to H_{2}S\left(g\right)$, acid = $HN{O}_3$, base = $NaSH$

Explanation of Solution

Concept:

The net ionic equation is a chemical equation that shows only the soluble and strong electrolytes reacting (ions) and omits the spectator ions which remain unchanged during the reaction.

According to Bronsted –Lowry acid base theory, acids are proton donors whereas bases are proton acceptors.

a) $2H^{+}\left(aq\right)+2N{O}_3^{-}\left(aq\right) +{Ca}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)+ {Ca}^{2+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)$

Therefore the net ionic equation is:

$H^{+}\left(aq\right)+{OH}^{-}\left(aq\right)\to H_{2}O\left(l\right)$

$HN{O}_3\left(aq\right)\to H^{+} \left(aq\right)+N{O}_3^{-}$ (aq)

So, $HN{O}_3$ is Lowry-Bronsted acid, because $Ca{\left(OH\right)}_2$ on ionization gives $O{H}^{-}$ ions which in turn combines with $H^{+}$ to form water, thus, Ca(OH)₂ is a Lowry-Bronsted base.

b) $2{Na}^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)+2H^{+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)\to 2{Na}^{+}\left(aq\right) +S{O}_4^{2-}\left(aq\right) +C{O}_2\left(g\right)+H_{2}O\left(l\right)$

Therefore the net ionic equation is:

$C{O}_3^{2-}\left(aq\right)+ 2H^{+}\left(aq\right)\to C{O}_2\left(g\right)+ H_{2}O\left(l\right)$

$H_2{SO}_4\left(aq\right)\to H^{+}\left(aq\right)+HS{O}_4^{-}\left(aq\right)H_{2}S{O}_4$ is Bronsted-Lowry acid

$C{O}_3^{2-}\left(aq\right)+ 2H^{+}\left(aq\right)\to C{O}_2\left(g\right)+ H_{2}O\left(l\right)N{a}_{2}C{O}_3$ is Bronsted-Lowry base.

c) $C{H}_{3}N{H}_2\left(aq\right)+H^{+}\left(aq\right)+{Br}^{-}\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)+{Br}^{-}\left(aq\right)$

Therefore the net ionic equation is:

$C{H}_{3}N{H}_2\left(aq\right)+ H^{+}\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)$

$HBr\left(aq\right)\to H^{+}\left(aq\right)+{Br}^{-}\left(aq\right)$

$HBr$ is Bronsted Lowry acid

$C{H}_{3}N{H}_2\left(aq\right)+ H^{+}\left(aq\right)\to C{H}_{3}N{H}_3^{+}\left(aq\right)$ CH₃NH₂ (aq) is Bronsted Lowry base.

d) $2C{H}_{3}COOH\left(aq\right)+ {Mg}^{2+}\left(aq\right)+ 2O{H}^{-}\to {Mg}^{2+}\left(aq\right) +2C{H}_{3}CO{O}^{-}\left(aq\right) +2H_{2}O\left(l\right)$

Therefore the net ionic equation is:

$C{H}_{3}COOH\left(aq\right)+ O{H}^{-}\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+ H_{2}O\left(l\right)$

$C{H}_{3}COOH\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)C{H}_{3}COOH$ is Bronsted-Lowry acid

$Mg(OH{)}_2\to {Mg}^{2+}\left(aq\right)+2O{H}^{-}(aq\left)Mg\right(OH{)}_2$ is Bronsted Lowry base.

e) $CaO\left(s\right)+ H_{2}O\left(l\right)\to {Ca}^{2+}\left(aq\right)+ 2O{H}^{-}\left(aq\right)$, This is the ionic equation since here all reactants and products are spectator ions.

$H_{2}O\left(l\right)\to H^{+} +O{H}^{-}$, $H_{2}O$ is Bronsted Lowry acid

$CaO\left(s\right)$ by accepting proton is converted into $Ca{\left(OH\right)}_2$ and thus is Bronsted-Lowry Base.

f) $LiH\left(aq\right)+ H_{2}O\left(l\right)\to {Li}^{+}\left(aq\right)+ O{H}^{-}\left(aq\right)+ H_2\left(g\right)$, This is the ionic equation since here all reactants and products are spectator ions.

$H_{2}O\left(l\right)\to H^{+} +O{H}^{-}$, $H_{2}O$ is  Bronsted Lowry acid

$LiH$ on dissociation produces $L{i}^{+}$ ion and $H^{-}$ ion where $H^{-}$ accepts $H^{+}$ to form $H_2$, thus $LiH$ acts as  Bronsted Lowry base

g) ${ 2H}^{+}\left(aq\right)+{SO}_4^{2-}\left(aq\right)+{Ba}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\to {Ba}^{2+}\left(aq\right)+{SO}_4^{2-}\left(aq\right)+2H_{2}O\left(l\right)$

Therefore the net ionic equation is:

$H^{+}\left(aq\right)+{OH}^{-}\left(aq\right)\to H_{2}O\left(l\right)$

$H_2{SO}_4\left(aq\right)\to H^{+}\left(aq\right)+HS{O}_4^{-}\left(aq\right)H_{2}S{O}_4$ is Bronsted-Lowry acid

$Ba{\left(OH\right)}_2$ on ionization produces $O{H}^{-}$ ion, which further accepts $H^{+}$ to form water. Thus $Ba{\left(OH\right)}_2$ is a Bronsted-Lowry base.

h) ${Na}^{+}\left(aq\right)+S{H}^{-}\left(aq\right)+H^{+}\left(aq\right)+{NO}_3^{-}\left(aq\right)\to {Na}^{+}\left(aq\right)+{NO}_3^{-}\left(aq\right)+H_{2}S\left(g\right)$

Therefore the net ionic equation is:

$S{H}^{-}\left(aq\right)+ H^{+}\left(aq\right)\to H_{2}S\left(g\right)$

$HN{O}_3\left(aq\right)\to H^{+}\left(aq\right)+N{O}_3^{-}\left(aq\right)$ So, $HN{O}_3$ is a Bronsted-Lowry acid.

$NaSH$ (aq) on ionization produces $S{H}^{-}$^- ion which further reacts with H⁺ to form $H_{2}O$.So $NaSH$ is a Bronsted-Lowry base.

Conclusion:

The net ionic equation is the equation that is left behind when all the spectator ions for both sides are cancelled out.