Chapter 15, Problem 15.110QA

Interpretation Introduction

To find:

The chemical and net ionic equation in each case

Also  identify the Bronsted-Lowry acid and base from a) to d)

Answer to Problem 15.110QA

Solution:

a) Chemical equation:

$2HCl\left(aq\right)+Ca{\left(OH\right)}_2\to Ca{Cl}_2\left(aq\right)+2H_{2}O\left(l\right)$

Net ionic equation:

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

acid = $HCl$, base = $Ca{\left(OH\right)}_2$

b) Chemical equation:

$H_{3}P{O}_4\left(aq\right)+3KOH\left(aq\right)\to K_{3}P{O}_4\left(aq\right)+3H_{2}O\left(l\right)$

Net ionic equation:

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

acid =$H_{3}P{O}_4$, base = $KOH$

c) Chemical equation:

$2HN{O}_3\left(aq\right)+{Na}_{2}C{O}_3\left(aq\right)\to 2NaN{O}_3\left(aq\right)+H_{2}C{O}_3\left(aq\right)$

Net ionic equation:

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

Acid = $O_3$, Base = ${Na}_{2}C{O}_3$

d) Chemical equation:

$2H_{3}P{O}_4\left(aq\right)+3Ca{\left(OH\right)}_2\left(aq\right)\to {Ca}_3{(PO}_4{)}_2\left(aq\right)+{6H}_{2}O\left(l\right)$

Net ionic equation:

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

Acid = $H_{3}P{O}_4$, Base = $Ca{\left(OH\right)}_2$

Explanation of Solution

1) Concept :

Chemical equation is the symbolic representation of a chemical reaction in which the number of atoms of each element on the left side is equal to the number of atoms on the right side.

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) Chemical equation:

$2HCl\left(aq\right)+Ca{\left(OH\right)}_2\to Ca{Cl}_2\left(aq\right)+2H_{2}O\left(l\right)$

This above chemical equation can be written as:

${2\mathrm{H}}^{+}\left(\mathrm{a}\mathrm{q}\right)+2\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{C}\mathrm{a}}^{2+}\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{O}\mathrm{H}}^{-}\left(\mathrm{a}\mathrm{q}\right)\to {\mathrm{C}\mathrm{a}}^{2+}\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{C}\mathrm{l}}^{-}\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

So the net ionic equation is:

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

$HCl\left(aq\right)\to H^{+}\left(aq\right)+C{l}^{-}\left(aq\right)HCl$ on dissociation gives $H^{+}$ ion, so it is a Bronsted-Lowry acid.

$O{H}^{-}$ ion is produced from the ionization of $Ca{\left(OH\right)}_2$.This $O{H}^{-}$ ion further accepts the proton $H^{+}$ to form water.So $Ca{\left(OH\right)}_2$ is a Bronsted-Lowry base.

b) Chemical equation:

$H_{3}P{O}_4\left(aq\right)+3KOH\left(aq\right)\to K_{3}P{O}_4\left(aq\right)+3H_{2}O\left(l\right)$

This above chemical equation can be written as:

${3H}^{+}\left(aq\right)+{PO}_4^{3-}\left(aq\right)+3K^{+}\left(aq\right)+3{OH}^{-}\left(aq\right)\to 3K^{+}\left(aq\right)+{PO}_4^{3-}\left(aq\right)+ 3H_{2}O\left(l\right)$

So the net ionic equation is:

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

$H_{3}P{O}_4\left(aq\right)\to {3\mathrm{H}}^{+}\left(\mathrm{a}\mathrm{q}\right)+{PO}_4^{3-}\left(aq\right)H_{3}P{O}_4($ on dissociation gives $H^{+}$ ion, so it is a Bronsted-Lowry acid.

OH^- ion is produced from the ionization of $KOH$.This ${OH}^{-}$ ion further accepts the proton $H^{+}$ to form water. So $KOH$ is a Bronsted-Lowry base.

c) Chemical equation:

$2HN{O}_3\left(aq\right)+{Na}_{2}C{O}_3\left(aq\right)\to 2NaN{O}_3\left(aq\right)+H_{2}C{O}_3\left(aq\right)$

This above chemical equation can be written as:

$2H^{+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)+2{Na}^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)\to 2N{O}_3^{-}\left(aq\right)+2{Na}^{+}\left(aq\right)+H_{2}C{O}_3\left(aq\right)$

So the net ionic equation is:

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

$H_{2}C{O}_3$ being a weak acid does not ionize completely into ions.

$HN{O}_3\left(aq\right)\to H^{+}\left(aq\right)+{NO}_3^{-}\left(aq\right)$ on dissociation gives $H^{+}$ ion, so it is a Bronsted-Lowry acid.

$C{O}_3^{2-}$ Ion is produced from the ionization of Na₂CO₃.This $C{O}_3^{2-}$ ion further accepts the proton $H^{+}$ to form carbonic acid H₂CO₃. So Na₂CO₃  is a Bronsted-Lowry base.

d) Chemical equation:

$2H_{3}P{O}_4\left(aq\right)+3Ca{\left(OH\right)}_2\left(aq\right)\to {Ca}_3{(PO}_4{)}_2\left(aq\right)+{6H}_{2}O\left(l\right)$

This above chemical equation can be written as:

${6H}^{+}\left(aq\right)+2{PO}_4^{3-}\left(aq\right)+{3Ca}^{+2}\left(aq\right)+6O{H}^{-}\left(aq\right)\to 3{Ca}^{2+}\left(aq\right)+2{PO}_4^{3-}\left(aq\right)+6H_{2}O\left(l\right)$

So the net ionic equation is:

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

$H_{3}P{O}_4\left(aq\right)\to {3\mathrm{H}}^{+}\left(\mathrm{a}\mathrm{q}\right)+{PO}_4^{3-}\left(aq\right)$ H₃PO₄ on dissociation gives $H^{+}$ ion, so it is a Bronsted-Lowry acid.

OH^- ion is produced from the ionization of $Ca{\left(OH\right)}_{2 }$ this ${OH}^{-}$ ion further accepts proton $H^{+}$ to form water. So $Ca{\left(OH\right)}_2$ is a Bronsted-Lowry base.

Conclusion:

The molecular equations are the balanced equations with reactants and products.

The net ionic reaction equation is the equation that is left behind after cancelling out the spectator ions for both sides.