Chapter 15, Problem 15.144QA

Interpretation Introduction

To determine:

a. The net ionic equation for the reaction of each of the given bases with aqueous $HCl$.

b. How substances sparingly soluble in water can act as effective antacids.

Answer to Problem 15.144QA

Solution:

a)

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

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

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

$2H^{+}\left(aq\right)+Mg{\left(OH\right)}_2\left(s\right) \to {Mg}^{+2}\left(aq\right)+2H_{2}O\left(l\right)$

b) The reaction of stomach acid with sparingly soluble substances causes only a small amount of the substance to dissolve and act as antacid.

Explanation of Solution

Aqueous $HCl$ reacts with the given bases to form a salt and carbonic acid or water. Only the aqueous species break down into ions. The net ionic equation is obtained by cancelling out the spectator ions from the total ionic equation.

a) Write the net ionic reaction for the given salts and HCl from total ionic reaction.

i) Reaction of $NaHC{O}_3$ with $HCl$:

It is given that $NaHC{O}_3$ has appreciable solubility in water.

$HCl \left(aq\right)+NaHC{O}_3\left(aq\right)\to NaCl \left(aq\right)+H_{2}O \left(l\right)+C{O}_2\left(g\right)$

Total ionic equation:

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

Net ionic equation:

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

ii) Reaction of $MgC{O}_3$ with $HCl$

$2 HCl \left(aq\right)+MgC{O}_3\left(s\right)\to MgC{l}_2\left(aq\right)+H_{2}O \left(l\right)+C{O}_2\left(g\right)$

Total ionic equation:

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

Net ionic equation:

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

iii) Reaction of $CaC{O}_3$ with $HCl$

$2 HCl \left(aq\right)+CaC{O}_3\left(s\right)\to CaC{l}_2\left(aq\right)+H_{2}O \left(l\right)+C{O}_2\left(g\right)$

Total ionic equation:

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

Net ionic equation:

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

iv) Reaction of $Mg{\left(OH\right)}_2$ with $HCl$

$2 HCl \left(aq\right)+Mg{\left(OH\right)}_2\left(s\right)\to MgC{l}_2\left(aq\right)+{2H}_{2}O \left(l\right)$

Total ionic equation:

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

Net ionic equation:

$2H^{+}\left(aq\right)+Mg{\left(OH\right)}_2\left(s\right) \to {Mg}^{+2}\left(aq\right)+2H_{2}O\left(l\right)$

b) Sparingly soluble substances do not dissociate in solution to a large extent.

For example:

$Mg{\left(OH\right)}_2\left(s\right)\rightleftharpoons M{g}^{+2}\left(aq\right)+2 O{H}^{-}\left(aq\right)$

When this salt reacts with the stomach acid, the $H^{+}$ ions from the acid are neutralized by the $O{H}^{-}$ ions present. This causes the equilibrium to shift to the right, and more of the substance dissociates. Thus, a small portion of the substance dissolves and neutralizes the stomach acid.

Hence, sparingly water soluble salts can act as effective antacids.

Conclusion:

The net ionic equation is written by writing the complete molecular and total ionic equations and cancelling out the spectator ions. A small portion of the sparingly soluble substance dissolves to neutralize the stomach acid and hence acts as an effective antacid.