Chapter 8, Problem 8.54QA

Interpretation Introduction

To write:

Balanced molecular and net ionic equations for following reactions:

a) Solid aluminum hydroxide reacts with a solution of hydrobromic acid   b) A solution of sulfuric acid reacts with solid sodium carbonatec) A solution of calcium hydroxide reacts with a solution of nitric acid 

Answer to Problem 8.54QA

Solution:a)

Balanced molecular equation: $Al{\left(OH\right)}_{3\left(s\right)}+3HB{r}_{\left(aq\right)}\to AlB{r}_{3\left(aq\right)}+3H_2{O}_{\left(l\right)}$

Net ionic equation: $Al{\left(OH\right)}_{3\left(s\right)}+{3H}_{\left(aq\right)}^{+}\to A{l}_{\left(aq\right)}^{3+}+3H_2{O}_{\left(l\right)}$

    b)

Balanced molecular equation: $H_{2}S{O}_{4\left(aq\right)}+N{a}_{2}C{O}_{3\left(s\right)}\to N{a}_{2}S{O}_{4\left(aq\right)}+H_2{O}_{\left(l\right)}+C{O}_{2\left(g\right)}$

Net ionic equation: ${2H}_{\left(aq\right)}^{+}+N{a}_{2}C{O}_{3\left(s\right)}\to 2N{a}_{\left(aq\right)}^{+}+H_2{O}_{\left(l\right)}+C{O}_{2\left(g\right)}$

     c)

Balanced molecular equation: $Ca{\left(OH\right)}_{2\left(aq\right)}+2HN{O}_{3\left(aq\right)}\to \mathbf{ }Ca{\left(N{O}_3\right)}_{2\left(aq\right)}+{2H}_2{O}_{\left(l\right)}$

Net ionic equation: $H_{\left(aq\right)}^{+}+O{H}_{\left(aq\right)}^{-}\to H_2{O}_{\left(l\right)}$

Explanation of Solution

Explanation:1)    Concept:

Using the chemical formula of the compounds given in each reaction, we can write the reaction.

Steps for writing net ionic equation:

i) Write down the complete balanced molecular equation with appropriate physical states.

ii) Write the total ionic equation.

iii) Cancel out spectator ions from the total ionic equation to get the net ionic equation.

2)    Calculations:a) The molecular equation for the reaction where solid aluminum hydroxide reacts with a solution of hydrobromic acid is

$Al{\left(OH\right)}_{3\left(s\right)}+HB{r}_{\left(aq\right)}\to AlB{r}_{3\left(aq\right)}+H_2{O}_{\left(l\right)}$

To balance the bromide ion and hydroxide ion, we place a coefficient of 3 in front of hydrobromic acid and water.

$Al{\left(OH\right)}_{3\left(s\right)}+3HB{r}_{\left(aq\right)}\to AlB{r}_{3\left(aq\right)}+3H_2{O}_{\left(l\right)}$

This is the balanced molecular equation.

Now write the total ionic equation, in which all the soluble salts separate into their component ions.$Al{\left(OH\right)}_{3\left(s\right)}+{3H}_{\left(aq\right)}^{+}+3B{r}_{\left(aq\right)}^{-}\to A{l}_{\left(aq\right)}^{3+}+3B{r}_{\left(aq\right)}^{-}+3H_2{O}_{\left(l\right)}$

Here, $B{r}^{-}$ is the spectator ion. After cancelling spectator ions, we get the net ionic equation.

$Al{\left(OH\right)}_{3\left(s\right)}+{3H}_{\left(aq\right)}^{+}\to A{l}_{\left(aq\right)}^{3+}+3H_2{O}_{\left(l\right)}$

b) The molecular equation for the reaction where solution of sulfuric acid reacts with solid sodium carbonate is

$H_{2}S{O}_{4\left(aq\right)}+N{a}_{2}C{O}_{3\left(s\right)}\to N{a}_{2}S{O}_{4\left(aq\right)}+H_{2}C{O}_{3\left(aq\right)}$

$H_{2}C{O}_3$ further gives water and carbon dioxide gas.

$H_{2}S{O}_{4\left(aq\right)}+N{a}_{2}C{O}_{3\left(s\right)}\to N{a}_{2}S{O}_{4\left(aq\right)}+H_2{O}_{\left(l\right)}+C{O}_{2\left(g\right)}$

Both sides have an equal number of atoms, so this is a balanced molecular equation.

Now write the total ionic equation, in which all the soluble salts separate into their component ions.

${2H}_{\left(aq\right)}^{+}+S{O}_{4\left(aq\right)}^{2-}+N{a}_{2}C{O}_{3\left(s\right)}\to 2N{a}_{\left(aq\right)}^{+}+ S{O}_{4\left(aq\right)}^{2-}+H_2{O}_{\left(l\right)}+C{O}_{2\left(g\right)}$

Here $S{O}_{4\left(aq\right)}^{2-}$ is the spectator ion. Deleting the spectator ion yields the net ionic equation.

${2H}_{\left(aq\right)}^{+}+N{a}_{2}C{O}_{3\left(s\right)}\to 2N{a}_{\left(aq\right)}^{+}+H_2{O}_{\left(l\right)}+C{O}_{2\left(g\right)}$

c) The molecular equation for the reaction where solution of calcium hydroxide reacts with a solution of nitric acid is

$Ca{\left(OH\right)}_{2\left(aq\right)}+HN{O}_{3\left(aq\right)}\to \mathbf{ }Ca{\left(N{O}_3\right)}_{2\left(aq\right)}+H_2{O}_{\left(l\right)}$

To balance the nitrate ion and hydroxide ion, we place a coefficient of 2 in front of nitric acid and water.

$Ca{\left(OH\right)}_{2\left(aq\right)}+2HN{O}_{3\left(aq\right)}\to \mathbf{ }Ca{\left(N{O}_3\right)}_{2\left(aq\right)}+{2H}_2{O}_{\left(l\right)}$

Now write the total ionic equation, in which all the soluble salts separate into their component ions.

$C{a}_{\left(aq\right)}^{2+}+2O{H}_{\left(aq\right)}^{-}+2H_{\left(aq\right)}^{+}+2N{O}_{3\left(aq\right)}^{-}\to C{a}_{\left(aq\right)}^{2+}+2N{O}_{3\left(aq\right)}^{-}+{2H}_2{O}_{\left(l\right)}$

Here, $C{a}^{2+}and N{O}_3^{-}$ are spectator ions; therefore, by cancelling them, we get the net ionic equation.

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

To reduce the coefficient, divide the equation by $2$.

$H_{\left(aq\right)}^{+}+O{H}_{\left(aq\right)}^{-}\to H_2{O}_{\left(l\right)}$

Conclusion:Using the steps of writing the net ionic equation, we can write the net ionic equation for each balanced reaction.