Write a net ionic equation for the precipitation reaction, if any, that occurs when aqueous solutions of the following ionic compounds are mixed: (a) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaCl}\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NiSO}_{4}\) (c) \(\mathrm{NaCl}\) and \(\mathrm{KNO}_{3}\)

Short Answer

Expert verified
(a) Net ionic equation: \( \mathrm{Pb}^{2+} \, (aq) + 2 \mathrm{Cl}^{-} \, (aq) \rightarrow \mathrm{PbCl}_{2} \, (s) \) (b) Net ionic equation: \( \mathrm{Ba}^{2+} \, (aq) + \mathrm{SO}_{4}^{2-} \, (aq) \rightarrow \mathrm{BaSO}_{4} \, (s) \) (c) No precipitation reaction occurs

Step by step solution

01

Write the molecular equation

Write the equation with all reactants and potential products: \[ \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \, (aq) + 2 \mathrm{NaCl} \, (aq) \rightarrow \mathrm{PbCl}_{2} \, (s) + 2 \mathrm{NaNO}_{3} \, (aq) \]
02

Write the complete ionic equation

Break down each compound into its respective ions: \[ \mathrm{Pb}^{2+} \, (aq) + 2 \mathrm{NO}_{3}^{-} \, (aq) + 2 \mathrm{Na}^{+} \, (aq) + 2 \mathrm{Cl}^{-} \, (aq) \rightarrow \mathrm{PbCl}_{2} \, (s) + 2 \mathrm{Na}^{+} \, (aq) + 2 \mathrm{NO}_{3}^{-} \, (aq) \]
03

Determine if a precipitation reaction occurs

Using the solubility rules, we can see that \(\mathrm{PbCl}_{2}\) is insoluble, so a precipitation reaction occurs.
04

Write the net ionic equation

Remove the spectator ions (\(\mathrm{Na}^{+}\) and \(\mathrm{NO}_{3}^{-}\)) and write the equation with only the ions involved in the precipitation reaction: \[ \mathrm{Pb}^{2+} \, (aq) + 2 \mathrm{Cl}^{-} \, (aq) \rightarrow \mathrm{PbCl}_{2} \, (s) \] (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NiSO}_{4}\) Repeat the steps for the second pair of compounds:
05

Write the molecular equation

Write the equation with all reactants and potential products: \[ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \, (aq) + \mathrm{NiSO}_{4} \, (aq) \rightarrow \mathrm{BaSO}_{4} \, (s) + \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \, (aq) \]
06

Write the complete ionic equation

Break down each compound into its respective ions: \[ \mathrm{Ba}^{2+} \, (aq) + 2 \mathrm{NO}_{3}^{-} \, (aq) + \mathrm{Ni}^{2+} \, (aq) + \mathrm{SO}_{4}^{2-} \, (aq) \rightarrow \mathrm{BaSO}_{4} \, (s) + \mathrm{Ni}^{2+} \, (aq) + 2 \mathrm{NO}_{3}^{-} \, (aq) \]
07

Determine if a precipitation reaction occurs

Using the solubility rules, we can see that \(\mathrm{BaSO}_{4}\) is insoluble, so a precipitation reaction occurs.
08

Write the net ionic equation

Remove the spectator ions (\(\mathrm{Ni}^{2+}\) and \(\mathrm{NO}_{3}^{-}\)) and write the equation with only the ions involved in the precipitation reaction: \[ \mathrm{Ba}^{2+} \, (aq) + \mathrm{SO}_{4}^{2-} \, (aq) \rightarrow \mathrm{BaSO}_{4} \, (s) \] (c) \(\mathrm{NaCl}\) and \(\mathrm{KNO}_{3}\) Repeat the steps for the third pair of compounds:
09

Write the molecular equation

Write the equation with all reactants and potential products: \[ \mathrm{NaCl} \, (aq) + \mathrm{KNO}_{3} \, (aq) \rightarrow \mathrm{NaNO}_{3} \, (aq) + \mathrm{KCl} \, (aq) \]
10

Write the complete ionic equation

Break down each compound into its respective ions: \[ \mathrm{Na}^{+} \, (aq) + \mathrm{Cl}^{-} \, (aq) + \mathrm{K}^{+} \, (aq) + \mathrm{NO}_{3}^{-} \, (aq) \rightarrow \mathrm{Na}^{+} \, (aq) + \mathrm{NO}_{3}^{-} \, (aq) + \mathrm{K}^{+} \, (aq) + \mathrm{Cl}^{-} \, (aq) \]
11

Determine if a precipitation reaction occurs

Using the solubility rules, we can see that all potential products are soluble, so no precipitation reaction occurs. Since no precipitation reaction occurs, we do not need to write a net ionic equation for (c). In summary: (a) Net ionic equation: \( \mathrm{Pb}^{2+} \, (aq) + 2 \mathrm{Cl}^{-} \, (aq) \rightarrow \mathrm{PbCl}_{2} \, (s) \) (b) Net ionic equation: \( \mathrm{Ba}^{2+} \, (aq) + \mathrm{SO}_{4}^{2-} \, (aq) \rightarrow \mathrm{BaSO}_{4} \, (s) \) (c) No precipitation reaction occurs

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