Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(\mathrm{HCl},(\mathrm{b}) \mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{CdSO}_{4}\).

First, we write the full ionic equation for the reactants: \(\mathrm{NaCH}_{3}\mathrm{COO} \rightarrow \mathrm{Na}^{+} + \mathrm{CH}_{3}\mathrm{COO}^{-}\), \(\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}\). When these ions combine, we have two possible products: \(\mathrm{NaCl}\) and \(\mathrm{HCH_3COO}\) (acetic acid). Using the solubility rules, we note that \(\mathrm{NaCl}\) is soluble since Na is an alkali metal. Acetic acid is soluble since it is a weak acid. Therefore, there is no precipitate formed in this case. Then, we write the balanced equation for the reaction: \(\mathrm{NaCH}_{3}\mathrm{COO} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{HCH_{3}COO}\).

First, we write the full ionic equation for the reactants: \(\mathrm{KOH} \rightarrow \mathrm{K}^{+} + \mathrm{OH}^{-}\), \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{NO}_3^-\). When these ions combine, we have two possible products: \(\mathrm{KNO}_{3}\) and \(\mathrm{Cu(OH)_2}\). Using the solubility rules, we note that \(\mathrm{KNO}_3\) is soluble since K is an alkali metal. However, \(\mathrm{Cu(OH)_2}\) is insoluble since hydroxides are generally insoluble, except when combined with alkali metals or ammonium. So, the precipitate that forms is \(\mathrm{Cu(OH)_2}\). Then, we write the balanced equation for the reaction: \(\mathrm{KOH} + \mathrm{Cu(NO}_{3})_{2} \rightarrow \mathrm{KNO}_{3} + \mathrm{Cu(OH)_{2}↓}\).

First, we write the full ionic equation for the reactants: \(\mathrm{Na}_{2}\mathrm{S} \rightarrow 2\mathrm{Na}^{+} + \mathrm{S}^{2-}\), \(\mathrm{CdSO}_{4} \rightarrow \mathrm{Cd}^{2+} + \mathrm{SO}_4^{2-}\). When these ions combine, we have two possible products: \(\mathrm{Na_2SO_4}\) and \(\mathrm{CdS}\). Using the solubility rules, we note that \(\mathrm{Na}_2\mathrm{SO}_4\) is soluble since Na is an alkali metal. However, \(\mathrm{CdS}\) is insoluble since sulfides are generally insoluble, except when combined with alkali metals or ammonium. So, the precipitate that forms is \(\mathrm{CdS}\). Then, we write the balanced equation for the reaction: \(\mathrm{Na}_{2}\mathrm{S} + \mathrm{CdSO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + \mathrm{CdS ↓}\).