Question

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

Short answer

The precipitates and balanced reactions are as follows: (a) Precipitate: \(\mathrm{Ni(OH)_{2}}\), Balanced Reaction: \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} + 2\mathrm{NaOH} \rightarrow \mathrm{Ni(OH)_{2}}\downarrow + 2\mathrm{NaNO}_{3}\) (b) No precipitate, Balanced Reaction: \(\mathrm{NaOH} + \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{KO} + \mathrm{Na}_{2}\mathrm{SO}_{4}\) (c) Precipitate: \(\mathrm{CuS}\), Balanced Reaction: \(\mathrm{Na}_{2}\mathrm{S} + \mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \rightarrow \mathrm{CuS}\downarrow + 2\mathrm{NaCH}_{3}\mathrm{COO} \)

Step-by-step solution

(a) Identify possible products and check solubility rules

When \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaOH}\) are mixed, the possible products are \(\mathrm{Ni(OH)_{2}}\) and \(\mathrm{NaNO}_{3}\). According to solubility rules, most hydroxides are generally insoluble, with some exceptions. \(\mathrm{Ni(OH)_{2}}\) is insoluble in water, and \(\mathrm{NaNO}_{3}\) is soluble as all nitrates are soluble. Therefore, a precipitate of \(\mathrm{Ni(OH)_{2}}\) will form.

(a) Write the balanced equation

The balanced equation for this reaction is: \[ \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} + 2\mathrm{NaOH} \rightarrow \mathrm{Ni(OH)_{2}}\downarrow + 2\mathrm{NaNO}_{3} \]

(b) Identify possible products and check solubility rules

When \(\mathrm{NaOH}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) are mixed, the possible products are \(\mathrm{KO}\) and \(\mathrm{Na}_{2}\mathrm{SO}_{4}\). Both potassium compounds (like \(\mathrm{KO}\)) and sodium compounds (like \(\mathrm{Na}_{2}\mathrm{SO}_{4}\)) are soluble in water, so there is no precipitate formed in this case.

(b) Write the balanced equation

Since there are no precipitates, the balanced equation will be: \[ \mathrm{NaOH} + \mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{KO} + \mathrm{Na}_{2}\mathrm{SO}_{4} \]

(c) Identify possible products and check solubility rules

Given, \(\mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\), the possible products are \(\mathrm{CuS}\) and \(\mathrm{NaCH}_{3}\mathrm{COO}\). According to the solubility rules, most sulfides are insoluble, with some exceptions. In this case, \(\mathrm{CuS}\) is insoluble, and \(\mathrm{NaCH}_{3}\mathrm{COO}\) is soluble given most acetates are soluble. Therefore, a precipitate of \(\mathrm{CuS}\) will form.

(c) Write the balanced equation

The balanced equation for this reaction is: \[ \mathrm{Na}_{2}\mathrm{S} + \mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \rightarrow \mathrm{CuS}\downarrow + 2\mathrm{NaCH}_{3}\mathrm{COO} \]