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Chapter 8: Problem 56

When an aqueous solution of \(\mathrm{H}_{2} \mathrm{~S}(a q)\) is mixed with an aqueous solution of copper(II) sulfate, a black precipitate of CuS forms. (a) Write and balance the full reaction equation for the reaction of \(\mathrm{H}_{2} \mathrm{~S}(a q)\) with \(\operatorname{CuSO}_{4}(a q)\). (b) Explain how your answer to (a) demonstrates that \(\mathrm{H}_{2} \mathrm{~S}(a q)\) is an acid. (c) Write the net ionic equation for the formation of \(\operatorname{CuS}(s)\)

Short Answer

Expert verified
(a) The balanced equation for the reaction is: \[\mathrm{H}_{2} \mathrm{~S} (a q) + \operatorname{CuSO}_{4} (a q) \rightarrow \operatorname{CuS} (s) + H_{2}SO_{4} (a q)\] (b) The acidic property of \(\mathrm{H}_{2} \mathrm{~S} (a q)\) is shown by its donation of two H⁺ ions to form sulfuric acid in the reaction. (c) The net ionic equation for the formation of \(\operatorname{CuS} (s)\) is: \[S^{2-} (a q) + Cu^{2+} (a q) \rightarrow \operatorname{CuS} (s)\]

Step by step solution

01

Write the unbalanced equation

The given reaction is between hydrogen sulfide (\(\mathrm{H}_{2} \mathrm{~S}\)) and copper(II) sulfate (\(\operatorname{CuSO}_{4}\)). When they react, they form copper(II) sulfide (CuS) and an acid. The unbalanced equation for this reaction is: \[\mathrm{H}_{2} \mathrm{~S} (a q) + \operatorname{CuSO}_{4} (a q) \rightarrow \operatorname{CuS} (s) + H_{2}SO_{4} (a q)\]
02

Balance the equation

To balance the equation, we need to ensure that the number of atoms of each element on both sides of the equation is equal. In this case, it's already balanced. The balanced equation is: \[\mathrm{H}_{2} \mathrm{~S} (a q) + \operatorname{CuSO}_{4} (a q) \rightarrow \operatorname{CuS} (s) + H_{2}SO_{4} (a q)\]
03

Identify the acidic property of \(\mathrm{H}_{2} \mathrm{~S} (a q)\)

To prove that \(\mathrm{H}_{2} \mathrm{~S}(a q)\) is an acid, we need to show it donates a proton (H⁺ ion) in the reaction. In the balanced equation, we can see that it combines with \(\operatorname{CuSO}_{4} (a q)\) to form \(\operatorname{CuS} (s)\) and \(H_{2}SO_{4} (a q)\). It donates two H⁺ ions to form sulfuric acid, which is a characteristic of acids.
04

Write ionic equation from balanced equation

First, we express the compounds in aqueous solution as ions: \[ \begin{split} \mathrm{H}_{2} \mathrm{~S} (a q) + \operatorname{CuSO}_{4} (a q) &\rightarrow \operatorname{CuS} (s) + H_{2}SO_{4} (a q) \\ 2\,H^+ (a q) + S^{2-} (a q) + Cu^{2+} (a q) &+ SO_{4}^{2-} (a q) \rightarrow \operatorname{CuS} (s) + 2\,H^+ (a q) + SO_{4}^{2-} (a q) \end{split} \]
05

Write net ionic equation

To write the net ionic equation, we remove the spectator ions, which are the ions that do not participate in the actual reaction. In this case, the \(2\,H^+\) and \(SO_{4}^{2-}\) ions are the spectator ions. The net ionic equation is: \[S^{2-} (a q) + Cu^{2+} (a q) \rightarrow \operatorname{CuS} (s)\]

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