Problem 39

Question

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Step-by-Step Solution

Verified
Answer
The net ionic equations for the given reactions are: (a) \(2\mathrm{H^{+}}(aq) + 2\mathrm{OH^{-}}(aq) \longrightarrow 2\mathrm{H_{2}O}(l)\) (b) \(\mathrm{Cu(OH)_{2}}(s) + 2\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{H_{2}O}(l)\) (c) \(\mathrm{Al(OH)_{3}}(s) + 3\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Al^{3+}}(aq) + 3\mathrm{H_{2}O}(l)\)
1Step 1: Reaction (a) - Write unbalanced molecular equation
The given unbalanced equation is: HBr(aq) + Ca(OH)2(aq) ->
2Step 2: Reaction (a) - Balance the equation
Let's balance the equation. We have \(2 \mathrm{HBr} (a q) + \mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{CaBr}_{2} (\mathrm{sq}) + 2\mathrm{H}_{2}\mathrm{O}(\ell)\)
3Step 3: Reaction (a) - Complete ionic equation
Now write the complete ionic equation: \(2\mathrm{H^{+}}(aq) + 2\mathrm{Br^{-}}(aq) + \mathrm{Ca^{2+}}(aq) + 2\mathrm{OH^{-}}(aq) \longrightarrow \mathrm{Ca^{2+}}(aq) + 2\mathrm{Br^{-}}(aq) + 2\mathrm{H_{2}O}(l)\)
4Step 4: Reaction (a) - Net ionic equation
Finally, the net ionic equation is: \(2\mathrm{H^{+}}(aq) + 2\mathrm{OH^{-}}(aq) \longrightarrow 2\mathrm{H_{2}O}(l)\) For Reaction (b):
5Step 5: Reaction (b) - Write unbalanced molecular equation
The given unbalanced equation is: Cu(OH)2(s) + HClO4(aq) ->
6Step 6: Reaction (b) - Balance the equation
Let's balance the equation. We have \( \mathrm{Cu}(\mathrm{OH})_{2}(s) + 2\mathrm{HClO}_4(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{ClO}_{4})_{2}(\mathrm{sq}) + 2\mathrm{H}_{2}\mathrm{O}(\ell)\)
7Step 7: Reaction (b) - Complete ionic equation
Now write the complete ionic equation: \(\mathrm{Cu(OH)_{2}}(s) + 2\mathrm{H^{+}}(aq) + 2\mathrm{ClO}_{4}^-(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{ClO}_{4}^-(aq) + 2\mathrm{H_{2}O}(l)\)
8Step 8: Reaction (b) - Net ionic equation
Finally, the net ionic equation is: \(\mathrm{Cu(OH)_{2}}(s) + 2\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{H_{2}O}(l)\) For Reaction (c):
9Step 9: Reaction (c) - Write unbalanced molecular equation
The given unbalanced equation is: Al(OH)3(s) + HNO3(aq) ->
10Step 10: Reaction (c) - Balance the equation
Let's balance the equation. We have \(\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{HNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Al}(\mathrm{NO}_{3})_{3}(\mathrm{sq}) + 3\mathrm{H}_{2}\mathrm{O}(\ell)\)
11Step 11: Reaction (c) - Complete ionic equation
Now write the complete ionic equation: \(\mathrm{Al(OH)_{3}}(s) + 3\mathrm{H^{+}}(aq) + 3\mathrm{NO}_{3}^-(aq) \longrightarrow \mathrm{Al^{3+}}(aq) + 3\mathrm{NO}_{3}^-(aq) + 3\mathrm{H_{2}O}(l)\)
12Step 12: Reaction (c) - Net ionic equation
Finally, the net ionic equation is: \(\mathrm{Al(OH)_{3}}(s) + 3\mathrm{H^{+}}(aq) \longrightarrow \mathrm{Al^{3+}}(aq) + 3\mathrm{H_{2}O}(l)\)