Writing Net Ionic Equations Write a net ionic equation for each of the following molecular equations. a. 2HClO4(aq) + Ca(OH)2(aq) → Ca(ClO4)2(aq) + 2H2O(l ) Perchloric acid, HClO4, is a strong electrolyte, forming H^+ and ClO4^- ions in solution. Ca(ClO4)2 is a soluble ionic compound. b.

Question

Writing Net Ionic Equations

Write a net ionic equation for each of the following molecular equations.

a. [latex]2HClO_4(aq) + Ca(OH)_2(aq) → Ca(ClO_4)_2(aq) + 2H_2O(l )[/latex]

Perchloric acid, [latex]HClO_4[/latex], is a strong electrolyte, forming [latex]H^+[/latex] and [latex]ClO_4^-[/latex] ions in solution. [latex]Ca(ClO_4)_2[/latex] is a soluble ionic compound.

b. [latex]HC_2H_3O_2(aq) + NaOH(aq) → NaC_2H_3O_2(aq) + H_2O(l )[/latex]

Acetic acid, [latex]HC_2H_3O_2[/latex], is a molecular substance and a weak electrolyte.

PROBLEM STRATEGY

You will need to convert the molecular equation to the complete ionic equation, then cancel spectator ions to obtain the net ionic equation. For each ionic compound in the reaction, use the solubility rules to determine if the compound will be soluble (in the solution as ions) or insoluble (present as an undissolved solid). For the complete ionic equation, represent all of the strong electrolytes by their separate ions in solution, adding (aq) after the formula of each. Retain the formulas of the other compounds. An ionic compound should have (aq) after its formula if it is soluble or (s) if it is insoluble.

Accepted Answer

a. According to the solubility rules presented in Table 4.1 and the problem statement, [latex]Ca(OH)_2[/latex] and [latex]Ca(ClO_4)_2[/latex] are soluble ionic compounds so they are strong electrolytes. The problem statement notes that [latex]HClO_4[/latex] is also a strong electrolyte. You write each strong electrolyte in the form of separate ions. Water, [latex]H_2O[/latex], is a nonelectrolyte (or very weak electrolyte), so you retain its molecular formula. The complete ionic equation is

[latex]2H^+(aq) + \cancel{2ClO_4^- (aq)} + \cancel{Ca2^{2+}(aq)} + 2OH^- (aq) → \cancel{Ca^{2+} (aq)} + \cancel{2ClO_4^- (aq)} + 2H_2O(l)[/latex]

After canceling spectator ions and dividing by 2, you get the following net ionic equation:

[latex]H^+(aq) + OH^-(aq) → H_2O(l)[/latex]

b. According to the solubility rules, NaOH and [latex]NaC_2H_3O_2[/latex] are soluble ionic compounds, so they are strong electrolytes. The problem statement notes that [latex]HC_2H_3O_2[/latex] is a weak electrolyte, which you write by its molecular formula. Water, [latex]H_2O[/latex], is a nonelectrolyte, so you retain its molecular formula also. The complete ionic equation is

[latex]HC_2H_3O_2(aq) + \cancel{Na^+(aq)} + OH^-(aq) → \cancel{Na^+(aq)} + C_2H_3O_2^- (aq) + H_2O(l)[/latex]

and the net ionic equation is

[latex]HC_2H_3O_2(aq) + OH^-(aq) → C_2H_3O_2^- (aq) + H_2O(l)[/latex]

Table 4.1

Solubility Rules for Ionic Compounds
Rule	Applies to	Statement	Exceptions
1	[latex]Li^+ , Na^+ , K^+ , NH_4^+[/latex]	Group IA and ammonium compounds are soluble.	—
2	[latex]C_2H_3O_2^-, NO_3^- [/latex]	Acetates and nitrates are soluble.	—
3	[latex]Cl^- , Br^- , I^- [/latex]	Most chlorides, bromides, and iodides are soluble.	[latex]AgCl , Hg_2Cl_2 , PbCl_2, AgBr, HgBr_2, Hg_2Br_2,\ PbBr_2, AgI, HgI_2, Hg_2I_2, PbI_2[/latex]
4	[latex]SO_4^{2-}[/latex]	Most sulfates are soluble.	[latex]CaSO_4, SrSO_4, BaSO_4, Ag_2SO_4, Hg_2SO_4,\ PbSO_4[/latex]
5	[latex]CO_3^{2-}[/latex]	Most carbonates are insoluble.	Group IA carbonates, [latex](NH_4)_2CO_3[/latex]
6	[latex]PO_4^{3-}[/latex]	Most phosphates are insoluble.	Group IA phosphates, [latex](NH_4)_3PO_4[/latex]
7	[latex]S^{2-}[/latex]	Most sulfides are insoluble.	Group IA sulfides, [latex](NH_4)_2S[/latex]
8	[latex]OH^-[/latex]	Most hydroxides are insoluble.	Group IA hydroxides, [latex]Ca(OH)_2, Sr(OH)_2,\Ba(OH)_2[/latex]

Question 4.2: Writing Net Ionic Equations Write a net ionic equation for e...

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