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Chapter 6: Problem 68

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

Short Answer

Expert verified
a. Balanced Formula Equation: \(3\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow 3\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Al}(\mathrm{NO}_{3})_{3}(a q)\) Complete Ionic Equation: \(3\mathrm{H^{+}}(a q) + 3\mathrm{NO}_{3}^{-}(a q)+ \mathrm{Al^{3+}}(s) + 3\mathrm{OH}^{-}(s) \rightarrow 3\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Al^{3+}}(a q) + 3\mathrm{NO}_{3}^{-}(a q)\) Net Ionic Equation: \(\mathrm{H^{+}}(a q) + \mathrm{OH}^{-}(s) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)\) b. Balanced Formula Equation: \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{KC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(a q)\) Complete Ionic Equation: \(\mathrm{H^{+}}(a q) + \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(a q) + \mathrm{K^{+}}(a q) + \mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{K^{+}}(a q) + \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(a q)\) Net Ionic Equation: \(\mathrm{H^{+}}(a q) + \mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)\) c. Balanced Formula Equation: \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{HCl}(a q) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CaCl}_{2}(a q)\) Complete Ionic Equation: \(\mathrm{Ca^{2+}}(a q) + 2\mathrm{OH}^{-}(a q) + 2\mathrm{H^{+}}(a q) + 2\mathrm{Cl}^{-}(a q) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Ca^{2+}}(a q) + 2\mathrm{Cl}^{-}(a q)\) Net Ionic Equation: \(2\mathrm{H^{+}}(a q) + 2\mathrm{OH}^{-}(a q) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\)

Step by step solution

01

Identify the products

In an acid-base reaction, the hydrogen ion from the acid (HNO3) will combine with the hydroxide ion from the base (Al(OH)3), forming water. The remaining elements (Al and NO3) will form a salt. So, the products of the reaction will be water and an aluminum nitrate salt.
02

Write the balanced formula equation

Now we will balance the equation: \[3\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow 3\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Al}(\mathrm{NO}_{3})_{3}(a q)\]
03

Write the complete ionic equation

To write the complete ionic equation, we break down soluble ionic compounds into their individual ions: \[3\mathrm{H^{+}}(a q) + 3\mathrm{NO}_{3}^{-}(a q)+ \mathrm{Al^{3+}}(s) + 3\mathrm{OH}^{-}(s) \rightarrow 3\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Al^{3+}}(a q) + 3\mathrm{NO}_{3}^{-}(a q)\]
04

Write the net ionic equation

To get the net ionic equation, we remove the spectator ions, which are the same ions on both sides of the complete ionic equation: \[\mathrm{H^{+}}(a q) + \mathrm{OH}^{-}(s) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)\] b. $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow$
05

Identify the products

For this reaction, the hydrogen ion from the acetic acid (HC2H3O2) will combine with the hydroxide ion from the potassium hydroxide (KOH), forming water. The remaining elements (K and C2H3O2) will form a salt. So, the products of the reaction will be water and potassium acetate salt.
06

Write the balanced formula equation

The balanced formula equation for this reaction is: \[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{KC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}(a q)\]
07

Write the complete ionic equation

The complete ionic equation is: \[\mathrm{H^{+}}(a q) + \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(a q) + \mathrm{K^{+}}(a q) + \mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{K^{+}}(a q) + \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}(a q)\]
08

Write the net ionic equation

The net ionic equation is: \[\mathrm{H^{+}}(a q) + \mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2}\mathrm{O}(l)\] c. $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow$
09

Identify the products

For this reaction, the hydrogen ion from the hydrochloric acid (HCl) will combine with the hydroxide ion from calcium hydroxide (Ca(OH)2), forming water. The remaining elements (Ca and Cl) will form a salt. The products of the reaction will be water and calcium chloride salt.
10

Write the balanced formula equation

The balanced formula equation is: \[\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2\mathrm{HCl}(a q) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CaCl}_{2}(a q)\]
11

Write the complete ionic equation

The complete ionic equation is: \[\mathrm{Ca^{2+}}(a q) + 2\mathrm{OH}^{-}(a q) + 2\mathrm{H^{+}}(a q) + 2\mathrm{Cl}^{-}(a q) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Ca^{2+}}(a q) + 2\mathrm{Cl}^{-}(a q)\]
12

Write the net ionic equation

Finally, the net ionic equation is: \[2\mathrm{H^{+}}(a q) + 2\mathrm{OH}^{-}(a q) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equations
In chemistry, balanced chemical equations represent chemical reactions, showing the relationship between reactants and products. The importance of balancing chemical equations cannot be overstated. This process ensures that the law of conservation of mass is respected, meaning the total mass of reactants equals the total mass of products.

This is done by adjusting the coefficients before the chemical formulas so that the number of each type of atom on both sides of the equation is equal. For example, when nitric acid reacts with aluminum hydroxide, the reaction is balanced as follows:
  • \[ 3\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow 3\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Al}(\mathrm{NO}_{3})_{3}(a q) \]
This balanced equation shows that each element and charge is conserved across the reaction, clearly representing the stoichiometry of the reaction. Balanced equations help chemists understand the proportions of substances involved in reactions and ensure accurate preparation of solutions and compounds.
Ionic Equations
Ionic equations take the concept of chemical equations a step forward by breaking soluble ionic compounds into their component ions. This emphasizes the ions that actually participate in the reaction, rather than simply presenting compounds.

In the case of soluble substances, such as strong acids, bases, and most salts, they dissociate into ions when dissolved in water. For instance, the complete ionic equation for the reaction between nitric acid and aluminum hydroxide is:
  • \[ 3\mathrm{H^{+}}(a q) + 3\mathrm{NO}_{3}^{-}(a q)+ \mathrm{Al^{3+}}(s) + 3\mathrm{OH}^{-}(s) \rightarrow 3\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Al^{3+}}(a q) + 3\mathrm{NO}_{3}^{-}(a q) \]
Notice how each compound is broken down into its ions, except for precipitates, liquids, and gases, which remain in their compound forms. This gives a clearer picture of the nature of the reaction in solution, highlighting what changes, and what stay the same.
Net Ionic Equations
Net ionic equations further refine the concept of ionic equations by isolating only the ions and molecules directly involved in the chemical reaction鈥攐ften referred to as the "core" action. First, it's necessary to identify spectator ions, which are ions that appear unchanged on both sides of the complete ionic equation, as they do not participate directly in the reaction.

Once these are identified, they can be removed from the equation, reducing it to its simplest form. In our example, the net ionic equation for the acid-base reaction between nitric acid and aluminum hydroxide is:
  • \[ \mathrm{H^{+}}(a q) + \mathrm{OH}^{-}(s) \rightarrow \mathrm{H}_{2}\mathrm{O}(l) \]
This equation indicates that the main event is the formation of water from hydrogen and hydroxide ions. Net ionic equations provide a focused insight into the chemistry process, simplifying the overall reaction and making it easier to understand the true chemical change.

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Most popular questions from this chapter

A \(10.00-\mathrm{g}\) sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead(II) nitrate to form \(21.75 \mathrm{g}\) of solid. Determine the mass percent of sodium chloride in the original mixture.

A 1.42 -g sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\) was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{g}\). Determine the atomic mass of \(\mathrm{M},\) and identify \(\mathrm{M}\).

A 100.0-mL aliquot of 0.200 \(M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

Using the general solubility rules given in Table \(6-1,\) name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion d. sulfate ion b. calcium ion e. mercury(I) ion, \(\mathrm{Hg}_{2}^{2+}\) c. iron(III) ion f. silver ion

A \(25.00-\mathrm{mL}\) sample of hydrochloric acid solution requires \(24.16 \mathrm{mL}\) of \(0.106 \mathrm{M}\) sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?
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