Problem 28 - Chemistry and Chemical Reactivity

Question

Complete and balance the equations for the following acid-base reactions. Name the reactants and products. (a) \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{KOH}(\mathrm{aq}) \rightarrow\) (b) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow\) (\(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is oxalic acid, an acid capable of donating two \(\mathrm{H}^{+}\)ions. See Study Question 23.)

Step-by-Step Solution

Answer

(a) Balanced equation: \( \mathrm{H}_{3} \mathrm{PO}_{4} + 3 \mathrm{KOH} \rightarrow \mathrm{K}_3\mathrm{PO}_4 + 3 \mathrm{H}_2\mathrm{O} \). (b) Balanced equation: \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} + \mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{CaC}_{2}\mathrm{O}_{4} + 2 \mathrm{H}_2\mathrm{O} \).

Step 1: Identify Reactants and Products for (a)

The given reaction involves phosphoric acid, \( \mathrm{H}_{3} \mathrm{PO}_{4} \), and potassium hydroxide, \( \mathrm{KOH} \). \( \mathrm{H}_{3} \mathrm{PO}_{4} \) will donate \( \mathrm{H}^{+} \) ions and \( \mathrm{KOH} \) will donate \( \mathrm{OH}^{-} \) ions. The product formed will be water and potassium phosphate, \( \mathrm{K}_3\mathrm{PO}_4 \).

Step 2: Balance the Equation for (a)

The balanced equation for the reaction is:\[ \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) + 3 \mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{K}_3\mathrm{PO}_4(\mathrm{aq}) + 3 \mathrm{H}_2\mathrm{O}(\mathrm{l}) \]

Step 3: Identify Reactants and Products for (b)

The given reaction involves oxalic acid, \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \), and calcium hydroxide, \( \mathrm{Ca}(\mathrm{OH})_{2} \). Oxalic acid donates \( \mathrm{H}^{+} \) ions and calcium hydroxide forms \( \mathrm{Ca}^{2+} \) and \( \mathrm{OH}^{-} \) ions. The product will be calcium oxalate, \( \mathrm{CaC}_{2}\mathrm{O}_{4} \), and water.

Step 4: Balance the Equation for (b)

The balanced equation for the reaction is:\[ \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) + \mathrm{Ca}(\mathrm{OH})_2(\mathrm{s}) \rightarrow \mathrm{CaC}_{2}\mathrm{O}_{4}(\mathrm{s}) + 2 \mathrm{H}_2\mathrm{O}(\mathrm{l}) \]

Step 5: Naming the Reactants and Products

For reaction (a), reactants are phosphoric acid and potassium hydroxide, products are potassium phosphate and water. For reaction (b), reactants are oxalic acid and calcium hydroxide, products are calcium oxalate and water.

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is a crucial skill in chemistry. It ensures that the same number of each type of atom appears on both sides of the equation. This aligns with the law of conservation of mass, which states that matter cannot be created or destroyed.
To balance an equation, follow these simple steps:

• List all atoms participating in the reaction for both reactants and products.
• Adjust coefficients (numbers before compounds) to achieve the same number of each atom type on both sides.
• Check that the coefficients allow for the smallest whole numbers possible.

In our example, the initial equation for reaction (a) involving phosphoric acid and potassium hydroxide starts unbalanced. By placing the coefficient 3 in front of potassium hydroxide and water, balance is achieved. The same procedure applied to reaction (b) ensures mass conservation.

Phosphoric Acid

Phosphoric acid, represented chemically as \(\mathrm{H}_3\mathrm{PO}_4\), is a triprotic acid. This means it can donate three hydrogen ions \(\mathrm{H}^+\). It reacts with bases to form salts and water, a classic example of an acid-base reaction.
In the reaction with potassium hydroxide (\(\mathrm{KOH}\)), each \(\mathrm{H}_3\mathrm{PO}_4\) molecule combines with three \(\mathrm{OH}^-\) ions to create water and the salt potassium phosphate (\(\mathrm{K}_3\mathrm{PO}_4\)). The balanced chemical equation for this reaction is:

• \(\mathrm{H}_{3} \mathrm{PO}_{4} + 3 \mathrm{KOH} \rightarrow \mathrm{K}_3\mathrm{PO}_4 + 3 \mathrm{H}_2\mathrm{O}\)

Such reactions show how phosphoric acid's multiple hydrogen ions can impact the stoichiometry of balancing.

Oxalic Acid

Oxalic acid, \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\), is a diprotic acid—meaning it can donate two hydrogen ions \(\mathrm{H}^+\). It often participates in reactions where a salt is formed alongside water.
When oxalic acid reacts with calcium hydroxide (\(\mathrm{Ca(OH)}_2\)), the products are calcium oxalate (\(\mathrm{CaC}_2\mathrm{O}_4\)) and water. In such acid-base reactions, the hydrogen ions from oxalic acid combine with the hydroxide ions from calcium hydroxide to form water. The balanced equation is:

• \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{CaC}_2\mathrm{O}_4 + 2 \mathrm{H}_2\mathrm{O}\)

Oxalic acid is not only relevant in chemistry; it's also present in various plant foods, giving a slightly sour taste.

Calcium Hydroxide

Calcium hydroxide, \(\mathrm{Ca(OH)}_2\), is a strong base. Known as slaked lime, it's used in a variety of applications from construction to water treatment. In chemistry, it readily reacts with acids to produce salts.
In our reaction involving oxalic acid, calcium hydroxide dissociates into \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^-\) ions in an aqueous solution. These ions interact with the acidic hydrogen ions \(\mathrm{H}^+\) from oxalic acid, leading to the creation of water and calcium oxalate:

• \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{CaC}_2\mathrm{O}_4 + 2 \mathrm{H}_2\mathrm{O}\)

Calcium hydroxide's strong basicity makes it a key player in neutralizing acidic conditions in both laboratory and practical applications.