Question

Write balanced net ionic equations for the following reactions, and label the conjugate acid-base pairs: (a) \(\mathrm{NaOH}(a q)+\mathrm{NaH}_{2} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{2} \mathrm{HPO}_{4}(a q)\) (b) \(\mathrm{KHSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{KHCO}_{3}(a q)\)

Short answer

(a) OH^- + H_2PO_4^- → H_2O + HPO_4^{2-}. Acid-base pairs: H_2PO_4^- and HPO_4^{2-}; OH^- and H_2O. (b) H^+ + CO_3^{2-} → HCO_3^+. Acid-base pairs: HCO_3^+ and CO_3^{2-}.

Step-by-step solution

- Write the full ionic equation (a)

Start by breaking down all aqueous compounds into their ions: \( \mathrm{NaOH}(aq) + \mathrm{NaH}_2 \mathrm{PO}_4(aq) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(l) + \mathrm{Na}_2 \mathrm{HPO}_4(aq) \) becomes \( \mathrm{Na}^+(aq) + \mathrm{OH}^-(aq) + \mathrm{Na}^+(aq) + \mathrm{H}_2 \mathrm{PO}_4^-(aq) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(l) + 2 \mathrm{Na}^+(aq) + \mathrm{HPO}_4^{2-}(aq) \)

- Identify and cancel out spectator ions (a)

The spectator ion here is \( \,\mathrm{Na}^+ \,\), which appears on both sides of the equation. Remove it to get:\( \mathrm{OH}^-(aq) + \mathrm{H}_2 \mathrm{PO}_4^-(aq) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(l) + \mathrm{HPO}_4^{2-}(aq) \)

- Identify conjugate acid-base pairs (a)

In the reaction:\(\mathrm{OH}^-(aq) + \mathrm{H}_2 \mathrm{PO}_4^-(aq) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(l) + \mathrm{HPO}_4^{2-}(aq)\)The acid-base pairs are: Conjugate Acid: \( \mathrm{H}_2 \mathrm{PO}_4^- \)\ and \( \mathrm{H}_2 \mathrm{O} \). Conjugate Base: \( \mathrm{OH}^- \) and \( \mathrm{HPO}_4^{2-} \).

- Write the full ionic equation (b)

Breaking down the second reaction into ions gives:\( \mathrm{KHSO}_4(aq) + \mathrm{K}_2 \mathrm{CO}_3(aq) \rightleftharpoons \mathrm{K}_2 \mathrm{SO}_4(aq) + \mathrm{KHCO}_3(aq)\) becomes \( \mathrm{H}^+(aq) + \mathrm{SO}_4^{2-}(aq) + 2 \mathrm{K}^+(aq) + \mathrm{CO}_3^{2-}(aq) \rightleftharpoons 2 \mathrm{K}^+(aq) + \mathrm{SO}_4^{2-}(aq) + \mathrm{K}^+(aq) + \mathrm{HCO}_3^+(aq)\).

- Identify and cancel out spectator ions (b)

The spectator ions here are \( \,\mathrm{K}^+ \,\) and \( \mathrm{SO}_4^{2-} \). Remove them to get:\( \mathrm{H}^+(aq) + \mathrm{CO}_3^{2-}(aq) \rightleftharpoons \mathrm{HCO}_3^+(aq) \)

- Identify conjugate acid-base pairs (b)

In the reaction:\( \mathrm{H}^+(aq) + \mathrm{CO}_3^{2-}(aq) \rightleftharpoons \mathrm{HCO}_3^+(aq) \)The acid-base pairs are: Conjugate Acid: \( \mathrm{HCO}_3^+ \). Conjugate Base: \( \mathrm{CO}_3^{2-} \).

Key concepts

Conjugate Acid-Base Pairs

In acid-base chemistry, conjugate acid-base pairs are very important. When an acid donates a proton (H+), it becomes a base. This new base is called the conjugate base of the acid. Similarly, when a base accepts a proton, it becomes an acid— known as a conjugate acid.
For example, let's look at reaction (a) from the exercise: OH⁻ + H₂PO₄⁻ ⇌ H₂O + HPO₄²⁻. Here, H₂O is the conjugate acid of OH⁻ because it's formed after OH⁻ accepts a proton. Likewise, HPO₄²⁻ is the conjugate base of H₂PO₄⁻ because it’s formed when H₂PO₄⁻ donates a proton.
This relationship helps in understanding how substances interconvert in acid-base reactions and illustrates the reversible nature of these reactions.

Balancing Chemical Equations

Balancing a chemical equation is essential in accurately representing a chemical reaction. It ensures that the number of atoms for each element is the same on both sides of the equation. For the net ionic equation to be balanced, both the mass and the charge must be balanced.
Take reaction (b) from the exercise:
KHSO₄ + K₂CO₃ ⇌ K₂SO₄ + KHCO₃.
Breaking them into ions gives us:
H⁺ + SO₄²⁻ + 2K⁺ + CO₃²⁻ ⇌ 2K⁺ + SO₄²⁻ + K⁺ + HCO₃⁺
Removing the spectator ions (K⁺ and SO₄²⁻), and simplifying, we get:
H⁺ + CO₃²⁻ ⇌ HCO₃⁻.
Balancing helps us to clearly see that the numbers of each type of atom and the charges are the same on both sides.

Acid-Base Reactions

Acid-base reactions are fundamental chemical processes where an acid reacts with a base. These reactions often form water and a salt. The driving force behind many of these reactions is the transfer of protons (H+).
In our exercise, both reactions (a) and (b) are examples of acid-base reactions. In reaction (a), OH⁻ acts as a base reacting with H₂PO₄⁻ (the acid), resulting in water and HPO₄²⁻. In reaction (b), H⁺ (an acid) reacts with CO₃²⁻ (a base) to form HCO₃⁻.
Understanding these reactions helps in predicting the products and the behavior of acids and bases in different chemical environments.