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Chapter 18: Problem 20

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

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

Step by step solution

01

- 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 \( \mathrm{Na}^+(aq) + \mathrm{OH}^-(aq) + \mathrm{Na}^+(aq) + \mathrm{H}_2 \mathrm{PO}_4^-(aq) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(l) + 2 \mathrm{Na}^+(aq) + \mathrm{HPO}_4^{2-}(aq) \)
02

- 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:\( \mathrm{OH}^-(aq) + \mathrm{H}_2 \mathrm{PO}_4^-(aq) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(l) + \mathrm{HPO}_4^{2-}(aq) \)
03

- 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 \( \mathrm{HPO}_4^{2-} \).
04

- 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 \( \mathrm{H}^+(aq) + \mathrm{SO}_4^{2-}(aq) + 2 \mathrm{K}^+(aq) + \mathrm{CO}_3^{2-}(aq) \rightleftharpoons 2 \mathrm{K}^+(aq) + \mathrm{SO}_4^{2-}(aq) + \mathrm{K}^+(aq) + \mathrm{HCO}_3^+(aq)\).
05

- Identify and cancel out spectator ions (b)

The spectator ions here are \( \,\mathrm{K}^+ \,\) and \( \mathrm{SO}_4^{2-} \). Remove them to get:\( \mathrm{H}^+(aq) + \mathrm{CO}_3^{2-}(aq) \rightleftharpoons \mathrm{HCO}_3^+(aq) \)
06

- 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: \( \mathrm{CO}_3^{2-} \).

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Key Concepts

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

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鈥� known as a conjugate acid.
For example, let's look at reaction (a) from the exercise: 翱贬鈦� + 贬鈧侾翱鈧勨伝 鈬� 贬鈧侽 + 贬笔翱鈧劼测伝. Here, 贬鈧侽 is the conjugate acid of 翱贬鈦� because it's formed after 翱贬鈦� accepts a proton. Likewise, 贬笔翱鈧劼测伝 is the conjugate base of 贬鈧侾翱鈧勨伝 because it鈥檚 formed when 贬鈧侾翱鈧勨伝 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:
KHSO鈧� + K鈧侰O鈧� 鈬� K鈧係O鈧� + KHCO鈧�.
Breaking them into ions gives us:
贬鈦� + SO鈧劼测伝 + 2K鈦� + 颁翱鈧兟测伝 鈬� 2K鈦� + SO鈧劼测伝 + K鈦� + HCO鈧冣伜
Removing the spectator ions (K鈦� and SO鈧劼测伝), and simplifying, we get:
贬鈦� + 颁翱鈧兟测伝 鈬� 贬颁翱鈧冣伝.
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), 翱贬鈦� acts as a base reacting with 贬鈧侾翱鈧勨伝 (the acid), resulting in water and 贬笔翱鈧劼测伝. In reaction (b), 贬鈦� (an acid) reacts with 颁翱鈧兟测伝 (a base) to form 贬颁翱鈧冣伝.
Understanding these reactions helps in predicting the products and the behavior of acids and bases in different chemical environments.

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Most popular questions from this chapter

How are the Arrhenius and Bronsted-Lowry acid-base definitions different? How are they similar? Name two BronstedLowry bases that are not Arrhenius bases. Can you do the same for acids? Explain.

Methoxide ion, \(\mathrm{CH}_{3} \mathrm{O}^{-}\), and amide ion, \(\mathrm{NH}_{2}^{-}\), are very strong bases that are "leveled" by water. What does this mean? Write the reactions that occur in the leveling process. What species do the two leveled solutions have in common?

Explain with equations and calculations, when necessary, whether an aqueous solution of each of these salts is acidic, basic, or neutral: (a) \(\mathrm{KBr} ;\) (b) \(\mathrm{NH}_{4} \mathrm{I} ;\) (c) KCN.

Which are Lewis acids and which are Lewis bases? (a) \(\mathrm{Cu}^{2+}\) (b) Cl" (c) \(\mathrm{SnCl}_{2}\) (d) \(\mathrm{OF}_{2}\)

A strong acid has a weak bond to its acidic proton, whereas a weak acid has a strong bond to its acidic proton. Explain.
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