Answer:
a. KOH (aq) + HCl (aq) → KCl (s) + H₂O (l)
b. NaOH (aq) + HCl (aq) → NaCl (s) + H₂O (l)
General Formulas and Concepts:
Chemistry
Unit 0
- Reading a Periodic Table
- Writing Compounds
- Polyatomic Ions
Aqueous Solutions
- States of matter
- Solubility Rules
Explanation:
We are given "Potassium hydroxide reacts with hydrochloric acid to form salt and water."
a)
From the question, we can figure out 1.) if the reaction exists and 2.) if it does exist, then reaction prediction.
- Potassium - K⁺
- Hydroxide (PAI) - OH⁻
- Potassium Hydroxide - KOH
- Hydro - H⁺
- Chloric (Chlorine) - Cl⁻
- Hydrochloric acid (strong) - HCl
Remembering solubility rules, we have:
- All common compounds of Group IA ions are soluble
- All common chlorides, bromides, and iodides are mostly soluble (exceptions are Ag⁺, Pb²⁺, Cu⁺, Hg₂²⁺)
Therefore, we can conclude that both potassium hydroxide and hydrochloric are both aqueous solutions:
KOH (aq) + HCl (aq) → ? + ?
We are also given from the problem that the reaction produces a salt and water.
- Molecular Formula of water - H₂O
We also use the double replacement reaction to do reaction prediction of the proposed salt:
KOH (aq) + HCl (aq) → KCl (s) + H₂O (l)
Hence, this is our answer to a.).
b.)
We are given that we switch the compound KOH (potassium hydroxide) with sodium hydroxide.
- Sodium - Na⁺
- Hydroxide (PAI) - OH⁻
- Sodium Hydroxide - NaOH
We can use the same steps as a.) to figure out our new reaction. Since we already know that Group IA compounds are soluble, we know that NaOH is soluble:
NaOH (aq) + HCl (aq) → ? + H₂O (l)
Following the same steps, we can simple do a double replacement reaction prediction to find the missing salt:
NaOH (aq) + HCl (aq) → NaCl (s) + H₂O (l)
We see that we would still be able to produce a [different] salt and the reaction does exist/run. And we have our final answer for b.).
Note:
While the salts produced (KCl and NaCl) are solids, they are soluble in water. Since we are running only a forward reaction, we know that we are producing a solid and liquid water. Usually, we would probably see these reactions in equilibrium (represented with a ⇆).
This also delves into the concepts of saturation.
