Write a balanced chemical equation for the aqueous reaction between potassium hydroxide and hydrobromic acid. Include physical states. Question options: KOH (s) + HBr (g) ® H2O (l) + KBr (aq) KOH (s) + HBr (g) ® H2O (l) + KBr (s) KOH (s) + HBr (aq) ® H2O (l) + KBr (s) KOH (aq) + HBr (aq) ® H2O (l) + KBr (aq)
The Correct Answer and Explanation is:
The correct balanced chemical equation for the aqueous reaction between potassium hydroxide (KOH) and hydrobromic acid (HBr) is:KOH (aq)+HBr (aq)→H2O (l)+KBr (aq)\text{KOH (aq)} + \text{HBr (aq)} \rightarrow \text{H}_2\text{O (l)} + \text{KBr (aq)}KOH (aq)+HBr (aq)→H2O (l)+KBr (aq)
Explanation:
- Reactivity of KOH and HBr: Potassium hydroxide (KOH) is a strong base, and hydrobromic acid (HBr) is a strong acid. When these two substances are mixed in aqueous solution, they undergo a neutralization reaction. The general form of a neutralization reaction between an acid and a base is: acid (aq)+base (aq)→water (l)+salt (aq)\text{acid (aq)} + \text{base (aq)} \rightarrow \text{water (l)} + \text{salt (aq)}acid (aq)+base (aq)→water (l)+salt (aq) In this case, the acid is HBr and the base is KOH.
- Ionization in aqueous solution:
- Potassium hydroxide dissociates into potassium ions (K⁺) and hydroxide ions (OH⁻) when dissolved in water: KOH (aq)→K+(aq)+OH−(aq)\text{KOH (aq)} \rightarrow \text{K}^+ (aq) + \text{OH}^- (aq)KOH (aq)→K+(aq)+OH−(aq)
- Hydrobromic acid dissociates into hydrogen ions (H⁺) and bromide ions (Br⁻) in water: HBr (aq)→H+(aq)+Br−(aq)\text{HBr (aq)} \rightarrow \text{H}^+ (aq) + \text{Br}^- (aq)HBr (aq)→H+(aq)+Br−(aq)
- Reaction:
The hydrogen ion (H⁺) from HBr combines with the hydroxide ion (OH⁻) from KOH to form water (H₂O): H+(aq)+OH−(aq)→H2O (l)\text{H}^+ (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O (l)}H+(aq)+OH−(aq)→H2O (l) The potassium ion (K⁺) pairs with the bromide ion (Br⁻) to form potassium bromide (KBr), which remains in the aqueous state because it is soluble in water: K+(aq)+Br−(aq)→KBr (aq)\text{K}^+ (aq) + \text{Br}^- (aq) \rightarrow \text{KBr (aq)}K+(aq)+Br−(aq)→KBr (aq) - Final balanced equation:
Considering the dissociation of both the acid and the base, the final balanced equation is: KOH (aq)+HBr (aq)→H2O (l)+KBr (aq)\text{KOH (aq)} + \text{HBr (aq)} \rightarrow \text{H}_2\text{O (l)} + \text{KBr (aq)}KOH (aq)+HBr (aq)→H2O (l)+KBr (aq) This equation represents the neutralization process, where water and a soluble salt (KBr) are formed, both of which are in the correct physical states (liquid and aqueous, respectively).
In this equation:
- KOH and HBr are in aqueous form because they dissociate completely in water.
- H₂O is liquid because it is formed as a product of the neutralization reaction.
- KBr remains in aqueous form because it is highly soluble in water.
This shows that the reaction happens in solution and is typical of strong acid-strong base reactions.
