Final answer:
NH₄Br(aq) reacts with LiOH(aq) to form LiBr(aq), water, and NH₃, where NH₄Br is a weak acid and LiOH is a strong base. The resulting lithium bromide is a salt, and the reaction releases ammonia gas.
Step-by-step explanation:
When a weak acid reacts with a strong base, the products are typically a salt and water. In the case of NH₄Br reacting with LiOH, we can write the reaction as follows:
NH₄Br(aq) + LiOH(aq) → LiBr(aq) + H₂O(l) + NH₃(aq)
This reaction shows that the weak acid (ammonium bromide) reacts with the strong base (lithium hydroxide) to produce lithium bromide and water, along with the release of ammonia gas. The bromide ion is a spectator ion and does not participate actively in the reaction.
As lithium hydroxide is a strong base, it will dissociate completely in solution to form Li⁺ ions and OH⁻ ions. Around 1% of dissolved ammonia will be present as NH₄⁺ ions due to its weak basic nature.