The question says write a reaction for the ionization of the following compound in water. Identify the acid, the base, the conjugate acid, and the conjugate base in each of them.
1. H2SO4
2. KOH
3. CH3COOH
4. NH3
5. HNO3
My guesses are:
H2SO4 + H2O -> H3O+ + HSO4-
KOH + H2O -> K+ + OH- + H2O
CH3COOH + H2O -> CH3COO- + H3O+
NH3 + H2O -> NH2- + H3O+
HNO3 + H2O -> NO3- + H3O+
Before I guess the acids and bases, are any of these equation right? Can someone explain the rules for writing these equations?
I don't see anything wrong with what you've done except the fourth one.
I would write
NH3 + HOH ==> NH4^+ + OH^- since NH3 is a stronger base than H2O.
Some would write the KOH as
KOH + H2O ==> K^+(aq) + OH^-(aq)
You need not guess at the acids and bases. For example, the first one.
H2SO4 + H2O ==> H3O^+ + HSO4^-
Now. Acids are proton donors. Which donated the proton?. Obviously the H2SO4 (since it is missing one on the right hand side). The base must be H2O for it accepted the proton (and it has an extra proton on the right side).
Yes, your equations are correct! Let's go through each reaction and identify the acid, base, conjugate acid, and conjugate base.
1. H2SO4 + H2O -> H3O+ + HSO4-
This is the ionization of sulfuric acid (H2SO4) in water. In this reaction, H2SO4 donates a proton (H+) to H2O, which acts as the base. H2SO4 is the acid, H2O is the base, H3O+ is the conjugate acid formed after accepting the proton, and HSO4- is the conjugate base remaining.
2. KOH + H2O -> K+ + OH- + H2O
This reaction represents the dissociation of potassium hydroxide (KOH) in water. In this case, OH- accepts a proton from H2O, which acts as the acid. KOH is the base, H2O is the acid, K+ is the conjugate base formed after losing OH-, and OH- is the conjugate acid remaining.
3. CH3COOH + H2O -> CH3COO- + H3O+
Here, acetic acid (CH3COOH) reacts with water. CH3COOH donates a proton (H+) to H2O, which acts as the base. CH3COOH is the acid, H2O is the base, CH3COO- is the conjugate base formed after losing the proton, and H3O+ is the conjugate acid remaining.
4. NH3 + H2O -> NH2- + H3O+
This equation represents the reaction of ammonia (NH3) with water. NH3 accepts a proton (H+) from H2O, which acts as the acid. NH3 is the base, H2O is the acid, NH2- is the conjugate base formed after accepting the proton, and H3O+ is the conjugate acid formed.
5. HNO3 + H2O -> NO3- + H3O+
In this reaction, nitric acid (HNO3) reacts with water. HNO3 donates a proton (H+) to H2O, which acts as the base. HNO3 is the acid, H2O is the base, NO3- is the conjugate base formed after losing the proton, and H3O+ is the conjugate acid formed.
To write these equations, you need to identify which compound donates a proton (acid) and which compound accepts a proton (base). The acid loses a proton, forming its conjugate base, while the base accepts a proton, forming its conjugate acid. Water (H2O) can act as both an acid and a base, depending on the reaction. The hydronium ion (H3O+) is formed when a proton (H+) is added to water.