In one experiment, 50.0ml of 0.10M CH3NH2, is mixed with 20.00ml of 0.10 M CH3NH3Cl. The kb of CH3NH2 is 3.70*10^-4.

a) write the chemical reaction for the equilibrium which becomes established, using H2O as a reactant.
b)Write the net-ionic equation that occurs when Ba(OH)2 is added to the mixture.

For A, would the equilibrium equation be H2O(l)+CH3NH2(aq)-->H30+(aq) + CH3NH-(aq)?

Please clarify and help with B.

What happens when we add NH3 to H2O?

NH3 + HOH ==< NH4^+ + OH^- and I'm sure you have seen this reaction many times. So if we replace one of the H atoms on NH3 with a methyl group (CH3), wouldn't the reaction be analogous?
CH3NH2 + HOH ==> CH3NH3^+ + OH^-

For part b, you can look at it two ways. One way is that adding Ba(OH)2 adds the common ion, OH^-, and by Le Chatelier's principle, that shifts the equilibrium to the left producing H2O and CH3NH3. The second way is to look at it as an acid/base reaction.
CH3NH3^+(acid) + Ba(OH)2(base) ==>

CH3NH3^+ + OH^- ==> CH3NH2 + HOH.
I suspect that Ba(OH)2 is a strong enough base that CH3NH2 could be broken up and NH3 might be evolved.

a) The chemical reaction for the equilibrium that becomes established can be represented as follows:

CH3NH2 (aq) + H2O (l) ⇌ CH3NH3+ (aq) + OH- (aq)

In this reaction, CH3NH2 acts as a weak base and accepts a proton (H+) from water (H2O), resulting in the formation of CH3NH3+ (the conjugate acid) and OH- (the hydroxide ion).

b) When Ba(OH)2 is added to the mixture, it will react with the OH- ions present in solution. The net-ionic equation representing this reaction is:

OH- (aq) + Ba2+ (aq) ⟶ Ba(OH)2 (s)

In this reaction, the hydroxide ion (OH-) reacts with the barium cation (Ba2+), forming solid barium hydroxide (Ba(OH)2). No additional spectator ions are involved in this net ionic equation.

For part A, the equilibrium equation you provided is incorrect. The reaction between CH3NH2 and H2O as reactants would not produce H3O+ and CH3NH- ions.

The correct equilibrium equation for the reaction of CH3NH2 (methylamine) and H2O (water) is:

CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)

In this reaction, water reacts with methylamine to form methylammonium cation (CH3NH3+) and hydroxide ion (OH-). This equilibrium occurs because methylamine (CH3NH2) is a weak base and reacts with water.

Moving on to part B, you are asked to write the net ionic equation that occurs when Ba(OH)2 (barium hydroxide) is added to the mixture.

The main reaction that occurs when Ba(OH)2 is added to the mixture is the neutralization reaction between the hydroxide ion (OH-) of Ba(OH)2 and the methylammonium cation (CH3NH3+) from part A. This reaction can be represented as:

CH3NH3+(aq) + OH-(aq) -> CH3NH2(aq) + H2O(l)

In this reaction, the hydroxide ion from Ba(OH)2 reacts with the methylammonium cation to form methylamine (CH3NH2) and water. This is a complete neutralization reaction.

The net ionic equation removes the spectator ions (ions that do not participate in the reaction) from the balanced equation. In this case, the spectator ions are Ba2+ and Cl-. The net ionic equation for the reaction is:

OH-(aq) + CH3NH3+(aq) -> CH3NH2(aq) + H2O(l)

This equation represents the actual species involved in the reaction, which is the hydroxide ion and the methylammonium cation.

Keep in mind that the net ionic equation only includes the species that are actually reacting and omits spectator ions.