What is the concentration of ammonia in a solution prepared by dissolving 0.10 mol of acetic acid and 0.10 mol of ammonium chloride in enough water to make 1.0 L of solution?

Verified step by step guidance

1. Identify the relevant chemical reaction: Ammonium chloride (NH4Cl) dissociates in water to form ammonium ions (NH4+) and chloride ions (Cl-).
2. Recognize that ammonium ions (NH4+) can react with water to form ammonia (NH3) and hydronium ions (H3O+). This is an example of an acid-base equilibrium.
3. Write the equilibrium expression for the reaction: NH4+ + H2O ⇌ NH3 + H3O+. The equilibrium constant expression (Ka) for this reaction is Ka = [NH3][H3O+]/[NH4+].
4. Use the initial concentrations: Initially, [NH4+] = 0.10 M and [NH3] = 0 M. Assume x is the change in concentration at equilibrium, so [NH3] = x and [NH4+] = 0.10 - x.
5. Substitute the equilibrium concentrations into the Ka expression and solve for x, which represents the concentration of ammonia (NH3) at equilibrium.