Final answer:
The pH of a 0.25 M solution of ammonium chloride is acidic and will be less than 7 because the ammonium ion reacts with water to form hydronium ions. This is calculated using the known K₁ value for the ammonium ion and applying an ICE table to determine the concentration of hydronium ions formed.
Step-by-step explanation:
To calculate the pH of a 0.25 M solution of ammonium chloride, we need to consider the ionization of ammonium chloride in water. Ammonium chloride (NH4Cl) is a salt derived from a weak base (NH3) and a strong acid (HCl). When NH4Cl dissolves in water, it produces NH4+ and Cl- ions. The NH4+ ion will then react with water to produce H3O+ and NH3:
NH4+ (aq) + H2O(l) → H3O+ (aq) + NH3 (aq)
Given the Ka value for NH4+ is 5.6 × 10-10, we can set up an ICE table to solve for the concentration of H3O+ produced. Assume x is the concentration of H3O+ produced, then at equilibrium, we would have [NH4+] = 0.25 - x, [H3O+] = x, and [NH3] = x. The equation for Ka will look like this:
Ka = [H3O+][NH3]/[NH4+] = (x)(x)/(0.25 - x)
Assuming x << 0.25, we can simplify the equation to:
Ka = x2/0.25 and solve for x: x = √(Ka × 0.25)
Then, we calculate the pH using the pH formula: pH = -log[H3O+]
Therefore, the pH will be less than 7, which corresponds to option (a) since the solution is acidic due to the formation of H3O+.