NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.049 M in NH4Cl at 25 °C?
.........NH4^+ + H2O ==> H3O^+ + NH3
I......0.049..............0.......0
C........-x...............x.......x
E.....0.049-x.............x.......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.049-x)
Solve for x = (H3O^+) and convert to pH.
To determine the pH of a solution containing NH4Cl, you need to consider the dissociation of NH4Cl and the subsequent reactions of NH4+ and Cl- ions with water.
NH4Cl dissociates into NH4+ and Cl- ions in water:
NH4Cl (s) -> NH4+ (aq) + Cl- (aq)
Since NH4Cl acts as a weak acid, NH4+ can react with water to produce H3O+ (hydronium) ions:
NH4+ (aq) + H2O (l) -> NH3 (aq) + H3O+ (aq)
The concentration of NH4Cl is given as 0.049 M, which means the concentration of NH4+ ions is also 0.049 M.
Since NH3 is a weak base, it can react with H3O+ ions:
NH3 (aq) + H3O+ (aq) -> NH4+ (aq) + H2O (l)
Given the Kb value of NH3 as 1.8 × 10^-5, we can observe that it's small, indicating that NH3 will not completely react with H3O+ ions. Therefore, we can assume that the concentration of NH4+ ions formed from NH3 and H3O+ reaction is negligible compared to the initial concentration of NH4+ ions.
Hence, we can consider the concentration of H3O+ ions to be equal to the concentration of NH4+ ions in the solution. Therefore, [H3O+] = 0.049 M.
The pH of a solution can be calculated using the equation:
pH = -log[H3O+]
By substituting the concentration of H3O+ ions into the equation, we get:
pH = -log(0.049)
Calculating this value gives us:
pH ≈ 1.31
Therefore, the pH of the solution containing 0.049 M NH4Cl at 25 °C is approximately 1.31.