Answer to Question #101863 in Chemistry for Julie ann

Question

The conjugate of ammonia NH4+, is a weak acid. If a 0.200 M NH4Cl solution has a pH of 5.0, what is the Ka of NH4+?

Answer 1

NH₄Cl dissociates as:

NH₄Cl <--> NH4⁺ + Cl^-

From here:

Ka = ([NH4⁺][Cl^-])/[NH₄Cl] = [NH4⁺]²/[NH₄Cl]

As pH = -log[H⁺]:

[H⁺] = 10^-pH

From here, the concentration of [NH4⁺] in the solution equals:

[H⁺] = 10^-5 M

As [H⁺] = [NH4⁺]= 10^-5 M, [NH₄Cl] = 0.200 M, Ka of NH4⁺ equals:

Ka = [NH₄⁺]²/[NH₄Cl] = (10^-5)² / 0.2 = 5 × 10^-10

Answer: 5 × 10^-10