Determine the ammonia concentration of an aqueous solution that has a pH of 11.00. The equation for the dissociation of NH₃ (K_b= 1.8 × 10^-5)is below: A)3.0 mol L^-1 B)0.056 mol L^-1 C)1.8 × 10^-2 mol L^-1 D)1.0 × 10^-3 mol L^-1

Question

Quizplus · Accepted Answer

Ammonia (NH3) is a weak base that reacts with water to produce NH4+ and OH-. The equilibrium expression for this reaction is:
NH3 + H2O ⇌ NH4+ + OH-
The dissociation constant (Kb) for NH3 is:
Kb = [NH4+][OH-]/[NH3]
In the given problem, the pH of the solution is 11.00. This means that [OH-] = 10^-3 mol/L (since pH + pOH = 14).
Using the above equilibrium expression and the given Kb value, we can solve for [NH3]:
Kb = [NH4+][OH-]/[NH3]
1.8 × 10^-5 = [NH4+][10^-3]/[NH3]
[NH3] = [NH4+][10^-3]/(1.8 × 10^-5)
At equilibrium, the concentration of NH4+ and NH3 are equal, so we can write:
[NH3] = [NH4+] = x (say)
Substituting in the above equation:
x = x * 10^-3/(1.8 × 10^-5)
x = 0.056 mol/L
Therefore, the ammonia concentration of the solution is 0.056 mol/L. Hence, the correct choice is B.

Determine the Ammonia Concentration of an Aqueous Solution That Has