The pH of a 0.55 M aqueous solution ammonia, NH₃, at 25.0 °C is 11.50. What is the value of K_b for NH₃?

Question

Quizplus · Accepted Answer

The equation for the dissociation of ammonia in water is:

NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant expression for this reaction is:

KS1U1B1bS1U1B0 = [NH4+][OH-] / [NH3]

To find the value of KS1U1B1bS1U1B0, we need to determine the concentrations of NH4+ and OH- in the solution.

Since the pH of the solution is 11.50, we can calculate the concentration of OH- as follows:

pH = -log[H+]

11.50 = -log[H+]

[H+] = 10-S1U1P11.50S1S1P = 3.16 × 10-S1U1P12S1S1P M

Since the solution is assumed to be a dilute solution of strong base, we can assume that [OH-] = [H+].

Therefore, [OH-] = 3.16 × 10-S1U1P12S1S1P M.

To calculate the concentration of NH4+, we can use the fact that ammonia is a weak base and thus undergoes partial dissociation in water. We can assume that the concentration of undissociated NH3 is equal to the initial concentration of ammonia, which is 0.55 M. We can then use the equation for the dissociation of ammonia to calculate the concentration of NH4+:

NH3 + H2O ⇌ NH4+ + OH-

Let x be the concentration of NH4+. Then, at equilibrium, [NH4+] = x and [OH-] = x.

The equilibrium constant expression becomes:

KS1U1B1bS1U1B0 = x2 / 0.55

Substituting the values of KS1U1B1bS1U1B0, [OH-], and [NH3]:

1.8 × 10-S1U1P15S1S1P0 = x2 / 0.55 × 3.16 × 10-S1U1P12S1S1P

Solving for x:

x = 1.8 × 10-S1U1P5.5S1S1P = 3.5 × 10-S1U1P3S1S1P M

Therefore, the concentration of NH4+ is 3.5 × 10-S1U1P3S1S1P M.

Finally, we can calculate KS1U1B1bS1U1B0 using the previously obtained values:

KS1U1B1bS1U1B0 = [NH4+][OH-] / [NH3] = (3.5 × 10-S1U1P3S1S1P) × (3.16 × 10-S1U1P12S1S1P) / 0.55 = 1.8 × 10-S1U1P5S1S1P0

Therefore, the answer is D) 1.8 × 10-S1U1P5S1S1P0.

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