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
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

Question

Quizplus · Accepted Answer

The pH of the solution is 11.00, which correspond to a pOH of 3.00 (since pH + pOH = 14).

Using the equation given, we can find the concentration of OH- ions:

NHS1U1B13S1U1B0(aq) + HS1U1B12S1U1B0O(l) ⇌ NHS1U1B14S1U1B0S1U1P1+ S1S1P0(aq) + OHS1U1P1-S1S1P0(aq)

Since this is a weak base equilibrium, we can make the assumption that [OH-] is equal to the concentration of NHS1U1B14S1U1B0 ions formed, since the dissociation of water can be neglected.

So, we can set up the equilibrium expression:

KS1U1B1b S1U1B0 = [NHS1U1B14S1U1B0][OHS1U1P1-S1S1P0] / [NHS1U1B13S1U1B0]

We are given that KS1U1B1b S1U1B0 = 1.8 × 10S1U1P1-5S1S1P0 and that the concentration of NHS1U1B13S1U1B0 is equal to the concentration of NHS1U1B14S1U1B0 that dissociated. Let's call this concentration x:

1.8 × 10S1U1P1-5S1S1P0 = x^2 / (1-x)

Solving for x using the quadratic formula, we get:

x = 0.0377 M

This is the concentration of [NHS1U1B14S1U1B0] and [OH-].

To find the concentration of NH3, we can use the fact that [NH3][OH-] = KS1U1B7b S1U1B0 = 1.8 × 10S1U1P1-5S1S1P0.

Plugging in the values we know, we get:

[NH3] = KS1U1B7b S1U1B0 / [OH-] = 1.8 × 10S1U1P1-5S1S1P0 / 0.0377 M = 0.056 M

Therefore, the answer is B.

Determine the Ammonia Concentration of an Aqueous Solution That Has