What is the pH of a 0.200 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

Question

What is the pH of a 0.200 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is NH₃(aq)+ H₂O(l)⇌ NH₄⁺(aq)+ OH^-(aq). A)2)02 B)2)72 C)11.98 D)11.28

Quizplus · Accepted Answer

First, we need to set up the expression for the dissociation constant, which is:
KS1U1B1bS1U1B0 = [NHS1U1B14S1U1B0S1U1P1+][OHS1U1P1-S1S1P0]/[NHS1U1B13S1U1B0]
Next, we can assume that the concentration of [OHS1U1P1-S1S1P0] can be neglected compared to the concentration of [NHS1U1B13S1U1B0], since the dissociation of NHS1U1B13S1U1B0 will be small compared to its initial concentration. This allows us to simplify the expression to:
KS1U1B1bS1U1B0 = [NHS1U1B14S1U1B0S1U1P1+]/[NHS1U1B13S1U1B0]
Next, we need to set up an ICE table to determine the concentration of [NHS1U1B14S1U1B0S1U1P1+]. Since NHS1U1B13S1U1B0 is a weak acid, we can use the following assumptions:

[NHS1U1B14S1U1B0S1U1P1+] = [H+] (from the dissociation equation)
[HS1U1B12S1U1B0O] = [NHS1U1B13S1U1B0] - [H+] (from the law of conservation of mass)

Using these assumptions, we can set up the ICE table as follows:

NHS1U1B13S1U1B0   H+   NHS1U1B14S1U1B0S1U1P1+
I        0.200            0    0
C       -x               +x   +x
E        0.200-x         x    x

Plugging in the equilibrium concentrations into the expression for the dissociation constant, we get:

1.8×10S1U1P1-5S1S1P0 = x^2 / (0.200 - x)

Since x is small compared to 0.200, we can make the approximation that 0.200 - x ≈ 0.200. This allows us to simplify the expression to:

1.8×10S1U1P1-5S1S1P0 = x^2 / 0.200

Solving for x gives us:

x = 2.83×10S1U1P1-4S1S1P0

Therefore, the pH of the solution is:

pH = -log[H+] = -log(x) = 11.28

So the best choice is D.

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