"The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb" for NH3

Kb = 1.77x10⁻⁵

Explanation:

When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

And the dissociation constant, Kb, for this equilibrium is:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

To find Kb you need to find the concentration of each species. The equilibrium concentrations are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate.

You can know [OH⁻] and, therefore, X, with pH of the solution, thus:

pH = -log [H⁺] = 11.612

[H⁺] = 2.4434x10⁻¹²

As 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]

4.0926x10⁻³ = [OH⁻] = X

Replacing, concentrations of the species are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

[NH₃] = 0.9459M

[NH₄⁺] = 4.0926x10⁻³M

[OH⁻] = 4.0926x10⁻³M

Replacing in Kb expression:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]

2b2t

explanation:

2b2t

the answer is c. nacl