A solution of ammonia has a pH of 10.50. what is the initial molarity of ammonia? Kb for ammonia is 1.8x10^-5

5.55 x 10⁻³ mol/L.

Explanation:

∵ pH + pOH = 14.

∴ pOH = 14 - pH = 14 - 10.50 = 3.5.

∵ pOH = - log[OH⁻].

∴ - log[OH⁻] = - 3.5.

∴ [OH⁻] = 3.16 X 10⁻⁴.

∵ [OH⁻] = √ (kb) (c)

kb = 1.8 x 10⁻⁵.

c is the initial concentration.

3.16 X 10⁻⁴ = √ (1.8 x 10⁻⁵) (c).

by squaring both sides:

(3.16 X 10⁻⁴) ² = (1.8 x 10⁻⁵) (c).

1.0 x 10⁻⁷ = (1.8 x 10⁻⁵) (c)

∴ c = (1.0 x 10⁻⁷) / (1.8 x 10⁻⁵) = 5.55 x 10⁻³ mol/L.