Asolution 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.

i think slash and

it is stable as it is in group 0 which have full outer shells of electrons