Determine the pH of 0.15M of ammonia with a kb of 1.8x10^-5

Answer 1

To determine the pH of the ammonia solution, we need to first calculate the concentration of hydroxide ions in the solution.

Ammonia, NH3, reacts with water to form the ammonium ion NH4+ and hydroxide ions OH- according to the following equilibrium reaction:

NH3 + H2O → NH4+ + OH-

The equilibrium constant for this reaction is the base dissociation constant (Kb) for ammonia, which is given as 1.8x10^-5.

Setting up an ICE table for the reaction:

NH3 + H2O ⇌ NH4+ + OH-
Initial: 0.15M 0 0
Change: -x +x +x
Equilibrium: 0.15-x x x

Using the Kb expression:
Kb = [NH4+][OH-] / [NH3]
1.8x10^-5 = x^2 / 0.15

Solving for x:
x = sqrt(1.8x10^-5 * 0.15) = 1.1x10^-3

Now, the concentration of OH- ions is x = 1.1x10^-3 M. To find the pOH:
pOH = -log(OH-) = -log(1.1x10^-3) = 2.96

Finally, to find the pH:
pH = 14 - pOH = 14 - 2.96 = 11.04

Therefore, the pH of a 0.15M ammonia solution is approximately 11.04.