Calculate the pH of a 1.5M solution of NH3 (Kb = 1.8 x 10-5 ).2.2811.724.579.43

To calculate the pH of a 1.5M solution of NH3, we first need to set up the equilibrium expression for the reaction of NH3 with water:

NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant expression for this reaction is:

Kb = [NH4+][OH-] / [NH3]

We can set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of the species at equilibrium.

Initial: [NH3] = 1.5 M, [NH4+] = 0, [OH-] = 0

Change: [NH3] = -x, [NH4+] = +x, [OH-] = +x

Equilibrium: [NH3] = 1.5 - x, [NH4+] = x, [OH-] = x

Substituting the equilibrium concentrations into the Kb expression gives:

1.8 x 10^-5 = x^2 / (1.5 - x)

Assuming x is much smaller than 1.5 (which is a valid assumption unless x is greater than about 5% of 1.5), we can simplify this to:

1.8 x 10^-5 = x^2 / 1.5

Solving for x gives:

x = sqrt(1.8 x 10^-5 * 1.5) = 0.0055 M

This is the concentration of OH- at equilibrium. We can then find the pOH of the solution using the formula:

pOH = -log[OH-] = -log(0.0055) = 2.26

Finally, we can find the pH of the solution using the formula:

pH = 14 - pOH = 14 - 2.26 = 11.74

So, the pH of the 1.5M solution of NH3 is approximately 11.74.