Calculate the pH of a 3.5M solution of NH3 (Kb = 1.8 x 10-5 ).11.92.14.29.8

To calculate the pH of a 3.5M 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 these species at equilibrium.

Initial: [NH3] = 3.5 M, [NH4+] = 0, [OH-] = 0

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

Equilibrium: [NH3] = 3.5 - x, [NH4+] = x, [OH-] = x

Substituting the equilibrium concentrations into the Kb expression gives:

1.8 x 10^-5 = x^2 / (3.5 - x)

Assuming x is much smaller than 3.5 (which is a reasonable assumption given the small value of Kb), we can simplify this to:

1.8 x 10^-5 = x^2 / 3.5

Solving for x gives the concentration of OH- at equilibrium.

x = sqrt(1.8 x 10^-5 * 3.5) = 0.0025 M

The pOH of the solution is then -log[OH-] = -log(0.0025) = 2.60

The pH of the solution can be found using the relationship pH + pOH = 14.00 (at 25 degrees Celsius), so:

pH = 14.00 - 2.60 = 11.40

So, the pH of the 3.5M solution of NH3 is approximately 11.40.