Kb for NH3 is 1.8 × 10-5. What is the pOH of a 0.15 M aqueous solution of NH4Cl at 25.0 °C?

To calculate the pOH of a 0.15 M aqueous solution of NH4Cl using the Kb for NH3 (1.8 × 10^-5), we first need to recognize that ammonium chloride (NH4Cl) dissociates in water as follows:

NH4Cl → NH4+ + Cl-

The ammonium ion (NH4+) acts as an acid, donating a proton (H+) and forming ammonia (NH3). Since ammonia is a weak base, its ionization can be represented by:

NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant for this reaction is the base dissociation constant (K_b):

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

Given that K_b = 1.8 × 10^-5 and we have a 0.15 M NH4Cl solution, initially, there would be no NH3 or OH-. When the ammonium ions react with water, some NH3 will form.

We can use the relationship between pH, pOH, and Ka or Kb:

pH + pOH = 14 (for neutral solutions)

Since we're interested in pOH, we'll solve for it:

pOH = 14 - pH

First, let's determine the pH. For a strong acid like HCl, the hydronium ion concentration ([H3O+]) would equal the initial concentration of NH4Cl (0.15 M) since NH4+ is essentially a spectator ion in this case. Therefore:

[H3O+] = 0.15 M

pH = -log[H3O+] = -log(0.15)

Now we can find pOH:

pOH = 14 - (-log(0.15))

Let's calculate these values.

pH = -log(0.15)
pOH = 14 - pH

Once we have the pH, we can then use the Kb value to find the actual pOH for the NH4Cl solution.