Determine the pH of a 0.28 M solution of pyridinium nitrate (C5H5NHNO3) at 25°C. [Pyridinium nitrate dissociates in water to give pyridinium ions (C5H5NH+), the conjugate acid of pyridine (Kb = 1.7 * 10-9), and nitrate ions (NO3-).]

The pH is 2.89

Explanation:

Step 1: Dara given

Molarity of C5H5NHNO3 = 0.28 M

Kb = 1.7*10^-9

Step 2: Calculate Ka

Ka = Kw / Kb

Ka = (10^-14) / (1.7*10^-9)

Ka = 5.88 * 10^-6

C5H5NH + in water C5H5N & H+

Ka = [C5H5N] [H30+] / [C5H5NH+]

Step 3: The molarity

The initial molarity of C5H5NH + (aq) = 0.28M

The initial molarity of C5H5N and H3O + = 0M

The mole ratio is 1:1 so there will be consumed X of C5H5NH + and there will be produced x of C5H5N and H3O+

The molarity at equilibrium for C5H5NH + is (0.28 - X) M

The molarity for C5H5N and H3O + = X M

Step 4: Calculate the pH

Ka = [C5H5N] [H30+] / [C5H5NH+]

5.88*10^-6 = X*X / (0.28-X)

5.88*10^-6 * (0.28-X) = X²

X = [H3O+] = [H+] = 0.00128 M

pH = - log[H+] = - log (0.00128)

pH = 2.89