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

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.