Answer:
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 pH of a 0.28 M solution of pyridinium nitrate (C5H5NHNO3) at 25°C is 2.89.
The pH of a solution can be calculated using the following procedure:
The data given in this question are as follows:
Next, we calculate Ka
Ka = Kw / Kb
Ka = (10-¹⁴)/(1.7 × 10-⁹)
Ka = 5.88 × 10-⁶
The molarity at equilibrium for C5H5NH+ is (0.28 - X)M
The molarity for C5H5N and H3O+ = X M
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
Therefore, the pH of a 0.28 M solution of pyridinium nitrate (C5H5NHNO3) at 25°C is 2.89.
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