Calculate the pH of each of the following solutions.
(a) 0.103 M HONH2 (Kb = 1.1 ✕ 10-8)
(b) 0.103 M HONH3Cl
(c) pure H2O
(d) a mixture containing 0.103 M HONH2 and 0.103 M HONH3C
(a) To calculate the pH of a weak base solution like HONH2, we first need to find the concentration of hydroxide ions (OH-).
HONH2 is a weak base, so it will partially dissociate in water according to the balanced chemical equation:
HONH2 + H2O ⇌ NH3OH+ + OH-
The Kb value for HONH2 is given as 1.1 x 10^-8. We can use this value to find the concentration of OH- ions.
Kb = [NH3OH+][OH-] / [HONH2]
Assuming the dissociation of HONH2 is small relative to its initial concentration (which is true for a 0.103 M solution), we can use x as the concentration of OH- ions formed.
Hence, Kb = x^2 / (0.103 - x)
Since x is much smaller compared to 0.103, we can ignore the -x term in the denominator.
Kb = x^2 / (0.103)
Rearrange the equation to solve for x:
x^2 = Kb * 0.103
x^2 = (1.1 x 10^-8)(0.103)
x^2 = 1.133 x 10^-9
Taking the square root of both sides:
x = √(1.133 x 10^-9)
x = 1.065 x 10^-5 M
The concentration of OH- ions is 1.065 x 10^-5 M.
To calculate the pH, we can use the equation:
pOH = -log[OH-]
pOH = -log(1.065 x 10^-5)
pOH ≈ 4.97
The pH of the solution can be found using the equation:
pH = 14 - pOH
pH = 14 - 4.97
pH ≈ 9.03
Therefore, the pH of the 0.103 M HONH2 solution is approximately 9.03.
(b) HONH3Cl is the conjugate acid of HONH2. We can calculate the pH of this solution using the same steps as above, except we need to consider that HONH3Cl will partially dissociate as a weak acid.
HONH3Cl + H2O ⇌ H3O+ + Cl-
The Kb value for HONH2 is given as 1.1 x 10^-8. We can use this value to find the concentration of H3O+ ions.
Kw = [H3O+][OH-]
Kw = (1.0 x 10^-14) = [H3O+][OH-]
[H3O+] = [OH-] = x (assume the dissociation of HONH3Cl is small relative to its initial concentration)
Kb = Kw / Ka (Ka is the acid dissociation constant)
1.1 x 10^-8 = (1.0 x 10^-14) / Ka
Ka = (1.0 x 10^-14) / (1.1 x 10^-8)
Ka ≈ 9.1 x 10^-7
Since [H3O+] = Ka, the concentration of H3O+ ions is 9.1 x 10^-7 M.
To calculate the pH, we can use the equation:
pH = -log[H3O+]
pH = -log(9.1 x 10^-7)
pH ≈ 6.04
Therefore, the pH of the 0.103 M HONH3Cl solution is approximately 6.04.
(c) Pure H2O is neutral, so its pH is 7.
(d) To calculate the pH of a mixture containing both HONH2 and HONH3Cl, we need to consider the concentrations of both substances and their respective acid-base properties. The pH of the resulting solution will depend on the relative concentrations and the strengths of their conjugate acids and bases.
In this case, we have equal concentrations (0.103 M) of HONH2 and HONH3Cl, and their conjugate acid-base pairs have similar strengths. Thus, the resulting pH will be close to neutral, around 7.
Therefore, the pH of the mixture containing 0.103 M HONH2 and 0.103 M HONH3Cl is around 7.