Calculate the [OH ‾ ], [H+], and the pH of 0.16 M solutions of each of the following amines.
(a) triethylamine [(C2H5)3N, Kb = 4.0 10-4]
(b) hydroxylamine (HONH2, Kb = 1.1 10-8)
Let's represent (C2H5O3N as B3N.
.............B3N + HOH ==> B3NH^+ + OH^-
initial......0.16...........0........0
change.......-x.............x........x
equil......0.16-x...........x........x
Kb = (B3BH^+)(OH^-)/(B3N)
Substitute Kb and from the ICE chart and solve for x = (OH^-). From that you can convert to pOH and to pH.
The second one is done the same way.
To calculate the [OH ‾ ], [H+], and the pH of a solution of these amines, we need to consider the acidity and basicity of the compounds.
(a) Triethylamine [(C2H5)3N, Kb = 4.0 x 10^-4]:
- Step 1: Identify the Kb of triethylamine. Kb is the base dissociation constant, which tells us the strength of the base. In this case, Kb = 4.0 x 10^-4.
- Step 2: Write the chemical equation for the reaction of the amine with water:
(C2H5)3N + H2O ⇌ (C2H5)3NH+ + OH ‾
- Step 3: Identify the initial concentration of the amine. In this case, it is 0.16 M.
- Step 4: Set up an ICE table to calculate the equilibrium concentrations. This table helps us track how the reactants and products change as the reaction reaches equilibrium.
Initial: (C2H5)3N: 0.16 M H2O: -
Change: (C2H5)3N: -x H2O: -x (C2H5)3NH+: +x OH ‾: +x
Equilibrium: (C2H5)3N: (0.16 - x) (C2H5)3NH+: x OH ‾: x
- Step 5: Write the expression for Kb and set up the equation:
Kb = ([C2H5)3NH+][OH ‾]) / [(C2H5)3N]
Kb = x * x / (0.16 - x)
- Step 6: Since the value of x, which represents the equilibrium concentration of [OH ‾], will be relatively small compared to 0.16 (initial concentration of (C2H5)3N), we can assume that the value of 0.16 - x is approximately equal to 0.16. Therefore, we can simplify the equation:
Kb = x^2 / 0.16 => x^2 = 0.16 * Kb
- Step 7: Solve for x:
x^2 = 0.16 * 4.0 x 10^-4
x^2 = 6.4 x 10^-5
x ≈ √(6.4 x 10^-5) ≈ 0.008
- Step 8: Calculate [OH ‾], [H+], and pH:
[OH ‾] = 0.008 M
To calculate [H+], we can use the expression Kw = [H+][OH ‾]:
Kw = [H+][OH ‾]
Kw = (1.0 x 10^-14) = [H+] * 0.008
[H+] = 1.0 x 10^-14 / 0.008 ≈ 1.25 x 10^-12 M
pH = -log([H+]) = -log(1.25 x 10^-12) ≈ 11.9
(b) Hydroxylamine (HONH2, Kb = 1.1 x 10^-8):
To calculate the [OH ‾ ], [H+], and the pH of hydroxylamine, we follow the same steps as above, but this time consider the Kb of hydroxylamine and adjust the equation accordingly.
Note: The pOH can be calculated using pOH = -log[OH ‾]. If you want to calculate the pOH, you can subtract the calculated pH value from 14.
Remember, these calculations assume ideal conditions and complete dissociation, so they may not be accurate in all cases.