To calculate the base dissociation constant, Kb, for NH3(aq) at 25°C, we can use the equation:
Kw = Ka * Kb
where Kw is the ionization constant of water (1.0 x 10^-14 at 25°C), and Ka is the acid dissociation constant of NH4+.
1. First, let's find the pOH of the solution using the measured pH.
pOH = 14 - pH
pOH = 14 - 11.12
pOH = 2.88
2. Next, let's convert pOH back to OH- concentration using the equation:
pOH = -log[OH-]
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.88)
[OH-] = 1.4 x 10^(-3) M
3. Since NH3 is a weak base, it reacts with water to form OH- ions. In the equilibrium equation:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
The concentration of OH- is given as 1.4 x 10^(-3) M.
The concentration of NH3 and NH4+ are equal to the initial concentration, which is 0.100 M.
4. Now, we can set up an ICE table to determine the equilibrium concentrations:
Initial:
NH3(aq): 0.100 M
NH4+(aq): 0 M
OH-(aq): 1.4 x 10^(-3) M
Change:
NH3(aq): -x
NH4+(aq): +x
OH-(aq): +x
Equilibrium:
NH3(aq): 0.100 - x
NH4+(aq): x
OH-(aq): 1.4 x 10^(-3) + x
5. Since Kb is equal to [NH4+][OH-] / [NH3], we need to determine the equilibrium concentrations.
Kb = ([NH4+][OH-]) / [NH3]
= (x) * (1.4 x 10^(-3) + x) / (0.100 - x)
6. We can assume that the x value (concentration of NH4+ and OH-) is small compared to 0.100 M (concentration of NH3) and can be ignored when subtracted.
Thus, we can simplify the equation to:
Kb = x * (1.4 x 10^(-3)) / 0.100
7. To solve for x, we need to assume that x is small enough that we can neglect it in the calculations.
Since x represents the concentration of NH4+ and OH-, let's assume that x is negligible compared to 0.100 M.
Therefore, we can simplify the equation further:
Kb = (1.4 x 10^(-3)) / 0.100
8. Finally, we can calculate Kb:
Kb = (1.4 x 10^(-3)) / 0.100
Kb ≈ 1.4 x 10^(-2)
Thus, the base dissociation constant, Kb, for NH3(aq) at 25°C is approximately 1.4 x 10^(-2).