NH4+ + OH- NH3 + H2O
What is the K for the reaction above in terms of Ka's, Kb's, Kw etc.?
NH4+ OH-=NH3+H20
To find the K value for the given reaction NH4+ + OH- -> NH3 + H2O, we need to relate it to the acid-base equilibrium constants, Ka and Kb.
1. Define the equilibrium constants:
- Ka: Acid dissociation constant, relating the concentration of the acid and its conjugate base in a solution.
- Kb: Base dissociation constant, relating the concentration of the base and its conjugate acid in a solution.
- Kw: Ion product of water, representing the autoprotolysis of water, in which water acts as both an acid and a base.
2. Write the chemical equation as an acid-base reaction:
NH4+ (acid) + OH- (base) -> NH3 (base) + H2O (acid)
3. Express the acid and base in terms of their respective conjugate acid/base pairs:
NH4+ (acid) + H2O (base) -> NH3 (base) + H3O+ (acid)
4. Now, we can relate the dissociation constants for this reaction using Kw:
Ka * Kb = Kw
5. Rearrange the equation to solve for K:
Ka = Kw / Kb
6. Substitute the known value of Kw at 25°C (298K), which is 1.0 x 10^-14 mol^2/L^2:
Ka = (1.0 x 10^-14 mol^2/L^2) / Kb
So, the K value for the given reaction NH4+ + OH- -> NH3 + H2O in terms of Ka's, Kb's, and Kw is Ka = (1.0 x 10^-14 mol^2/L^2) / Kb.
You don't have an arrow. I can't tell where the reactants stop and products begin.
For the reaction,
Keq = (NH3)/(NH4^+)(OH^-) = 1/Kb for NH3