Calculate ⌈H3O +⌉ ,[OH −] and [NH4 +] for 0.10MNH3 solution where
1.3% of NH3 dissociates.
To calculate the concentration of H3O+, [OH-], and [NH4+] in a 0.10M NH3 solution where 1.3% of NH3 dissociates, you need to determine the equilibrium concentrations using an ICE (Initial, Change, Equilibrium) table.
Let x represent the amount dissociated (in moles). Since 1.3% of NH3 dissociates, x = 0.013 * 0.10M = 0.0013M.
The dissociation reaction of NH3 is as follows:
NH3 + H2O ⇌ NH4+ + OH-
Using the equilibrium concentrations, we can set up the ICE table:
NH3 + H2O ⇌ NH4+ + OH-
Initial 0.10M 0M 0M 0M
Change -0.0013M -0.0013M +0.0013M +0.0013M
Equilibrium 0.10 - 0.0013M 0 - 0.0013M 0.0013M 0.0013M
Since NH3 is a weak base, it reacts with water to produce equal amounts of NH4+ and OH- ions.
Thus, the equilibrium concentration of [NH4+] and [OH-] is 0.0013M.
The concentration of H3O+ is equal to the concentration of [NH4+], so [H3O+] = 0.0013M.
Therefore, the calculated concentrations are:
[H3O+] = [NH4+] = 0.0013M
[OH-] = 0.0013M