Consider the titration of 100.0 mL of 0.500 M NH3 (Kb = 1.8 x 10-5) with 0.500 M HCl. At the
stoichiometric point of this titration, the [H+] is:
NH3 + HCl ==> NH4Cl
(NH4Cl) at eq. point = 0.500 x (100/200) = 0.25M
The pH is determined by the hydrolysis of the salt, NH4Cl.
.......NH4^+ + H2O ==> H3O^+ + OH^-
initial.0.25M..........0........0
change...-x.............x.......x
equil....0.25-x.........x........x
Ka for NH4^+ = (Kw/Kb for NH3) = (H3O^+)(OH^-)/(NH4^+)
Substitute into the equil equation from the equil line in the ICE chart and solve for x = (OH^-) and convert to pH.
To determine the concentration of [H+] at the stoichiometric point of this titration, we need to consider the reaction that takes place between NH3 and HCl.
The balanced chemical equation for the reaction is:
NH3 + HCl -> NH4+ + Cl-
In this reaction, NH3 (ammonia) acts as a base, and HCl (hydrochloric acid) acts as an acid.
At the stoichiometric point of the titration, the number of moles of NH3 and HCl reacting are equivalent. This means that all of the NH3 has reacted with HCl, resulting in the formation of NH4+ and Cl- ions.
Since the reaction is 1:1 between NH3 and HCl, the concentration of NH4+ ions formed at the stoichiometric point will be equal to the initial concentration of NH3. In this case, the concentration of NH3 is given as 0.500 M.
Now, to find the [H+], we need to understand that NH4+ is the conjugate acid of NH3. So, when NH3 reacts with HCl, NH4+ acts as an acid and donates a proton (H+) to the solution.
Since NH4+ is a weak acid, we can assume that it only partially dissociates. At equilibrium, we can use the equilibrium constant expression for the dissociation of NH4+:
Kb = [NH4+][OH-] / [NH3]
Since we are interested in [H+], we can use the relationship that [OH-] * [H+] = Kw (where Kw is the ion product constant for water) to rewrite the expression:
Kb = [NH4+][H+] / [NH3] = Kw / [OH-]
Rearranging the equation to solve for [H+], we get:
[H+] = (Kw * [NH3]) / Kb
Given that Kb = 1.8 x 10^(-5), we can use this value along with the concentration of NH3 (0.500 M) and the value of Kw (1.0 x 10^(-14)) to calculate the [H+].
[H+] = (1.0 x 10^(-14) * 0.500) / (1.8 x 10^(-5))
Calculating this expression will give you the concentration of [H+] at the stoichiometric point of the titration.