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

Question

stoichiometric point of this titration, the [H+] is:

Answer 1

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.