A solution of 25.0mL of 0.10mol/L hydroiodic acid is titrated with
0.10mol/L ammonia.Determine the pH at equivalence.
HCl + NH3 ==> NH4Cl
millimols HCl = mL x M = 25 x 0.1 = 2.5
mmols NH3 = 2.5 so volume must be 25.0 mL.
Total volume - 50 mL.
M NH3 = mmols/mL = 2.5/50 = 0.05 M.
The pH at equivalence point is determined by the hydrolysis of the NH4Cl.
.......NH4^+ + H2O ==> NH3 + H3O^+
I.....0.05.............0......0
C......-x..............x......x
E....0.05-x............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.05-x)
Solve for x = (H3O^+) and convert to pH.