Question

a. What volume of 0.085 M HCl is required to titrate 25.00 mL of a 0.100 M NH3 solution to the equivalence point?

b. What is the pH at the equivalence point?

Answer 1

a)
at equivalence point,
mol of HCl = mol of NH3
0.085 M * V = 0.100 M * 25.00 mL
V= 29.41 mL

b)
both NH3 and HCl will react to form NH4+
mol of NH4+ = mol of NH3 reacted= 0.100 M * 25 mL = 2.5 mmol

total volume = 25.00 + 29.41 = 54.41 mL

[NH4+] = number of mol / volume
= 2.5 mmol / 54.41 mL
=0.0459 M

Kb of NH3 = 1.8*10^-5 M

Ka = 10^-14/Kb
   = 10^-14/ (1.8*10^-5)
   = 5.566*10^-10

for simplicity lets write acid ion as BH+

BH+      + H2O ----->     BOH   +   H+
0.0459                    0         0
0.0459-x                  x         x

Ka = [H+][BOH]/[BH+]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.566E-10)*0.0459) = 5.05E-6

pH = -log [H+] = -log (5.05E-6) = 5.30

Answer: 5.30