You have an unknown concentration of ammonia. When you add 15.0mL of a 0.400 M HCl solution to 50.0mL of the ammonia solution, the pH of the solution becomes 8.75. What is the concentration of the original ammonia solution? [Kb (NH3) = 1.8 x 10^-5]

Answer 1

Use the Henderson-Hasselbalch equation.
15.0 mL x 0.400M = 6.0 millimols HCl
50 mL x ?M NH3 = ??

...........NH3 + HCl ==> NH4^ + Cl^-
...........x.......0.......0
add.............6.0.........
change...-6.0..-6.0.......6.0
equil....x-6.0...0.........6.0
pH = pKa + log (base)/(acid)
8.75 = 9.26 + log(x-6)/(6)
Solve for x = millimols NH3
M NH3 = millimols/50 mL

Answer 2

.1132 M @ equilibrium