Posted by Martha on .
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]

Chemistry  Equilibrium 
DrBob222,
Use the HendersonHasselbalch 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....x6.0...0.........6.0
pH = pKa + log (base)/(acid)
8.75 = 9.26 + log(x6)/(6)
Solve for x = millimols NH3
M NH3 = millimols/50 mL 
Chemistry  Equilibrium 
Shaniqua,
.1132 M @ equilibrium