A 1M solution (200ml) of NH3 (Kb of ammonia=1.8*10-5) is added to 200ml of 0.5M HCl. Calculate:

Question

a) the resulting pH?
b) The delta ph due to addition of 15ml of 1M HNO3 to solution obtained in part a
c) I am stuck on this:
The delta ph due to addition of 15ml of 1M NaOH to the solution obtained in part a

Answer 1

NH3 + HCl --> NH4Cl + H2O
millimols NH3 = 200 x 1 = 200
mmols HCl added = 200 x 0.5 = 100
So you have 100 mmols NH3 in 400 mL solution. I assume from your post that you worked the remainder of part a.

For part b we have the soln from part a and add 15 mmols (15x1) NaOH.
So the OH^- in the soln consists of the OH^- from the ionization of NH3 plus the OH^- from the NaOH. You should note that the OH^- from NaOH will suppress the ionization of NH3.
The OH^- from NaOH = 15 mmols/415 mL = approx 0.0361 but you should confirm that and work it more accurately if needed.
To this you need to add the OH^- from the NH3. I suspect this is negligible (but not zero) but you must go through and see.
NH3 + H2O ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
(NH4^+) = x
(OH^-) = x + 0.0361 and you assume x is small so OH^- is 0.0361M
(NH3) = 100/415 = about 0.241M
Solve for x = about 1E-4 which shows x is small in comparison to the OH^- from NaOH. Calculate pH from this and then delta pH.