Question

Ammonia is a weak base with a Kb of 1.8 x 10-5. A 100.0 mL sample of 0.13 M aqueous solution of ammonia (NH3) is mixed with 87 mL of 0.15 M solution of the strong acid HCl, calculate the pH of the final solution.
A) 2.9
B) 11.1
C) 8.8
D) 5.2
E) None of the above

Answer 1

millimols NH3 = mL x M = 100 x 0.1 = 13
mmols HCl = 87 x 0.15 = 13.05
My best guess is that these are to be considered the same since the 87 has two s.f. and 13 does too.
........NH3 + HCl ---> NH4Cl
I.......13....13........0
C......-13...-13.......+13
E........0.....0.........0

I'm assuming the HCl exactly neutralizes the NH3 in which case the pH is determined by the hydrolysis of the salt, NH4Cl, which has a concn of 13mmols/187 mL = about 0.07 and you need to clear up that number a little as well as all that folllow.
........NH4^+ + H2O ==> NH3 + H3O^+
I.......0.07............0......0
C.......-x..............x......x
E......0.07-x...........x......x

Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.07-x) and solve for x = (H3O^+ and convert to pH. NH4^+ salt have a pH of about 5.2