What is the pH of a solution resulting from the addition of 10.0 mL of 0.17 M HCl to 50.0 mL of 0.16 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 ✕ 10-5.
millimols HCl = mL x M = 10 x 0.17 = 1.7
mmols NH3 = 50 x 0.16 = 8
..... HCl + NH3 ==> NH4Cl
I.....1.7....8.......0
C....-1.7..-1.7......1.7
E......0....6.3.....1.7
Therefore, the resulting solution consists of 6.3 mmols NH3 and 1.7 mmols NH4^+ in 60 mL.
(NH3) = mmols/mL = 6.3/60 = approx 0.1 but you need to do it more accurately.
(NH4^+) = 1.7/60 = 0.03 (approximately)
........NH3 + H2O --> NH4^+ + OH^-
I.......0.1.........0.03.......0
C.......-x.............x.......x
E......0.1-x.......0.03+x......x
Kb = (NH4^+)(OH^-)/(NH3)
Substitute the E line into the Kb expression and solve for x = OH^- and
convert to pH.