A solution is prepared using ammonia and ammonium chloride. The concentration of the ammonia equals the fourth and fifth digits of your Student ID Number divided by 100. The concentration of the ammonium chloride equals the first two digits of your Student ID Number divided by 100. Find the pH of the solution.
A typical equilibrium problem: write the relevant reactions, write the mass action expression, set up an ICE table, and then solve for the unknown.
Let [NH3]init = Cam = 4th5th/100 and [NH4Cl]init = 1st2nd/100 = 0.10 M for all students
NH4+(aq) + Cl–(aq) ← NH4Cl(aq)
NH4+(aq) + H2O(l) → ← H3O+(aq) + NH3(aq)
Ka = [H3O+]e[NH3]e [NH4+]e = 5.6×10–10
Initial0.10 0 Cam
Change– x +x – x
Equilibrium0.10 – x x Cam + x
Approximate? 0.10/5.6×10–10 = 1.8×108 > 100, Yes
5.6×10–10 = (x)(Cam)/(0.10)
x = [H3O+]e = (5.6×10–10)(0.10)/(Cam)
pH = –log[H3O+]e, all to two places past the decimal.
Since Cam ~ 0.5 - 0.65 for all answers, [H3O+]e is about 10–10 M, so the pH answers are all around 10.
[NH3]init (M) [NH4Cl]init (M) [H3O+]e (M) pH
0.50 0.10 1.1×10–10 9.95
0.52 0.10 1.1×10–10 9.95
0.54 0.10 1.0×10–10 10.00
0.55 0.10 1.0×10–10 10.00
0.56 0.10 1.0×10–10 10.00
0.57 0.10 9.8×10–11 10.01
0.58 0.10 9.7×10–11 10.02
0.59 0.10 9.5×10–11 10.02
0.60 0.10 9.3×10–11 10.03
0.61 0.10 9.2×10–11 10.04
0.62 0.10 9.0×10–11 10.04
0.63 0.10 8.9×10–11 10.05
0.65 0.10 8.6×10–11 10.06