Consider a titration of 33.8 mL of 1.64 M solution of aqueous ammonia [Kb(NH3) =
ID: 823156 • Letter: C
Question
Consider a titration of 33.8 mL of 1.64 M solution of aqueous ammonia [Kb(NH3) = 1.8x10^-5] with a 1.24 M solution of hydrochloric acid. Calculate:
a) The pH of the aqueous ammonia solution before the titration,
b) The pH of the solution at half-equivalence point,
c) The pH after 7.5 mL of HCL were added.
d) The pH of the solution at the equivalence point.
e) The pH of the solution when 1.00 mL of the HCl have been added after the equivalence point was reached.
If you could please show me each step I would greatly appreciate it.
Explanation / Answer
At the beginning, you have a solution that is 1.64 M in NH3
(a). NH3 reacts with water:
NH3 + H2O --> NH4+ + OH-
Kb = [NH4+][OH-]/ [NH3] = 1.8 X 10^-5
Let x = [OH-] = [NH4+]. then
Kb = x^2/1.64 = 1.8 X 10^-5
x = 5.43 X 10^-3 =[OH-]
pOH = 2.26
pH = 14 - 2.26 = 11.74
(b). at half equivalence point, [NH3] = [NH4+].
Ka = 1 X 10^-14 / Kb
So, pKa = 9.26
So, with H-H equation:
pH = pKa + log [NH3]/[NH4+] = 9.26 + log 1 = 9.26
(c) HCl reacts with NH3 to form NH4+
moles NH3 initially = 1.64 mol/L X 0.0338 L = 5.54 X 10^-2 moles NH3
moles HCl added = 1.24 mol/L X 0.0075 L = 9.3 X 10^-3 moles HCl added
moles NH4+ formed = 9.3 X 10^-3
moles NH3 remaining = 4.61 X 10^-2
Ka = 1X10-14 / Kb = 5.56 X 10^-10
pKa = 9.26
pH = pKa + log [NH3]/[NH4+]
pH = 9.26 + log (4.61X10^-2 / 9.3X10^-3)
So, pH = 9.95
(d). ) at the equivalence point, the solution contains only NH4+
NH4+ --> NH3 + H+
Ka = [H+][NH3]/ [NH4+]
[NH4+] = 5.54 X 10^-2 mol / (0.0338*2) L = 0.82 M
let x = [H+] = [NH3]
Ka = [H+][NH3] / [NH4+]
Ka = 5.56 X 10^-10 = x^2 / 0.82
x = [H+] = 2.135 X 10^-5
So, pH = 4.67