Calculate the ph of 0.10 m ammonia solution calculate the ph after 50 ml of this solution is treated with 25 ml of 0.10 hcl
Answers
NH3 + H2O ----------> NH4+ + OH-
Kb = [NH4+ ] [OH- ] / [NH3] = 1.77 x 10^-5
Before neutralization,
[NH4+ ] = [OH-] = x
Thus
Kb = x2 / 0.10 = 1.77x 10-5
Hence x = 1.33 x 10-5 = [OH-]
Now Kw = [H+][OH-]
So [H+] = Kw/ [OH-] = 10-14 / 1.33x 10-12 = 7.51 x 10^-12
pH = - log(7.51 x 10^-12) = 11.12
On expansion of 25 ml of 0.1 M HCl solution for 50 ml of 0.1 M smelling salts arrangement , 2.5 mol of alkali are killed. The subsequent 75 ml arrangement contains the remaining un- neutralized 2.5 mol of NH3 particles and 2.5 mol of NH4+.
NH3 + HCl -------> NH4+ + Cl-
2.5 2.5 0 0
At equilibrium 0 0 2.5 2.5
The resulting 75 ml of solution contains 2.5 mol of ammonium ions and 2.5 mol of un - neutralized ammonia molecules. This ammonia exists in the following equilibrium.
NH4OH --------> NH4+ + OH-
0.033 M- y y y
Where y = [OH-] = [NH4+]
The final 75ml solution after neutralization already contains 2.5 mol NH4+ ions, thus total concentration of NH4+ ions is given as:
[NH4+] = 0.033 + y
As y is small, [NH4OH] = 0.033 M and [NH4+] = 0.033M
Kb = [NH4+] [OH-] / [NH4OH] = y (0.033) / (0.033) = 1.77x 10-5 M
Thus , y = 1.77 x 10^-5 = [OH-]
[H+] = 10-14 / 1.77 x 10^-5 = 0.56 x 10^-9
pH = 9.24