Calculate the ph of 0.1m ammonia solution calculate the ph after 50ml of this solution is treated with 25 ml of 0.1m hcl. The dissociation constant of ammonia kb is 1.77*10^-5
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
Explanation:
The solution is given in the photo
