Calculate the pH of a 0.10 M ammonia solution.
Answers
NH3 + H2O → NH4+ + OH-
Kb = [NH4+ ] [OH- ] = 1.77 x 10-5
[NH3]
Before neutralization,
[NH4+ ] = [OH-] = x
Thus
Kb = x2 = 1.77x 10-5
0.10
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
p H = - log(7.51 x 10-12) = 11.12
On addition of 25 ml of 0.1 M HCl solution to 50 ml of 0.1 M ammonia solution , 2.5 mol of ammonia are neutralized. The resulting 75 ml solution contains the remaining unneutralised 2.5 mol of NH3 molecules 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 NH4+ ions and 2.5 mol of uneutralised NH3 molecules. This NH3 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-] = y (0.033) = 1.77x 10-5 M
[NH4OH] (0.033)
Thus , y = 1.77 x 10-5 = [OH-]
[H+] = 10-14 = 0.56 x 10-9
1.77 x 10-5
Hence p H = 9.24
Answer:
Explanation:
NH3 + H2O → NH4+ + OH-
Kb = [NH4+ ] [OH- ] / [NH3] = 1.77 x 10-5
Before neutralization,
[NH4+ ] = [OH-] = x
Thus
Kb = x2 = 1.77x 10-5
0.10
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
p H = - log(7.51 x 10-12) = 11.12
-----Naveen.J
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