Question

calculate the pH of a 0.10M ammonia solution​

Answer 1

Answer:

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

Hope you understand

Please make it a brainlist answer

Answer 2

Explanation:

hope that's help you!!!