What's the pH for this solution?

Question

What is the pH of a 0.6M solution of NH3? (a weak base)

Answer 1

pH will be 11.51 or very close to it.

Kb=[NH4+]*[OH-]/[NH3]

[NH4+] = [OH-] = x
[NH3] = 0.6 M - x

x << 0.6 M thus [NH3] = 0.6 M

this leaves us with:
Kb = x*x/0.6
Kb = 1.77e-5

x = sqroot( 1.77e-5 * 0.6) = 0.0032

pH = 14 - log[OH-] = 14 - 2.49 = 11.51

Answer 2

You need the Ka value for ammonia, which in solution forms:

NH3 + H2O ===> NH4+ + OH-

So, ammonia gas in solution forms ammonium ion (NH4+) and hydroxide ion (OH-). This means that a solution of ammonia gas in water will be alkaline (basic), with a pH greater than 7.

Your problem states that you have a 0.6M solution of ammonia in water. If **all** the ammonia were converted to ammonium, then the solution would be easy, because [OH-] would equal [NH3], and then you'd calculate pOH, subtract it from 14, and voila.

But ammonia is a weak acid, so it doesn't completely dissociate, and we have a more challenging task.

For this reaction, we need to use the following:

Kb = Kw/Ka = [OH-]*[NH4+]/[NH3]

where Kb = base dissociation constant (NH3)
Kw = dissociation constant of water = 10E-14
Ka = dissociation constant of the conjugate acid (NH4+) = 5.62E-10

Kb = 1E-14/(5.62E-10) = 1.78E-05

Kb = 1.78E-05 = [OH-]*[NH4+]/[NH3]

We know the starting concentration of NH3 is 0.6M, but only some of that has dissociated into NH4+, so the ending concentration is 0.6-x. And since the dissociation of one molecule of ammonia produces one each of ammonium ion and hydroxide ion, [OH-] = [NH4+] = x, so:

1.78E-05 = (x^2)/(0.6-x)

converting this into quadratic form:

0 = x^2 + 1.78E-05x - 1.07E-05

and I will leave it to you to do the arithmetic of the quadratic equation to solve for x.

Once you have x, which is equal to [OH-], you can calculate pH by:

10E-14 = [H+]*[OH-], or

[H+] = 10E-14/[OH-], and

pH = -log[H+]

Answer 3

between 7 and 10

Answer 4

nh3 is ammonium so just check the ph for ammonium on Wikipedia .com

Answer 5

13......guess