Question

What is the pH of a 0.5M NH4HCO3 solution?

Answer 1

pH = -log \(\frac{[HA]}{[A]}\)

Answer 2

This only applies for the HCO3 I believe

Answer 3

For the NH4 you have to work it out on a seperate equilibrium

Answer 4

What I did was find out [H+] formed from NH4 dissociation.

Then I used the following formula to get the [H+] from the HCO3 dissociation and association (because there are 2 equilibria involved with this):
[H+] = \sqrt{(Ka1*Ka2)}

Answer 5

And then add up both these [H+] quantities (ie. from NH4, and that from HCO3), and then take the -log([H+]) to get pH.

Answer 6

@abb0t

Answer 7

What you need to do is to write down Henderson's equations for both couples implied in this problem, namely :

NH\(_4^+\)/NH\(_3\) and H\(_2\)CO\(_3\)/HCO\(_3^-\)

because, the main reaction that can take place will be:

NH\(_4\)HCO\(_3\) —> NH\(_4^+\) + HCO\(_3^-\) —> NH\(_3\) + H\(_2\)CO\(_3\)

Once this is done, add these equations and simplify.

You will find about pH = 7.8 , according to values you can find in the pKa tables.

Answer 8

How do you know that is the main reaction that will take place, and not, say, HCO3- acquiring a proton to form H2CO3?

Answer 9

There are no extra protons in the solution (no added strong acid).

The only reactions that can take place are between water, ammonium ions and hydrogencarbonate ions.

The most likely one is between ammonium ions and hydrogencarbonate ions, bacause:
a) the ammonium ion is a stronger acid than water itself
b) the hydrogencarbonate ion is a stronger base than water itself

Answer 10

For some reason, I cannot see the math code in its proper form...

Answer 11

@Vincent-Lyon.Fr
So I made an ICE table and solved the equilibrium using:

NH4(+) + HCO3(-) <=> NH3 + H2CO3
0.5-x, 0.5-x, x, x, respectively
Used Ka of NH4(+)

Ka = (x)^2/(0.5)^2

Solved this and retrieved 0.000011853 M = [NH3] = [H2CO3]

Then, solved H-H equation seperately for the pH:

NH4(+)/NH3
pKa used is of NH4(+)

H2CO3/HCO3-
pKa used is of H2CO3

I added obtained pH values from both equations and got 13.0199...
What did I do incorrectly?

Answer 12

Why don't you follow the steps I mentioned ? It is much easier than your method.

(For your information, your error is that the equilibrium constant of your first reaction is not Ka of NH4+)

Answer 13

Are you saying to do the following reactions?

1. NH4+ <=> NH3 + H(+)

Ka = [NH3][H+]/[NH4+]

Ka used here is of NH4+

2. H(+) + HCO3- <=> H2CO3

Ka = [H2CO3]/[H+][HCO3-]

Ka used here is of H2CO3


... take the -log([H+]) obtained, and that is the pH?

Answer 14

Almost.

Rather Ka = [H+][HCO3-]/[H2CO3] Ka used here is of H2CO3

Now multiply those equations and simplify. You will get [H+] = ...
which will lead you to the pH.

Answer 15

So need to take the equations:

[1]: Ka1 = [NH3][H+]/[NH4+]
[2]: Ka2 = [H2CO3]/[H+][HCO3-]

^- And do [1]*[2], which gives:

(Ka1)*(Ka2) = [NH3][H+]*[H2CO3] / [H+][NH4+][HCO3-]
The [H+] cancel out though....

(Ka1)*(Ka2) = [NH3][H2CO3]/[NH4+][HCO3-]

...

How were you able to get [H+] = .... ?

Answer 16

=============================================
Ok instead, let us proceed by your method:

I use the H-H equation of HCO3-/H2CO3, and NH4+/NH3:

pH = pKa1 + log([HCO3-]/[H2CO3])
pH = pKa2 + log([NH3]/[NH4+])

Then, I add these two equations:

2pH = pKa1 + pKa2 + log([HCO3-]/[H2CO3]) + log([NH3]/[NH4+])
2pH = pKa1 + pKa2 + log((HCO3-][NH3])/([H2CO3][NH4+])

But then, what is the concentration of [H2CO3] and [NH3]?

Answer 17

Ok nevermind I believe they will cancel each other out. So then, the log-term will ultimately be 0.

Answer 18

@Vincent-Lyon.Fr

Answer 19

Great! You did it!

So finally pH = (pK1 + pK2)/2

Tables give 9.2 for ammonium/ammonia and 6.4 for the first acidity of H2CO3.

Hence pH = 7.8

Answer 20

I just realized that the pKa table given in my hand-out was 3.77 for H2CO3, 25C. I believe this is a mistake: it should be 6.4, which is in agreement with what you said.

Therefore, pH = (9.25 + 6.4)/2 = 7.825 = 7.8.