# Acid dissociation constant

1. Dec 1, 2006

### alias_grace

A 0.1 mol/L aqueous solution of weak monoprotic acid (contains one ionizable hydrogen atom) has a hydrogen ion concentration of 0.001mol/L. The value of Ka is:
a) 10-6
b) 10-2
c) 10-3
d) 10-5

I took 0.001 and divided by 0.1 to get 10-2 and I also multiplied them together to get 10-5. That was the only thing I could think of to do. So I know that b) or d) is right. I don't know why though.

2. Dec 1, 2006

### GCT

The concentration of the acid that they gave you is the formal concentration, you need the equilibrium concentration for the Ka equation, you can deduce this based on the fact that the acid must have dissociated to produce the hydronium ions.

3. Dec 12, 2006

### alias_grace

I'm sorry, I don't know how to do that. I understand what you are saying though. I just want to fully understand how to do the question. I know you are saying that from knowing the hydrogen ion concentration, I will be able to figure out how much the acid is dissociated. I am just having trouble seeing the connection.

4. Dec 13, 2006

### Staff: Mentor

As a first approximation assume that there is only one source of H+ - dissociated acid, and that concentration of conjugated base is identical to H+:

HA <-> H+ + A-

[H+] = [A-]

Check pH calculation lectures for many other examples.

5. Dec 13, 2006

### GCT

0.1 M was the formal concentration of the acid, HA

As Borek has mentioned, the dynamics of the dissociation can be presented as HA<-->A- + H+,

Thus the equilibrium concentration of HA, [HA] is

0.1 M - .001 M

since the proton (hydronium ion) was formed through the dissociation
of HA, the equilibrium concentration of hydronium ions (H+, proton) is the value that has to be subtracted from the formal concentration.

6. May 25, 2007

### Styx

I am doing this question as well but I am not sure whether or not I am correct...

What I did was:

HA = 0.10 - 0.001 = 0.099
H+ = 0.001
[H+] = [A-]
Therefore A- = 0.001

k = [H+][A-] / [HA]
k = (0.001)(0.001) / (0.099) = 10^-5

7. May 25, 2007

perhaps an ICE table would help