# How Do You Calculate the Acid Dissociation Constant?

• alias_grace
In summary, Borek said that the equilibrium concentration of hydronium ions is the value that has to be subtracted from the formal concentration. He then calculated the equilibrium concentration using the Ka equation.
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.

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.

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.

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.

alias_grace said:
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.

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.

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

perhaps an ICE table would help

## 1. What is the definition of acid dissociation constant?

The acid dissociation constant, denoted by Ka, is a quantitative measure of the strength of an acid in aqueous solution. It represents the equilibrium constant for the dissociation reaction of an acid into its conjugate base and a hydrogen ion.

## 2. How is the acid dissociation constant related to pH?

The acid dissociation constant is directly related to pH through the equation pH = -log(H+), where H+ represents the concentration of hydrogen ions in solution. The lower the value of Ka, the weaker the acid and the higher the pH.

## 3. What factors influence the value of acid dissociation constant?

The value of acid dissociation constant is influenced by the nature of the acid, temperature, and the ionic strength of the solution. In general, stronger acids will have larger Ka values, and higher temperatures and ionic strengths will decrease Ka values.

## 4. What is the difference between Ka and pKa?

Ka is the actual numerical value of acid dissociation constant, while pKa is the negative logarithm of Ka. pKa is often used as a more convenient way to express the strength of an acid, as it simplifies calculations and allows for direct comparison between different acids.

## 5. How is the acid dissociation constant experimentally determined?

The acid dissociation constant can be experimentally determined through various methods, such as titration, spectrophotometry, and potentiometry. These methods involve measuring the concentration of the acid and its conjugate base at equilibrium and plugging these values into the Ka expression.

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