How Do You Calculate the Acid Dissociation Constant?

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Discussion Overview

The discussion revolves around calculating the acid dissociation constant (Ka) for a weak monoprotic acid based on given concentrations of hydrogen ions and the formal concentration of the acid. The scope includes mathematical reasoning and conceptual clarification related to equilibrium concentrations in acid-base chemistry.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant calculates Ka using the hydrogen ion concentration and the formal concentration of the acid, arriving at two potential answers (10-2 or 10-5) but expresses uncertainty about the reasoning behind these values.
  • Another participant points out the need to use equilibrium concentrations for the Ka calculation, suggesting that the acid must have dissociated to produce the hydronium ions.
  • A participant acknowledges difficulty in understanding how to determine the degree of dissociation from the hydrogen ion concentration and seeks clarification.
  • One participant proposes a simplified model where the concentration of the conjugate base is assumed to be equal to the hydrogen ion concentration, suggesting a basic equilibrium expression.
  • Another participant reiterates the need to adjust the formal concentration of the acid by subtracting the concentration of dissociated hydrogen ions to find the equilibrium concentration.
  • A different participant shares their own calculation process, arriving at a value for Ka of 10^-5 and expressing uncertainty about their correctness.
  • One participant suggests that using an ICE table might be beneficial for organizing the information and calculations.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to calculating Ka, with no consensus on the correct method or final answer. Multiple competing views and methods remain present in the discussion.

Contextual Notes

Participants highlight the importance of distinguishing between formal and equilibrium concentrations, but the discussion does not resolve the specific assumptions or steps needed for the calculations.

alias_grace
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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.
 
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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
 

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