Ka of strong acids

[591489]

Would the Ka of 1M HCl be 1 ? since its (1 mol H3O x 1 mol Cl)/(1mol HCL)

Similarly, would the Ka of 2M HCl be 2?

Thanks!

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[inasensegone]

Would the Ka of 1M HCl be 1 ? since its (1 mol H3O x 1 mol Cl)/(1mol HCL)

Similarly, would the Ka of 2M HCl be 2?

Thanks!

It seems like you may not be understanding what Ka means. It is the dissociation constant of the acid at equilibrium. That means it represents the ratio of concentration of products to concentration of reactants at equilibrium. Because a strong acid like HCl nearly dissociates 100%, you could expect the Ka to be a big a** number (it forms almost all products and leaves almost no reactant, at equilibrium).

 

[591489]

Oh I see, thank you for your explanation. I was thinking it was the concentration of HCl that we started with. Thanks! I'm not sure why I got this mixed up. Is there a term where we use the (initial concentration of prod)/(the final conc of reactants)?

 

[Czarcasm]

Oh I see, thank you for your explanation. I was thinking it was the concentration of HCl that we started with. Thanks! I'm not sure why I got this mixed up. Is there a term where we use the (initial concentration of prod)/(the final conc of reactants)?

Ka is the equilibrium constant for the dissociation of any acid at equilibrium. It's value is derived from equilibrium conditions at a given temperature (since all equilibrium values are temperature dependent). We typically use Ka to compare the strength of different acids and/or to find out how much of a given acid will dissociate.

HCl is one of those extremely strong acids that for all intents and purposes, will dissociate to completion. Therefore, if you have 1M HCl, you can simply conclude that it'll dissociate entirely -- that is, be fully consumed to produce 1M H3O+ and 1M Cl- (the products of this dissociation). Knowing the Ka in this case, isn't exactly necessary.

On the other hand, consider a weak acid: Let's say you have 1M Acetic Acid. Because it's weak, it's hard to know how much of the intial acid will react to produce product. In this case, it's absolutely necessary to know the Ka value, because with this information, you can calculate the concentration of H3O+ produced, which in turn allows you to calculate the pH of a solution.

Typically, this entails creating an ICE Chart but for the MCAT, we can simply assume that the amount reacting is negligibly small compared to the starting concentration (1M). In other words, you can substitute your unknown concentrations of product as "X" and maintain that you have 1M acetic acid at final conditions (rather than 1M-x, where x is the amount that dissociated to product). Then you can solve for the unknown variable by substituting these variables into the Ka expression. In this way, you can find the value of "X," which in turn allows you to calculate the pH of the solution if necessary.

Hope this makes sense.

 

[591489]

Thanks! That was very thorough and clear 🙂

 

[Teleologist]

If Ka of HCl were 1, then 1 molar HCl would only dissociate to an extent of about 62%.

In other words, HCl would stop being a strong acid.

 

[Czarcasm]

@Teleologist I'm sorry but everything you just said made absolutely no sense. The % dissociation of HCl = 100%. 62% would imply it only partially dissociates (a characteristic of weak acids), and that's simply not true. I'm not following how you even arrived at that percentage either.

 

[591489]

That was a hypothetical calculation. Basically, pointing out how it wouldn't make sense for the Ka of HCl to be 1

 

[Teleologist]

@Teleologist I'm sorry but everything you just said made absolutely no sense. The % dissociation of HCl = 100%. 62% would imply it only partially dissociates (a characteristic of weak acids), and that's simply not true. I'm not following how you even arrived at that percentage either.

Actually the dissociation of HCl depends on concentration. In extremely concentrated solutions (e.g. 18 M) HCl dissolution is not approx. 100%.

Also I was just testing the OP's thought that the Ka of HCl = 1. I wished to show that through a simple calculation, one can easily see that a Ka of 1 for HCl contradicts the definition of a strong acid (nearly complete or complete dissociation at most, read "not extremely extremely high" molarities).

 

[Czarcasm]

Actually the dissociation of HCl depends on concentration. In extremely concentrated solutions (e.g. 18 M) HCl dissolution is not approx. 100%.

Also I was just testing the OP's thought that the Ka of HCl = 1. I wished to show that through a simple calculation, one can easily see that a Ka of 1 for HCl contradicts the definition of a strong acid (nearly complete or complete dissociation at most, read "not extremely extremely high" molarities).

Right, I see what you're saying but honestly though, I think if you emphasize too much on rare exceptions like this, it does more to hurt you than do you good. Sorry, but I just can't imagine anyone doing well or even finishing on time with that mind set. For the MCAT, you should be okay to assume that HCl along with the other notoriously strong acids dissociate 100%. I've done tons of problems and passages and never once have I encountered a scenario where I had to consider otherwise -- at least in regards to its concentration (your example above).

 

[Blueprint MCAT Tutor]

Is there a term where we use the (initial concentration of prod)/(the final conc of reactants)?

EDIT: wait I just re-read what you posted and I misinterpreted. The specific thing you're asking [prod]i / [rxt]f doesn't exist, as far as I know. I'll leave the rest of my post as-is since it's all still true stuffs.

Skimming quickly through the thread it looks like nobody answered that question.

Reaction Quotient = Q = [Prod] / [Rxt] at any given point in time
Equlibrium Constant = Keq (or Ka, Kb, etc) = Q @ equilibrium

So if Keq < Q then you have "too much" products and the reaction will shift backwards
If Keq = Q then you're @ equil. and rxn goes nowhere
And if Keq > Q then you have "too much" reactants and the reaction will shift forwards

(notice the way I wrote it - the arrow points in the direction the reaction will shift)

Finally, for salt dissocations:

Ion Product = IP = [Ions] at any given concentration
Solubility Product = Ksp = IP @ equilibrium

If Ksp < IP then you've got too much ions and you're supersaturated and you'll get ppt until you get down to saturation
If Ksp = IP then you're at the saturation (equilibrium) point and nothing happens
If Ksp > IP then you're unsaturated and you could, if you wanted, dissolve more ions in sol'n.

Hope this helps! 🙂

-Bryan