Ionization of Strong Acids: Does Concentration Matter?

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SUMMARY

The discussion centers on the ionization of strong acids, specifically hydrochloric acid and sulfuric acid, and whether concentration affects their complete ionization in water. It concludes that while strong acids typically ionize completely, the concentration can influence the extent of ionization due to equilibrium dynamics. When a strong acid is neutralized, the removal of H+ ions can induce further dissociation until a new equilibrium is established. The Bronsted-Lowry theory is referenced to explain the acid-base interactions involved in this process.

PREREQUISITES
  • Understanding of strong acids, specifically hydrochloric acid and sulfuric acid.
  • Familiarity with chemical equilibrium concepts and equilibrium constants.
  • Knowledge of the Bronsted-Lowry acid-base theory.
  • Basic principles of acid-base neutralization reactions.
NEXT STEPS
  • Study the principles of chemical equilibrium in detail.
  • Learn about the Bronsted-Lowry acid-base theory and its applications.
  • Investigate the effects of concentration on the ionization of weak acids.
  • Explore equilibrium constants and their significance in acid-base reactions.
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Chemistry students, educators, and professionals interested in acid-base chemistry, particularly those studying the behavior of strong acids in various concentrations and their ionization dynamics.

JonnyG
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I know that a "strong acid" is one that ionizes completely in water. For example, hydrochloric acid, sulfuric acid, nitric acid, etc are considered strong acids because they all completely neutralize. Now, this leads me to two questions:

1) Is the concentration of a strong acid irrelevant in determining whether or not it completely ionizes? So for example, would 95% sulfuric acid ionize completely?

2) This question is applicable only if the answer to the above question is no. So suppose that the concentration IS a determining factor in ionization %, even if the acid is strong. Would neutralizing the acid induce further ionization? Let me give a concrete example. Suppose we had 1 mole of concentrated hydrochloric acid. Suppose that at this concentration, 50% of the HCl ionized. Then 0.50 moles of HCl is ionized and the 0.50 moles exists in solution as molecular HCl. Now suppose we added 0.50 moles of NaOH. This completely neutralizes the 0.50 moles of ionized HCl, correct? So now we have, in solution, 0.50 moles of NaCL and 0.50 moles of HCl. Now, does more HCl ionize since the concentration of HCl has decreased? Does the NaCl in the solution effect how much further the remaining HCl ionizes? (if it does actually ionize)
 
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Question 1: "Yes."
 
Have you heard about equilibrium reactions? Equilibrium constants? If not, follow the Bystander answer. If not - there is a lot of chemistry that you have to learn yet.
 
Borek said:
Have you heard about equilibrium reactions? Equilibrium constants? If not, follow the Bystander answer. If not - there is a lot of chemistry that you have to learn yet.

I have learned about different types of equilibrium. For example, every liquid has an equilibrium vapor pressure. I am guessing there is a type of equilibrium involved with my question? There is a chapter in my textbook that is titled "chemical equilibrium". I will be at that section within a couple of days. Maybe this section will answer my question for me? In any case, could you please give me a brief answer?
 
Start here:

https://en.wikipedia.org/wiki/Chemical_equilibrium

In short: for an acid neutralizing H+ produced during neutralization will induce further dissociation, till the system gets to a new equilibrium. And in general there are no acids strong enough to be always 100% dissociated (although for some acids that's quite a good approximation), so dissociation is always an equilibrium process.
 
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It seems to me that equilibrium is reached in the following way: the acid donates a proton to a water molecule, forming a hydronium ion. However, the hydronium ion can also donate the proton back to the conjugate base of the acid, which reproduces the acid molecule. This back and forth will continue until equilibrium is reached and both processes are continuing at equal rates.
 
The only thing left for you to notice is that in both cases it is the same scheme - an acid donates its proton to a base. After the reaction what was an acid becomes a base (minus proton), what was a base becomes an acid (protonated).

You have just discovered what the Bronsted-Lowry acid and base are :wink:
 
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