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If a reaction requires a "strong" acid - what does this mean? Does it mean any acid of a specific concentration or pH? Are only certain acids classed as "strong"? I have read on Wikipedia that it needs to ionise completely in an aqueous solution - in this case, is a strong acid (for example, HCl) still classed as a strong acid if it is in a 0.00001M solution?

Mithoron
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Ell
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2 Answers2

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The strength or weakness of an acid is independent of its concentration. A strong acid dissociates completely in solution, so $\ce{HCl}$ at any molarity would be considered strong, and $\ce{HCOOH}$ (for example) at any molarity would be weak. It is more enlightening to consider $K_a$ values, which are essentially the equilibrium constants for the ionization of an acid $\ce{HA}$. $$K_a=\frac{[\ce{H}^+][\ce{A}^-]}{[\ce{HA}]}$$ This value will either be very small (<<1) or very large (>>1000). Acids with very small $K_a$ are considered weak, and acids with very large $K_a$ are considered strong (the $K_a$ of strong acids is usually not stated and assumed to be infinite).

Addendum: The $K_a$ value is concentration-independent because you are only interested in how well the acid dissociates from its molecular form. If you had $0.0000001\ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCl}$, practically every molecule of $\ce{H-Cl}$ would dissociate into $\ce{H^+}$ and $\ce{Cl^-}$ no matter how many there are. But if you had $12 \ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCOOH}$ (methanoic acid), only a small proportion of the $\ce{HCOOH}$ would dissociate into $\ce{H^+}$ and $\ce{HCOO}^-$, irrespective of how many you had to start.

Community
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Zen
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  • Would it not be simpler to just say that a high concentration of hydronium causes acidity? And, would it matter which definition of acid we are using (Arrhenius and Bronsted-Lowry)?

    (I am still learning, so this is a legit question that I have)

     – Carlos Carlsen Jun 07 '16 at 16:13
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    Nitric acid Ka is about 24, but it is considered a strong acid. Instead of "assumed to be infinite", a strong acid is more like Ka > 1 – DavePhD Jun 07 '16 at 16:18
  • Also, having a $K_\mathrm{a} > 1$ is not equal to ‘fully dissociated in solution’. It is ‘fully dissociated in aquaeous solution.’ Tiny, but important difference. – Jan Jun 07 '16 at 16:45
  • A Ka>1 does not mean an acid is fully dissociated, even in water. If you start with 1 M of nitric acid, the equilibrium concentrations will become (1-a) for nitric acid, and a for the nitrate-ion and hydronium ion (with a the dissociation between 0-1). Solving Ka (value=24, given in the comments above) for a results in a value for a of about 0.96, meaning 96% dissociation for nitric acid, not 100%! Ka>1 just means the equilibrium is shifted towards the products. – Stikke Jun 28 '19 at 10:55
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What makes an acid strong is also dependent on the context. E.g. lactic acid is considered strong in body fluids because lactate is completely ionized in body pH 7.4.

Jesh
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