4
$\begingroup$

For example, nitric acid or sulfuric acid in an equal-molar solution of perchloric acid. Won't perchloric protonate a significant concentration of nitrate ions?

$\endgroup$
4
  • 2
    $\begingroup$ You’re in aquaeous solution? Yes, as long as the overall concentration isn’t too high. $\endgroup$ – Jan Nov 27 '15 at 21:09
  • 3
    $\begingroup$ Nitrate ions, not nitrite ions. Nitric acid -> nitrate; nitrous acid -> nitrite. But what Jan said. Unless you make the acids really concentrated, it's not likely to happen. $\endgroup$ – orthocresol Nov 27 '15 at 21:21
  • $\begingroup$ What if the solution was highly concentrated acid? $\endgroup$ – user32134 Dec 2 '15 at 18:13
  • 1
    $\begingroup$ As far as solutions in highly concentrated acid, Nitric acid acts like a base in concentrated sulfuric acid (it forms the nitronium ion). $\endgroup$ – AlaskaRon Dec 3 '15 at 5:19
2
$\begingroup$

Suppose you have nitric acid (about Ka=20) and perchloric acid (Ka > 100,000).

$\frac{[A-][H+]}{[HA]} = K_a$

In 1M nitric acid alone, let x be [$\ce{NO3-}$], that is dissociated nitric acid

$\frac{x^2}{(1-x)} = 20$

x = 0.954M

the nitric acid would be about 95% dissociated.

If the solution is 1M nitric and 1M perchloric, with all the perchloric acid dissociating, the nitric acid equilibrium would be:

$\frac{x(1 + x)}{(1-x)} = 20$

x= 0.913

the nitric acid would be about 91% dissociated.

For 2M nitric acid alone:

$\frac{x^2}{(2-x)} = 20$

x= 1.83M

or 92% dissociated.

For 2M nitric and 2M perchloric:

$\frac{x(2 + x)}{(2-x)} = 20$

x = 1.69M

or 84% dissociated.

So yes it can make a difference and it is more apparent at higher concentration.

$\endgroup$
1
  • 1
    $\begingroup$ It might be worth mentioning the "leveling effect," e.g. something in concentrated sulfuric or acetic acid. $\endgroup$ – Geoff Hutchison Dec 2 '15 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.