There are two choices. One is Nitric acid, and the other is phosphoric acid.

The solution of each type of these acids has a concentration of 0.1 mole. I answer would be phosphoric acid, because each molecule contains more than 1 hydrogen ion. Thus it would be more conductive.

Does strong/ weak affect conductivity?


Whether the acid is a strong or weak acid does affect its conductivity. Weaker acids are less dissociated and generate fewer ions in the equilibrium.

Nitric acid $\ce{K_{a}=24}$

Phosphoric acid $\ce{K_{a}=7.1 x 10^{-3}}$

link to Ka reference

Nitric acid is more dissociated than phosphoric acid

Electrical Conductivity κ in mS/cm at 1 Mass Percent Concentration

link to conductivity reference

Nitric acid ~ 56.1

Phosphoric acid ~ 10.1

Nitric acid is more dissociated and more conductive than phosphoric acid.

  • $\begingroup$ Do u memorize these things, which is dissociated more? And the constant of ionization for each acid? $\endgroup$ May 28 '14 at 0:57
  • $\begingroup$ But then in one mole of phosphoric acid, there are are three moles of hydrogen ions? Which is significantly more than what sulfuric acids have. $\endgroup$ May 28 '14 at 1:01
  • $\begingroup$ "Do u memorize these things" - No way, just google them; "one mole of phosphoric acid, there are are three moles of hydrogen ions" - Doesn't really matter if it doesn't ionize (dissociate) very much. $\endgroup$
    – ron
    May 28 '14 at 1:07
  • $\begingroup$ Why but more ions means a stable current of electricity? $\endgroup$ May 28 '14 at 1:16
  • $\begingroup$ Understood, but there are not more ions with phosphoric acid because it doesn't dissociate. Above I gave the value for the first ionization constant for phosphoric acid and it was 10^-3 times smaller than the dissociation constant for nitric acid. Look at how much SMALLER the second and third dissociation constants are for phosphoric acid;. There is just a LOT LESS H+ present with phosphoric acid compared to nitric acid(chem.wisc.edu/deptfiles/genchem/sstutorial/Text12/Tx125/…) $\endgroup$
    – ron
    May 28 '14 at 1:22

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