How to explain the increase in current with the increase in the concentration of responsible ions?

I have conducted an experiment wherein I have used different concentrations of $\ce{H2SO4}$, $\ce{HNO3}$ and $\ce{H3PO4}$ as the electrolytes. I filled the acid in glass tube and used a 9V battery, wires and two graphite electrodes to complete the circuit. I also attached an ammeter to check the current variation during a time period of 1 minute.  
 I observed that the current was higher with more concentrated solutions, although my query is that when I add more of these acids, the concentration of the $\ce{OH-}$ ions reduces; it is the $\ce{OH-}$ ions and the $\ce{H+}$ ions that dissociate.  
 How exactly do I explain the increase in current due to increase in concentration?




The electrical current is carried by **all** ions in the solution, including the $\ce{SO\_4^{2-}}$/ $\ce{HSO\_4^-}$, $\ce{NO\_3^-}$, and $\ce{PO\_4^{3-}}$/ $\ce{HPO\_4^{2-}}$/ $\ce{H\_2PO\_4^-}$ ions, and not just by $\ce{H+}$ and $\ce{OH-}$. So, even though the $\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.


**EDIT** (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system. For example, in $\ce{NaOH}$ the $\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\ce{OH-}$:


* $\ce{H2O}$ is reduced to $\ce{H2}$ and $\ce{OH-}$ at the cathode, increasing the $\ce{OH-}$ concentration there
* $\ce{OH-}$ is oxidized to $\ce{O2}$ and $\ce{H+}$ at the anode, decreasing the $\ce{OH-}$ concentration there
* Thus, $[\ce{OH-}]$ changes from high at the cathode to low at the anode
* In order for the solution to maintain [electroneutrality](http://www.chem1.com/acad/webtext/elchem/ec1.html "electroneutrality"), something has to balance the higher concentration of negative charge at the cathode, and in this system $\ce{Na+}$ is the major contributor. (Water self-ionization can't provide enough $\ce{H+}$ to maintain the charge balance.)


In other systems such as $\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:


* Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$
* Water is oxidized at the anode, producing $\ce{H+}$, which reacts with $\ce{HPO4^{2-}}$ to form $\ce{H2PO4-}$
* Water is reduced at the cathode, producing $\ce{OH-}$, which reacts with $\ce{H2PO4^{-}}$ to form $\ce{HPO4^{2-}}$ and $\ce{H2O}$
* So, $\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode. These concentration differences result in a net transfer of charge between the two electrodes.
* All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\ce{Na+}$, similar to the $\ce{NaOH}$ case.


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