Transimpedance amplifier circuit for water conductivity
Assuming that:


You connect the ground as shown below.
You use a dual supply.




Then, thanks to the the negative feedback, the inverting input will be at a virtual ground. Therefore the voltage applied to the "water" will be the same value as the battery.

What you will measure is:

$$V_{OUT}=-R_1 \cdot I$$

Where \$I\$ is the current flowing out of the battery, as indicated.

As suggested in the comments, do not use a DC voltage, use AC instead. You can either:


put an AC voltage in place of the battery
OR you can remove the battery (replace it with a direct connection to ground) and put an AC signal to the non inverting input (instead of ground). The negative feedback will have the inverting input to assume the same (AC) voltage as the non inverting one.