How to measure EC, pH, 10k thermistor, DO, and Ion-selective probes

My guess is the \$ Rds_{ on }\$ for the internal mosfets are not matched which is the case for most GPIO's, they have wide tolerances on the GPIO circuity and since the ESP8266 doesn't have any documentation or testing (its a Chinese product and 95% of them have poor documentation and testing).

I'm assuming your using the GPIO for a low side switch, if you are, I would buy some mosfets and use them as a low side switch in the place of the GPIO ports so you know what your \$ Rds_{ on }\$ is going to be.

Either way I know that the conductivity of a liquid is going to be from 1 Ohm to a milli-Ohm, and if you put a mosfet in series with that that has 10's of milli-Ohms of resistance (and almost certainly varying from port to port). Since this is an unknown, why not buy some cheap mosfets and make it a known source of error.

So try this:
Make Vin go to your gpio, get some mosfets with the \$ Rds_{ on }\$ that you require and make sure the voltage of the GPIO (probably 3.3) will turn them on. Your load would be the connection to the water. You may need a dual stage with your 12V to get mosfets with a low enough \$ Rds_{ on }\$ for your application.

I would also look at the tolerances on your resistors, 330Ω at 1% would be 3.3Ω. So if your plugging in 330Ω in your software, the actual resistor could be anywhere from 333.3Ω to 326.7Ω. Same thing with the capacitors as some ceramics have 10% tolerances.

R1 (in addition to the resistance of the probe) limits the Q1 base current to a safe value, and probably is there to prevent killing Q1 with an accidental short to ground.

R2 pulls the base up to Vcc when the input is open or a high resistance. This assures that the transistor is turned off unless the base current exceeds a minimum value. In round numbers, this is 22 uA (0.6 V Vbe / 27,000).

If you know what Vcc is, you can calculate the equivalent probe resistance needed to cause the output to change. The transistor probably has a gain of around 100, but for decent saturation with a light load assume a value of 20-30.

The 240 ohms in series with the LED indicates to me that Vcc = 5V, so the base voltage is 4.4 V. The external resistance needed for 22 uA is (Ohm's Law) 4.4 / 22 uA = 200K. Minus the 1K already there = 199K. That is the point at which the transistor starts to conduct. But wait, there's more.

There's about 12.5 mA of LED current, plus 5 mA of R3 current, we'll guess another 1.5 mA of output current. That adds up to 19 mA of collector current. Round up to 20 mA, divide by a transistor gain of 20, and you need 1 mA of base current. Back to Ohm's Law, Rbase = 4.4 V / .001 A - 4.4K. Minus R1 = 3.4 K external sensor resistance for a firmly saturated transistor.