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It is known that good DI water have resistance ~20 MΩ·cm But how can I measure that? Using good vanilla ohmmeter (with 2000 MΩ range) showed crazy results (too low, not much dependent from distance between probes). I need that to compare DI water from 2 sources.

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if you got too low results it means you might have either a short circuit somewhere that is leaking current. the bulk and surface resistivities might be a lot different so you might have part of the electrodes near the surface. –  lurscher May 19 '11 at 18:21
    
as other posters mentioned, your measurement is meaningless unless you specify the geometry of the conductor and the points where the probes are relative to the geometry –  lurscher May 19 '11 at 18:30
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5 Answers 5

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That DI water has some specific resistance, not resistance! To measure that, You need a apropriate probe, inserting the tips of a vanilla ohmmeter into the water, excuse, is ridiculous! Conductivity measurements have to be made using alternating currents. The frequency is about some kHz to some dozen kHz.

For this purpose You can buy special meters, called conductometers, (In electrochemistry one deals with conductance, not resistance) at a range of cheap to luxurious.

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I see... Will I get proper results if I build bridge-circuit which will "swap" probes at 10kHz and will use known water geometry? I have some kinda-pure silicon (10 Ohm/sq), will it help if I make electrodes out of it instead of usual copper? –  BarsMonster May 20 '11 at 9:32
    
As I said already, buy a ready-made Conductometer including a probe. Self-made probes need some calibration, how will You do that? –  Georg May 20 '11 at 10:26
    
I've checked prices - they are around some 500$+, which is a bit of overkill for my one-off measurements. Can I measure this myself using approach in earlier comment provided that I can tolerate 80% error? :-) –  BarsMonster May 23 '11 at 9:45
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Two things come to mind. One a matter of experimental technique, and the other one of definitions.

  • Firstly: how clean were/are the probes? If you introduce a water soluble material from the surface of the probes you will have true screwed up your measurement.

  • Secondly: Are you sure that value is not for the resistivity of water. Resistivity is generally symbolized with $\rho$ and is related to the resistance by $$ R = \int \frac{\rho}{A} dr $$ where $A$ is the cross-sectional area presented along the path element $dr$. This implies that the resistance is dependent on the geometry of the measurement. Just plunging the probes into a glass of water isn't going to cut it.

    The units you give are proper for resistivity, and not for resistance.

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Tips of a ohmmeter working on DC will not work at all, because polarisation voltages will lead to erratic results. –  Georg May 19 '11 at 18:36
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Resistance (that's what your meter reads) is related to resistivity (that's the 20M Ohm x cm term) via the geometry of the problem. If you have a body of the measured substance with UNIFORM CROSS SECTION (doesn't matter what shape) between two parallel and highly conductive plate electrodes (one plate on each end of the measured body, and in full contact with it), then:

Resistance = Resistivity x (L/A), where A is the area of the uniform cross section, and L is the distance between the electrodes.

So, to bring the resistance down to something within the range of you multimeter's accuracy, simply construct the measuring set up to have A >> L, then do the math from your resistance measurement to get back to resistivity. For instance, if you pick a circular cross seciton, with D = 2r = 20 cm (about 8 inches), then A = 315 cm^2. If you have 1.5 cm of liquid between the two plates, then A/L = 210. If your meter reads 100K Ohms, then you know the resistivity is 210 x 100K = 21M Ohms.

By the way, the conductivity I found for "ultra-pure" water is 18.2M (Ohm x cm). I forget if that is before or after exposure to air. The CO2 in the air dissolves into the H2O, changing its Ph and conductivity. Don't be surprised, if you are working with highly purified water, when its conductivity changes after exposure to air.

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""between two parallel and highly conductive plate electrodes (one plate on each end of the measured body, and in full contact with it), then:"" Wrong! Since the 30ties at least one uses capacitive electrodes, where the metal is not in contact with the sample at all. –  Georg May 19 '11 at 18:38
    
My answer is not a treatise on the most advanced methods. For some materials there is a polarization if DC current is used. I don't think DI water is one of those materials. Furthermore, a discussion of the virtues of AC vs. DC measures is beyond where the OP is at. Big corporations with big labs and big budgets can afford the gear to do AC measurements and squeeze a bit more accuracy out of their readings. The OP doesn't have $5K to spend. he has a multimeter. The set up I recommend uses his multimeter, and also clearly shows the relationship between resistivity and resistance. –  Vintage May 19 '11 at 18:49
    
Ridiculous! Even the cheapest conductometers since long work on AC! –  Georg May 19 '11 at 19:36
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If you're looking to measure conductivity on a budget, try the project that this person created.

It includes a good explanation of how to do it, plus the circuit and some other goodies.

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Very nice project, although I've already bought Chinese conductivity meter for ~35$. –  BarsMonster May 11 '13 at 13:03
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Try instead measuring the conductivity. You'll need to use a conductivity meter. Once you know the conductivity, you'll know the resistivity, as it is the reciprocal: $\rho = \frac{1}{\sigma}$

You might want to take a look at this thread.

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