Need to calculate the resistance accross membrane in water

AI Thread Summary
The discussion revolves around calculating the resistance across an ionic membrane in a homemade acrylic cell containing 0.1M NaCl. The user applied a DC voltage of 9.6V but measured only 4.45V across the water, with a current of 0.93mA, leading to a calculated resistance of approximately 4785 Ohms. Concerns were raised about the low voltage reading and the placement of voltmeter leads, suggesting that the experimental setup may need clarification or adjustment. The importance of understanding the physical setup and measurement points was emphasized to ensure accurate calculations. Overall, the discussion highlights the need for precise methodology in electrical measurements within ionic solutions.
M. Hammad Khan
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Hello everybody,
I am chemist. and I prepare ionic membrane and placed in a home made acrylic cell containing 0.1M NaCl separated into two portions by ionic membrane. I applied DC voltage of 9.6V (using 9V battery). Using multi-meter the measured DC voltage in water was 4.45V. The distance between the electrodes is 20cm. The measured amperes were 0.93mA. Can I calculate the resistance from voltage drop? and how? Please help or guide if I m doing wrong.
 
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It's not clear to me what you are doing, but a diagram might help. My best guess is that the resistance is 4.45V divided by 0.93 mA which is 4785 Ohms. But without an understanding of what you are doing, that caculcation might not be the right one, or the experimental method could be fundamentally flawed.

The thing that seems strange is the battery is loaded down to 4.45 V with less than 1 mA of current. Not sure why this is.
 
M. Hammad Khan said:
Using multi-meter the measured DC voltage in water was 4.45V.
Voltage is always between two points.
"voltage in water" is just three words that convey no information.

Where were the two voltmeter leads physically placed ?
 

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