From: Ronald Wong (ronwong@inreach.com)
Date: Thu Sep 04 2003 - 13:28:12 PDT
Message-Id: <l03102800bb7ca5cc510b@[209.209.18.41]> Date: Thu, 4 Sep 2003 13:28:12 -0700 From: Ronald Wong <ronwong@inreach.com> Subject: Re: measuring conductance
Sally Seebode asked:
>Hi,
>I am trying to do an AP chemistry lab from Ebbing. The lab has students
>measure teh conductance of different solutions. The outcome to see that
>molecular do not conduct and ionic do, that strong and weak acids conduct
>differently, that concentration affects conductivity, and that mixtures
>conductivity is not always additive.
>
>It requires a conductance cell or conductance meter. I have voltmeters and
>ammeters. Are these the same?
No. These meters are not the same. The voltmeter measures the potential
difference between two places (in units of volts - V) and the ammeter
measures the current flowing from one place to another (in units of amperes
- A).
I know nothing about measuring conductance in an AP chem class but,
qualitatively, Geoff's suggestion involving a battery and a buzzer would
seem just fine.
If you want to quantify things then keep the following in mind:
Conductivity is the reciprocal of resistivity.
Resistivity (in ohm-meters) depends on the resistance of the conductor (R
in ohms), it's length (L in meters), and it's cross-sectional area (A in
square meters - i.e. m^2).
Mathematically, Resistivity = A * R /L.
Thus, Conductivity = L/(A * R) = (L/A) * (1/R).
Two metal plates of known area with their faces separated a known distance
apart in your solutions would lead to a fixed value for L/A in the above
expression.
So Conductivity = K * (1/R) where K = constant = L/A (in m^-1)
The value of R can be determined by measuring the current (I) that flows
between the plates when you apply a known voltage (V) across the plates.
Ohm's Law says R = V/I so (1/R) = (I/V)
and we end up with Conductivity = K(I/V).
It all comes down to applying a voltage across the plates of your cells
(using a battery or some other source of potential difference) and
measuring the voltage across the plates with your voltmeter while, at the
same time, measuring the current flowing through the circuit with your
ammeter. Plugging the values into the last formula for I and V will give
you the conductance.
The unit for conductance is 1/(ohm-m) but technically it should be written
as ohm^-1/m based on the accepted standard for this quantity. When I was
going through school, it was called a mho (ohm spelled backwards) but I
suspect that would be frowned on today.
******************
If you don't have to determine the actual values of conductance
in ohm^-1/m but just the relative values so that you can compare
the conductance between one solution you are investigating
and another, you can use the information above to simplify your
students' work.
Just get a whole series of plates of the same size (you don't even
have to measure their areas) and mount them in pairs at opposite
ends of a class set of non-conducting rods of exactly the same
length and cross-sectional area.
Large fender washers secured to the ends of a long plastic rod
might do (you may have to drill holes in the ends of the rod to
avoid splitting them when you attach the washers to them by means
of a screw passing through the center of the washer). The
electrical connection to these plates is made using alligator
clips at the end of the wires coming from your source of electricity.
The ratio of L/A is the same for all of these "probes" so you can
arbitrarily consider their value to be a unit of measurement. You
can even give the unit a name if you wish - like one "Sally"
(this would be a good time to review the arbitrary nature of
measurement in general. You might even ask your students how the
conversion factor for converting a "Sally" to a m^-1 can be
arrived at).
Now all they have to do in their lab is determine the value of I/V.
This ratio will be equal to the conductance in Sally/ohm.
******************
Things to keep in mind if you decide to take this approach to measuring
conductance:
1. Make sure that the metal plates are completely submerged
in the solution (if not, the effective cross-sectional area
will be less than the area of the plates).
2. Make sure the distance between the plates remains unchanged.
3. If you are using variable voltage power supplies, start each
run with very small voltages first. I suspect that some of
your solutions will have very little resistance no matter
how far apart the plates are (or how small their area).
4. Since you've never done this lab before, do the lab with all
kinds of variations to find out what kind of problems
your students might run into BEFORE you spring it on
them.
Have fun doing it.
ron
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