Relationship between concentration and resistance of aqueous solutions

Question

I'm a senior physics/chemistry student working on a practical assignment where I am trying to identify the resistance of CuSO4 in solution (distilled water). I have recorded my data and determined it is an inversely proportional relationship (res=1/conc.).

My working hypothesis was that the relationship would be directly proportional, though I wasn't really sure at the time. I was hoping someone would be able to explain the reasoning behind this in more detail, or even reference some sources that go into detail that I could use in my report.

Answer 1

Be aware of multiple nonlinear processes, when it comes to electrolytes and electrodes.

For molar conductivity there is nonlinear Kohrausch equation:

$$\Lambda_\text{m} =\Lambda_\text{m}^\circ - K\sqrt{c}$$ where $\Lambda_\text{m}^\circ$ is the molar conductivity at infinite dilution (or "limiting molar conductivity").

Conductivity is then $\kappa = \Lambda_\text{m} \cdot c=\Lambda_\text{m}^\circ \cdot c - K \cdot c^{3/2}$

Answer 2

Why is simple: When you dissolve more of any salt that ionizes in water, there are more ions to "shuttle" energy carrying electrons across at a time. How is a bit more complicated. You see when you dissolve a salt in water, the molecules, due to some phenomenon, dissociates and each atom is charged either positively or negatively due to less or more electrons respectively. When you insert two electrodes in the solution to complete a circuit, the anode terminal is positively charged due to holes concentration, and cathode terminal negatively charged due to electrons concentration. The anode attracts the negatively charged anions and on contact with the anode, they loose electron(s) to the electrode. Likewise the cathode attracts positively charged cathions, and they gain electron(s) on contact with the electrode. The atoms become their neutral self as they drift away from the electrodes. These atoms, due to the inherent cahotic random motion at the atomic scale, collide, bond, and ionize almost simultaneously and instantaneously to restart the cycle again. Now the more of these ions available to transport electrons and also create a dense cluster of themselves increasing the chance of quicker collision to restart the cycle, the more conductive or less resistant a salt solution is. The above explained process explains why a warmer solution is to some degree more conductive than a cooler one, because the increased kinetic energy in the solution helps increase the frequency of collision of neutral atoms.

Answer 3

When you dissolve a salt in water, it dissociates into two ions (one positively-charged particle and an equally negatively charged one). For the sake's of simplicity, we call $PN$ the salt and $P$ and $N$ respectively the positive and negative part.

A fraction of salt molecules, when dissolved in water, dissociates into ions. When two ions meet each other, they can join an become salt again. At equilibrium, there is going to be a fraction $f$ of salt molecules which are dissociated while the remaining part $(1-f)$ will still be in salt form. So, if you dissolve a concentration $c$ of salt in water, you end up with $fc=[P]$ positive carriers (where $[\cdot]$ indicates the concentration), $fc=[N]$ negative carriers and $(1-f)c=[PN]$ still in salt form.

(Why it has to be a fraction $fc$ and not a more complicated function of $c$ is due to the kind of chemical reaction $PN<=>P+N$)

Now, conductivity is defined as $J=\sigma E$ (where $J$ is the current density and $E$ the electric field). The current is proportional to the number of carriers and therefore so is the conductivity. These are simple computations you should have done in physics - just google them, but I think it makes sense: the more carrriers are around, the more charge moves per unit time!

Now, in this case, the number of carriers depends on the salt concentration $[P]+[N] \propto fc$ and therefore $\sigma \propto c$. Finally, because the resistivity is defined as $\sigma={1\over \rho}$ you end up with $${1\over\rho}\propto c$$ which is what you measured.

Summing up, by definition the conductivity depends on the number of charges proportionally, thus the resistivity inversely. Then, because of the reaction causing the dissociation of the salt in ions, the number of charges depends on the initial concentration $c$ also proportionally, and therefore also the conductivity, and therefore the resistivity inversely.

As you can see, the main reason is the kind of chemical reaction involved (dissociation), which is the reason for linear dependence on the concentration. Indeed, should you choose a different stronger salt or go to very high concentration regimes, that behavior would change.

Answer 4

at lower concentration of ions the resistance will decrese due to the incese in the charge carriers but at high concentration the resistance will increse as inter ionic attration will increse and hence it will increse the resistance. you can verify it using Kohlrausch's law of independent migration of ions.

https://en.wikipedia.org/wiki/Molar_conductivity