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Suppose we have two 0.1M solutions in water. What is their conductivity and how will it change when we mix them, and why?

A) NaCl and KCl:

Okay, so both are strong electrolytes and conduct well but what happens when they get mixed and why?

B) CH3COOH and NH3*H2O:

Both are weak electrolytes and will conduct electricity. Again, I don't know what happens after mixing and why?

C) HCl and NaOH

Both strong electrolytes, they neutralize each other in a reaction. So, the result is no conductance (or very weak?)

Thank you for your answers.

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closed as off-topic by MaxW, A.K., Jon Custer, Mithoron, Todd Minehardt Nov 6 '18 at 22:50

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The first case is pretty straight forward. As you mentioned, both $\ce{NaCl}$ and $\ce{KCl}$ are good electrolytes. Assuming the $\ce{NaCl}$ solution is there in a beaker, and you are adding $\ce{KCl}$ solution from a burette, the conductivity of the solution will not change significantly. This is because although two strong electrolytes are added, the number of ions per unit volume remains same as both solutions have the same concentration. However, none of $\ce{NaCl}$ and $\ce{KCl}$ will dissociate completely because of the common ion effect. This effect will, however, not be significant.

The second and third cases are best analysed using conductometric titration curves.

Let's consider the case of $\ce{HCl}$ and $\ce{NaOH}$ first. I will assume $\ce{HCl}$ is present in the beaker and $\ce{NaOH}$ is being added from the burette. The $\ce{OH-}$ ions being added will neutralise the $\ce{H+}$ ions from the solution immediately and will form the weakly ionising electrolyte $\ce{H2O}$. However the $\ce{Na+}$ ions and $\ce{Cl-}$ ions will remain in solution. So, effectively, $\ce{H+}$ is being replaced by $\ce{Na+}$. Since the ionic mobility of $\ce{Na+}$ is lower than the ionic mobility of $\ce{H+}$, the conductivity of the mixture decreases. However, the phenomenon reverses on reaching the equivalence point. After the equivalence point, there is no more neutralisation of $\ce{H+}$ and $\ce{OH-}$ and they remain dissociated in solution on further addition of $\ce{NaOH}$. So, the conductivity increases from this point until $\ce{NaOH}$ is added. Here is the conductometric titration curve enter image description here

The final case is that of $\ce{CH3COOH}$ and $\ce{NH4OH}$. Again, I will assume that $\ce{CH3COOH}$ is present in the beaker and the $\ce{NH4OH}$ is added from the burette. This case is a little different from the previous case. The neutralisation of $\ce{H+}$ and $\ce{OH-}$ will occur in this case too. There will be an initial decrease in conductivity as there is decreased in dissociation $\ce{CH3COOH}$ due to common ion effect. However, after that, there will be a rapid increase in the conductivity till the equivalence point is reached. This is because $\ce{NH4OH}$ converts the weak electrolyte $\ce{CH3COOH}$ to $\ce{CH3COONH4}$ which is a very strong electrolyte and ionised almost completely. After the equivalence point is reached, the conductivity of the mixture becomes indifferent to further addition of $\ce{NH4OH}$ as $\ce{NH4OH}$ is a weak electrolyte and hence, does not contribute to conductivity on further addition. Here is the conductometric titration curve -

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I think: A) When you mix both you have better conductivity by increment of ions in solutions.

B) none conductivity When you mix both you have a amine, then When you mix both the conductivity decrease.

C) When you mix both the conductivity increase (little) because you have complete disossiated ions (Na+ and Cl-) in the solution.

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    $\begingroup$ I object. In (A), when you mix two 0.1M solutions, you don't obtain a solution with higher concentration of ions, hence no increase. (B) is wrong because you won't be getting an amine in ambient conditions. (C) is wrong for more subtle reasons. $\endgroup$ – Ivan Neretin Nov 6 '18 at 12:13

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