# Net conductivity of mixtures

If we have a solution, of conductivity k1, and another solution of same volume having conductivity k2, then what will be the conductivity if I mix the 2? I think that net conductance should be the sum of individual conductance of the 2 solutions in the final beaker (where volume is doubled). But then won't the individual conductance of the 2 solutions in double the volume be different from that in the initial volume?

Also, can we say that conductivity is the conductance of unit volume of solution?

This concept of electrolytic solution conductivity is complex and dependent on a variety of factors aside from mere concentration of electrolytes. There are separate laws used to describe the relationship between solution conductance and concentration dependent on the dissociative strength of electrolytes and the extent of overall solution saturation.

In your simple case, just recognize that the net conductance of the final solution may not work out to be the mere sum of conductances in the original solutions

From Wikipedia:

The specific conductance of a solution containing one electrolyte depends on the concentration of the electrolyte. Therefore, it is convenient to divide the specific conductance by concentration. This quotient, termed molar conductivity, is denoted by Λm $$\Lambda _{m}={\frac {\kappa }{c}}$$

Hence, since conductivity is proportional to concentration, and concentrations aren't additive, you cannot expect conductivities to be additive.

• First let me remark that at low concentrations the conductivity will be proportional to concentration as noted. However if you extrapolate this line to higher concentration then the linear relationship overestimates the conductance. In general the greater the concentration the worse the relative error. – MaxW Sep 3 '16 at 21:24
• Formula for centration mixing - letting $c_m$ be the concentration of the mixture then: $$c_m = \dfrac{c_1*v_1 + c_2*v_2}{v_1 + v_2}$$ – MaxW Sep 3 '16 at 21:25