# How to use Kohlrausch's law with ionic equivalent conductances? [closed]

My text says that for an ionic compound like AxBy
if ionic equivalent conductance of Ay+ is a and that of Bx- ion is b,
then the equivalent conductance for AxBy would be a/y + b/x

Now my question is: as the unit of equivalent conductance is S cm2 (gm equivalent)-1 so if i multiply it by a quantity x with unit (gm equivalent) mol-1, then i will get a quantity with unit S cm2 mol-1. This should be the Molar conductance of the ion granted that the value of x was correct. Now for example in Al2(SO4)3 for Al3+ ion the value of x would be 3 because one mole of Al3+ ions would contains 3 gram equivalents. So if equivalent conductance of Al3+ ion was a, then its molar conductance would be 3a and similarly for SO42- ion with equivalent conductance b, the molar conductance would be 2b.
Now from all that i can tell that the molar conductance of Al2(SO4)3 would be,
2(3a) + 3(2b) = 6a + 6b

Now from the formula of equivalent conductance and molar conductance
(Molar conductance)/n = Equivalent conductance
where n is the valence factor for desired compound. So in our case n=6 for Al2(SO4)3

∴ equivalent conductance = (6a + 6b)/6 = a + b
but according to my text it would be a/3 + b/2
Tell me where am i going wrong. And the result I ended up on works for some questions and the thing that my text provided works for some questions, and then there are some questions in which nothing works. I am really confused...
And sorry for any grammatical atrocities.

EDIT: I am just trying to ask that when equivalent conductance of two ions are given to me as a and b, then what would be the equivalent conductance of the compound to which these ions belong? Would it be
a + b
or would it be
a/(charge on first ion) + b/(charge on second ion)
or is it something totally else?

## closed as unclear what you're asking by A.K., Jon Custer, Todd Minehardt, Tyberius, MithoronMar 5 at 17:35

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You textbook writer has explained it in a very complicated way. I think you are confused by the concept of equivalent. Gram equivalent is an obsolete unit now but it is still used in conductance measurements.

First understand the definition of gram equivalent weight for a salt (and it applies to salts only). Acids/ Bases/Redox agents have different definitions.

Gram Equivalent weight of a salt or an ion = gram molecular weight/charge on the ion

This is the definition of gram equivalent.

From the context of your book, ionic equivalent conductance is same as "molar conductance" of aluminum ion. Here a and b refer to molar conductance. This is the mistake you are making. Thus, a 1 Molar Al3+ solution has 3 gram equivalents of Al/ 1 L or it is a 3 N solution (N = Normal).

So if you are given ionic equivalent conductance of Al2(SO4)3 (at infinite dilution) then your textbook is right = (ionic equiv. conductance Al)/ charge on Al + (ionic equiv. conductance sulfate)/ charge on sulfate = a/3 + b/2

By the way, equivalent conductance can only be added at infinite dilution. Whatever you see in tables are usually given at "infinite dilution".

• But aren't molar conductances and equivalent conductances different? Because it is clearly mentioned in my book that a and b are the equivalent conductances. And yes I am really confused about this whole equivalent thing. Thanks for replying! – My_chemistry_teacher_sucks Mar 5 at 11:22
• Molar conductance and equivalent conductance are same for monovalent ions but not for mutivalent ions, acids or bases. I think it will be easier for you if you realize that molar conductance = ionic equivalent conductance in the context of your textbook. "Equivalent" is a way of expressing amount like mole. Just like 1 M solution of X contains 1 mol/L, a 1 Normal solution of X contains 1 gram equivalent/ 1 L. The author has done a poor job in explaining it and using poor terminologies. en.wikipedia.org/wiki/Molar_conductivity. – M. Farooq Mar 5 at 14:09
• Ok thanks! I guess i will try another book. – My_chemistry_teacher_sucks Mar 6 at 12:09